A gas sample occupies a volume of 155 mL at a temperature of 316 K and a pressure of 0. 989 atm. How many moles of gas are there? 


2Points


Show your work

Answers

Answer 1

There are approximately 0.00614 moles of gas in the sample.

To find the number of moles of gas in the sample, we will use the Ideal Gas Law formula: PV = nRT.

Given:
Volume (V) = 155 mL = 0.155 L (converted to liters)
Temperature (T) = 316 K
Pressure (P) = 0.989 atm
Gas constant (R) = 0.0821 L atm / K mol

We need to find the number of moles (n).

Rearranging the formula for n: n = PV / RT


1. Convert the volume to liters: 155 mL = 0.155 L
2. Plug in the given values into the formula: n = (0.989 atm) x (0.155 L) / (0.0821 L atm / K mol) x (316 K)
3. Simplify the equation and solve for n: n ≈ 0.00614 mol

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Related Questions

What quantity in moles of hydrogen gas at 150. 0 °C and 23. 3 atm would occupy a vessel of 8. 50 L?

Answer ASAP

Answers

The number of moles of hydrogen gas comes out to be 5.700 that can be calculated using the ideal gas equation.

Using ideal gas equation,

PV = nRT ......(1)

It is given that,

T = 150.0 °C

P = 23.3 atm

V = 8.50 L

To calculate the number of moles, substitute the known values in equation (1).

PV = nRT

23.3 atm x 8.50 L  =  n x 0.0821 L atm/mol/K x 423.15 K

n = (23.3 atm x 8.50 L) / (0.0821 L atm/mol/K x 423.15 K)

  =  198.05 / 34.74 mole

  = 5.700 moles

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Why is the answer a not d?

Answers

The correct answer is therefore A, -1.66 V.

The given information includes the standard reduction potential of the half-reaction:

Ag(aq) + e- → Ag(s) E° = +0.80 V

We can use this information along with the standard cell potential equation to find the standard reduction potential of the half-reaction:

E°cell = E°reduction + E°oxidation

where E°cell is the standard cell potential, E°reduction is the reduction potential for the half-reaction being reduced, and E°oxidation is the oxidation potential for the half-reaction being oxidized.

In this case, the two half-reactions involved are:

M3+(aq) + 3e- → M(s) (reduction)

3Ag(aq) → 3Ag+(aq) + 3e- (oxidation)

The reduction half-reaction needs to be flipped and its potential sign changed to obtain the oxidation potential:

M(s) → M3+(aq) + 3e- (oxidation)

The standard cell potential is the difference between the reduction and oxidation potentials:

E°cell = E°reduction + E°oxidation

E°cell = E°(M3+(aq) + 3e- → M(s)) + E°(M(s) → M3+(aq) + 3e-)

E°cell = E°(M(s) → M3+(aq) + 3e-) + (-E°(3Ag(aq) → 3Ag+(aq) + 3e-))

E°cell = E°(M(s) → M3+(aq) + 3e-) - E°(Ag(aq) → Ag(s))

E°cell = E°(M3+(aq) + 3e- → M(s)) - E°(Ag(aq) → Ag(s))

E°cell = -2.46 V - 0.80 V = -3.26 V

Therefore, the standard reduction potential for the half-reaction M3+(aq) + 3e- → M(s) is:

E°(M3+(aq) + 3e- → M(s)) = E°cell + E°(Ag(aq) → Ag(s))

E°(M3+(aq) + 3e- → M(s)) = -3.26 V + 0.80 V = -2.46 V

The correct answer is therefore A, -1.66 V.

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Which of these ionization processes requires the highest amount of
energy?
(a) na(g) --> na*(g) + e;
(b) mg(g) --> mg (g) + e;
(c) al(g) --> alt(g) + e;
(d) ca(g) --> ca*(g) + e;

Answers

The ionization process that requires the highest amount of energy is (d) ca(g) --> ca*(g) + e, as calcium has a higher ionization energy than the other elements listed.

To answer this question, we need to consider the ionization energy for each element involved. Ionization energy is the amount of energy required to remove an electron from an atom or ion in the gaseous state. The ionization processes mentioned are:

(a) Na(g) --> Na+(g) + e-
(b) Mg(g) --> Mg+(g) + e-
(c) Al(g) --> Al+(g) + e-
(d) Ca(g) --> Ca+(g) + e-

Comparing the first ionization energies for these elements:
Na: 496 kJ/mol
Mg: 738 kJ/mol
Al: 577 kJ/mol
Ca: 590 kJ/mol

Process (b) Mg(g) --> Mg+(g) + e- requires the highest amount of energy, as magnesium has the highest ionization energy among the given elements.

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Who was the first person that was to attend to arrange the element in what year

Answers

It is unclear which specific element you are referring to in your question. However, if we are talking about the periodic table of elements, the first person to attend to arrange the elements was Dmitri Mendeleev in the year 1869.

Mendeleev was a Russian chemist who noticed patterns in the properties of elements and arranged them in order of increasing atomic weight. He left gaps in his periodic table for elements that had not yet been discovered, and even predicted the properties of these yet-to-be-discovered elements based on their position in the table.

Mendeleev's work revolutionized the field of chemistry and led to a better understanding of the nature of elements and their relationships to one another. Today, the periodic table is an essential tool for scientists and students alike in understanding the properties and behavior of chemical elements.

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To begin the experiment, 1. 65g of methane CH4 is burned in a bomb calorimeter containing 1000 grams of water. The initial temperature of water is 18. 98oC. The specific heat of water is 4. 184 J/g oC. The heat capacity of the calorimeter is 615 J/ oC. After the reaction the final temperature of the water is 36. 38oC.

5. The total heat absorbed by the water and the calorimeter can be by adding the heat calculated in steps 3 and 4. The amount of heat released by the reaction is equal to the amount of heat absorbed with the negative sign as this is an exothermic reaction. (2pts)
a. Using the formula ΔH = - (qcal + qwater ) , calculate the total heat of combustion. Show your work.
b. Convert heat of combustion (answer from part a) from joules to kilojoules. Show your work. 6. Evaluate the information contained in this calculation and complete the following sentence: (2pts) This calculation shows that burning _______ grams of methane [TAKES IN] / [GIVES OFF] energy (Choose one).

7. The molar mass of methane is 16. 04 g/mol. Calculate the number of moles of methane burned in the experiment. Show your work. (2pts)

8. What is the experimental molar heat of combustion in KJ/mol? Show your work. (2pts)

9. The accepted value for the heat of combustion of methane is -890 KJ/mol. Explain why the experimental data might differ from the theoretical value in 2-3 complete sentences. (2pts)

10. Given the formula: % error= |(theoretical value - experimental value)/theoretical value)| x 100 Calculate the percent error. Show your work. (2pts)​

Answers

The heat of combustion of methane is -802.41 kJ/mol, indicating that the combustion of methane is an exothermic reaction that releases heat energy.

To calculate the heat of combustion of methane, we can use formula:

q = (m_water x C_water x ΔT) + (C_cal x ΔT)

Plugging in the values, we get:

q = (1000 g x 4.184 J/g°C x 17.4°C) + (615 J/°C x 17.4°C)

q = 21997.45 J

Next, we need to calculate the number of moles of methane burned:

moles [tex]CH_4[/tex] = mass [tex]CH_4[/tex] / molar mass [tex]CH_4[/tex]

moles [tex]CH_4[/tex] = 65 g / 16.04 g/mol

moles [tex]CH_4[/tex] = 4.05 mol

Finally, we can calculate the heat of combustion per mole of methane:

ΔH = q / moles [tex]CH_4[/tex]

ΔH = -802.41 kJ/mol

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--The complete Question is, 1. 65g of methane CH4 is burned in a bomb calorimeter containing 1000 grams of water. The initial temperature of water is 18. 98oC. The specific heat of water is 4. 184 J/g oC. The heat capacity of the calorimeter is 615 J/ oC. After the reaction the final temperature of the water is 36. 38oC.--

Based on the equation and the enthalpies of formation shown, what is the AH of the reaction? A.-5335.8 B.-2815.8 C. -580.7 D.580.7

Answers

The AH of the reaction is given as C. -571.6 kJ/mol

How to solve

The enthalpy change of a reaction (∆H) can be calculated using the formula:

∆H = Σn ∆Hf°(products) - Σn ∆Hf°(reactants),

where n is the stoichiometric coefficient of each substance in the balanced equation.

If we apply this to the reaction of H2(g) and O2(g) forming H2O(l), we get ∆H = -571.6 kJ/mol, where ∆Hf°(H2(g)) = 0 kJ/mol and ∆Hf°(O2(g)) = 0 kJ/mol.

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Given the following reaction equation and the enthalpies of formation (∆Hf°) for each substance, what is the ∆H of the reaction?

2 H2(g) + O2(g) → 2 H2O(l)

∆Hf°(H2(g)) = 0 kJ/mol

∆Hf°(O2(g)) = 0 kJ/mol

∆Hf°(H2O(l)) = -285.8 kJ/mol

A. -5335.8 kJ/mol

B. -2815.8 kJ/mol

C. -571.6 kJ/mol

D. 580.7 kJ/mol

You are given 10. 34 grams of C7H14O7. How many moles of the compound do you have?

Answers

There are 0.0492 moles of the compound C7H14O7 when given 10.34 grams.

To determine how many moles of the compound C7H14O7 you have when given 10.34 grams, you need to follow these steps:

1. Calculate the molar mass of the compound C7H14O7:
- For carbon (C), there are 7 atoms, each with a molar mass of 12.01 g/mol.
- For hydrogen (H), there are 14 atoms, each with a molar mass of 1.01 g/mol.
- For oxygen (O), there are 7 atoms, each with a molar mass of 16.00 g/mol.

2. Add up the molar masses:
- Molar mass of C7H14O7 = (7 * 12.01) + (14 * 1.01) + (7 * 16.00) = 84.07 + 14.14 + 112.00 = 210.21 g/mol.

3. Use the formula to convert grams to moles:
- Moles = mass (grams) / molar mass (g/mol)

4. Plug in the values and solve for moles:
- Moles of C7H14O7 = 10.34 grams / 210.21 g/mol = 0.0492 moles.

So, you have 0.0492 moles of the compound C7H14O7 when given 10.34 grams.

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After 45 days a radioactive material has decayed 55. 1%, after an additional 45 days, what percent of the original amount will it have decayed to

Answers

The material has decayed to 37.2% of the original amount after 90 days (with an additional 45 days after decaying 55.1%).

The amount of radioactive material remaining after time t can be calculated using the formula:

[tex]N(t) = N0 * (1/2)^(t/T)[/tex]

where N0 is the initial amount, t is the time elapsed, and T is the half-life of the radioactive material.

In this problem, the initial amount of radioactive material has decayed by 55.1% after 45 days. This means that:

[tex]N(45) = N0 * (1 - 0.551) = 0.449 * N0[/tex]

After an additional 45 days, the total time elapsed is now 90 days. We can use the same formula to calculate the amount of radioactive material remaining after 90 days:

[tex]N(90) = N0 * (1/2)^(90/T)[/tex]

We can then use the two equations to solve for the percentage of the original amount that has decayed after 90 days:

[tex]N(90) = 0.449 * N0 * (1/2)^(90/T)0.449 = (1/2)^(45/T)[/tex]

Taking the natural logarithm of both sides:

[tex]ln(0.449) = ln(1/2)^(45/T)[/tex]

ln(0.449) = -(45/T) * ln(2)

T = -(45/ln(2)) * ln(0.449) = 86.5 days (to the nearest tenth)

Now that we know the half-life of the material, we can use the original formula to calculate the amount of material remaining after 90 days:

[tex]N(90) = N0 * (1/2)^(90/86.5) = 0.372 * N0[/tex]

Therefore, the material has decayed to 37.2% of the original amount after 90 days (with an additional 45 days after decaying 55.1%).

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Read the chemical equation. N2 + 3H2 – 2NH3 Using the volume ratio, determine how many liters of NH3 is produced if 3. 6 liters of H2 reacts with an excess of N2, if all measurements are taken at the same temperature and pressure? 5. 4 liters 2. 4 liters 1. 8 liters 1. 2 liters​

Answers

Using this volume ratio, we can determine that (b) 2.4 liters of ammonia are produced when 3.6 liters of hydrogen reacts with an excess of nitrogen.

The given chemical equation represents the reaction between nitrogen and hydrogen to produce ammonia. The balanced equation shows that for every 3 volumes of hydrogen, 2 volumes of ammonia are produced.

According to the balanced chemical equation N₂ + 3H₂ → 2NH₃, for every 3 volumes of H₂, 2 volumes of NH₃ are produced.

Therefore, if 3.6 liters of H₂ reacts, the amount of NH₃ produced can be calculated as follows:

3.6 L H₂ * (2 L NH₃ / 3 L H₂) = 2.4 L NH₃

Therefore, 2.4 liters of NH₃ would be produced if 3.6 liters of H₂ reacts with an excess of N₂. The correct answer is option (b) 2.4 liters.

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16. If the difference in electro-negativities of the combining atoms is zero, then the bond formed is a
(a) covalent bond
(b) electrovalent bond
(c) non-polar covalent bond
(d) polar covalent bond

Answers

C. Non-polar bonds have similar electronegativities, so when you subtract them, the answer will be zero.

A 31. 4 gg wafer of pure gold initially at 69. 7 ∘C∘C is submerged into 64. 1 gg of water at 26. 8 ∘C∘C in an insulated container. The specific heat capacity for gold is 0. 128 J/(g⋅∘C)J/(g⋅∘C) and the specific heat capacity for water is 4. 18 J/(g⋅∘C)J/(g⋅∘C)? Part A What is the final temperature of both substances at thermal equilibrium?

Answers

A 31.4 g gold wafer initially at 69.7°C is submerged into 64.1 g of water at 26.8°C. The final temperature at which both substances reach thermal equilibrium is 31.9°C.

The final temperature of both substances at thermal equilibrium needs to be determined.

We can use the principle of conservation of energy. Since the system is insulated, the heat lost by the gold will be equal to the heat gained by the water.

The heat lost by the gold can be calculated using:

Q = mcΔT

where Q is the heat lost, m is the mass of the gold, c is its specific heat capacity, and ΔT is the change in temperature.

Similarly, the heat gained by the water can be calculated using:

Q = mcΔT

where Q is the heat gained, m is the mass of the water, c is its specific heat capacity, and ΔT is the change in temperature.

Setting these two equations equal to each other and solving for the final temperature, we get:

[tex]m_{\text{gold}} \cdot c_{\text{gold}} \cdot (T_{\text{final}} - T_{\text{initial\_gold}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{initial\_water}})[/tex]

where [tex]$m_{\text{gold}}$[/tex] is the mass of the gold, [tex]c_{\text{gold}}[/tex] is its specific heat capacity, [tex]T_{\text{initial\_gold}}[/tex] is its initial temperature, [tex]m_{\text{water}}[/tex] is the mass of the water, [tex]$c_{\text{water}}$[/tex] is its specific heat capacity, and [tex]T_{\text{initial\_water}}[/tex] is its initial temperature.

Plugging in the values we get:

[tex]31.4 \, \text{g} \times 0.128 \, \text{J/(g} \cdot \text{°C)} \times (T_{\text{final}} - 69.7^\circ\text{C}) = 64.1 \, \text{g} \times 4.18 \, \text{J/(g} \cdot \text{°C)} \times (T_{\text{final}} - 26.8^\circ\text{C})[/tex]

Solving for [tex]$T_{\text{final}}$[/tex], we get:

[tex]T_{\text{final}} = \frac{(31.4 \, \text{g} \times 0.128 \, \text{J/(g} \cdot \text{°C)} \times 69.7^\circ\text{C}) + (64.1 \, \text{g} \times 4.18 \, \text{J/(g} \cdot \text{°C)} \times 26.8^\circ\text{C})}{(31.4 \, \text{g} \times 0.128 \, \text{J/(g} \cdot \text{°C)}) + (64.1 \, \text{g} \times 4.18 \, \text{J/(g} \cdot \text{°C)})}[/tex]

[tex]$T_{\text{final}}$[/tex] = 31.9°C

Therefore, the final temperature of both substances at thermal equilibrium is 31.9°C.

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0. 008 moles of C3H7OH contains how many atoms of carbon?

Answers

0.008 moles of C₃H₇OH contains 1.44528 x 10^22 atoms of carbon.

To find the number of carbon atoms in 0.008 moles of C₃H₇OH, follow these steps:

1. Identify the number of carbon atoms in one molecule of C₃H₇OH. In this case, there are 3 carbon atoms.
2. Calculate the total number of molecules in 0.008 moles of C₃H₇OH by multiplying the number of moles by Avogadro's constant (6.022 x 10^23 molecules/mol).

0.008 moles * (6.022 x 10^23 molecules/mol) = 4.8176 x 10^21 molecules

3. Multiply the total number of molecules by the number of carbon atoms in each molecule to find the total number of carbon atoms:

4.8176 x 10^21 molecules * 3 carbon atoms/molecule = 1.44528 x 10^22 carbon atoms

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What is the correct equilibrium expression for the dissociation of the base pyridine:


C5H5N + H2O â C5H5NH+ + OH-



A. Kb = [C5H5NH+][OH-] / [C5H5N]


B. Kb = [C5H5N][OH-] / [C5H5NH+][H2O]


C. Kb = [C5H5NH+][OH-] / [C5H5N][H2O]


D. Kb = [C5H5NH+][C5H5N] / [OH-]


E. Kb = [C5H5N][OH-] / [C5H5NH+]

Answers

The correct equilibrium expression for the dissociation of the base pyridine is: C₅H₅N + H₂O ↔ C₅H₅NH+ + OH- is A. Kb = [C₅H₅NH+][OH-] / [C₅H₅N]. The correct option is A.

The equilibrium expression for the reaction of a weak base with water is Kb = [BH+][OH-] / [B], where BH+ is the conjugate acid of the weak base B. In this case, pyridine (C₅H₅N) is the weak base, and its conjugate acid is C₅H₅NH+.

The concentration of water is assumed to be constant and is not included in the equilibrium expression. Therefore, the equilibrium expression for the dissociation of pyridine is Kb = [C₅H5₅H+][OH-] / [C₅H₅N].

Option A is the correct expression since it follows the correct form for the equilibrium expression of a weak base with water. Option B has the concentrations of water and the conjugate acid of the weak base in the denominator, which is incorrect. Option C has the concentration of water in the denominator, which is incorrect.

Option D has the concentration of hydroxide ions (OH-) in the denominator, which is incorrect. Option E has the concentrations of the weak base and its conjugate acid in the denominator, which is also incorrect. Hence option A is the correct option.

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Why might your value be different from absolute zero? (HINT: Think errors in the lab. )

Answers

Value might be different from absolute zero due to several factors like Measurement errors, External factors, Non-ideal conditions.

"Why might your value be different from absolute zero?" we need to understand the following terms:

1. Value: Refers to a quantity or numerical measurement in a specific context.
2. Absolute zero: The lowest possible temperature, at which all molecular motion stops. It is 0 Kelvin (K) or -273.15 degrees Celsius (°C) or -459.67 degrees Fahrenheit (°F).

Your value might be different from absolute zero due to several factors, such as:
1. Measurement errors: If you are measuring a temperature, there could be inaccuracies in your measuring device, leading to a value different from absolute zero.
2. External factors: The presence of heat or energy in your system can cause the value to deviate from absolute zero.
3. Non-ideal conditions: In real-world situations, reaching absolute zero is practically impossible due to quantum effects and other factors, causing your value to be higher than absolute zero.

By understanding these factors, you can identify why your value may differ from absolute zero.

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How many grams of protein are needed to produce 23,000 cal of energy? Every gram of protein can produce 17 KJ of energy

Answers

A total of 96,320 kJ / 17 kJ per gram of protein = 5,670 grams of protein.

To determine the grams of protein needed to produce 23,000 calories of energy, we need to convert the calories to kilojoules (kJ) and then divide by the energy produced by each gram of protein.

23,000 calories = 96,320 kJ (1 calorie = 4.184 kJ)
Each gram of protein produces 17 kJ of energy.


Protein is an important nutrient for our bodies, as it provides the building blocks for our muscles, bones, and other tissues. It also plays a role in many cellular functions and processes. One of the functions of protein is to provide energy for our bodies, although this is not its primary role.

When we eat protein, our bodies break it down into amino acids, which can then be used for various purposes. One of these purposes is to produce energy.

Every gram of protein contains 4 calories, or 17 kilojoules, of energy. This is less than the amount of energy provided by a gram of fat (9 calories or 37 kilojoules) or a gram of carbohydrate (4 calories or 17 kilojoules), but it is still significant.

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What is the name of this branched alkene? Please help me as fast as possible I need to study, please! ILL MARK AS BRAINLIEST PLEASE HELP MEE

Answers

The name of the branched alkene given in the question is:

6-ethyl-8-methyl-5-propyl-2-nonene or 6-ethyl-8-methyl-5-propylnon-2-ene

How do i determine the mane of the branched alkene?

The naming of compound is obtained by the of IUPAC standard. This is illustrated below:

Identify the parent chain. In this case, the longest chain is carbon 9. Thus, the parent name is nonene.Identify the substituent groups attached. In this case the substituent groups attached are: CH₃, CH₂CH₃ and CH₂CH₂CH₃ Identify the position of the substituents by considering the double bond. In this case, the double bond is at carbon 2, CH₂CH₃ is located at carbon 6, CH₃ is located at carbon 8 and CH₂CH₂CH₃ is located at carbon 5.Combine the above to obtain the IUPAC name for the compound.

Thus, the IUPAC name for the branched alkene is:

6-ethyl-8-methyl-5-propyl-2-nonene or 6-ethyl-8-methyl-5-propylnon-2-ene

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A person uses 500kcal of energy to run a race. convert the energy used for the race to the following energy units:
joules(j)

kilojoules (kj)


1 calorie= 4.184 joules

Answers

Answer: Look at the image I attached - I drew what you should write.

Will give brainliest no fake answers please

determine the empirical formula of a compound containing 48. 38 grams of carbon, 6. 74 grams of hydrogen, and 53. 5 grams of oxygen.


in an experiment, the molar mass of the compound was determined to be 180. 15 g/mol. what is the molecular formula of the compound?


for both questions, show your work or explain how you determined the formulas by giving specific values used in calculations

Answers

a. The empirical formula of the compound is [tex]CH_2O.[/tex] b. Moles of oxygen is 3.344 mol and c. The molecular formula of the compound is [tex]C_6H_12O_6[/tex].

To determine the empirical formula of the compound:

Convert the mass of each element to moles using its molar mass:

Moles of carbon = 48.38 g / 12.011 g/mol = 4.030 mol

Moles of hydrogen = 6.74 g / 1.008 g/mol = 6.690 mol

Moles of oxygen = 53.5 g / 15.999 g/mol = 3.344 mol

Divide each number of moles by the smallest number of moles to get the simplest whole-number ratio of atoms:

Carbon: 4.030 mol / 3.344 mol = 1.205 ≈ 1

Hydrogen: 6.690 mol / 3.344 mol = 1.999 ≈ 2

Oxygen: 3.344 mol / 3.344 mol = 1

Therefore, the empirical formula of the compound is [tex]CH_2O.[/tex]

To determine the molecular formula of the compound:

Calculate the empirical formula mass:

Mass of  [tex]CH_2O.[/tex] = 12.011 g/mol + 2(1.008 g/mol) + 15.999 g/mol = 30.026 g/mol

Empirical formula mass x n = Molar mass

n = Molar mass / Empirical formula mass = 180.15 g/mol / 30.026 g/mol = 6.000

Multiply each subscript in the empirical formula by n to get the molecular formula:

Molecular formula = [tex](CH_2O)_6[/tex] =  [tex]C_6H_12O_6[/tex]

Therefore, the molecular formula of the compound is  [tex]C_6H_12O_6[/tex]

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Find the percent composition of a sample containing 1.29 grams of carbon and


1.71 grams of oxygen.

Answers

The percent composition of the sample containing 1.29 grams of carbon and 1.71 grams of oxygen is 43% carbon and 57% oxygen.

The percent composition of a sample can be calculated by dividing the mass of each element in the sample by the total mass of the sample and then multiplying by 100%.

To find the percent composition of a sample containing 1.29 grams of carbon and 1.71 grams of oxygen, we need to calculate the total mass of the sample first.

Total mass of the sample = mass of carbon + mass of oxygen
= 1.29 grams + 1.71 grams
= 3 grams

Now, we can calculate the percent composition of carbon and oxygen in the sample:

Percent composition of oxygen = (mass of oxygen / total mass of the sample) x 100%
= (1.71 grams / 3 grams) x 100%
= 57%

Percent composition of carbon =  (mass of carbon / total mass of the sample) x 100%

=(1.29 grams / 3 grams) x 100%

= 43%

Therefore, the sample contains 43% carbon and 57% oxygen by mass.

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A mixture of 33. 6 g of Cr(NO3)2 and 60. 5 g of CuSO4 is dissolved in sufficient water to make 98 mL of solution, where the cations react. In the reaction, copper metal is formed and each chromium ion loses one electron. How many electrons are transferred in the balanced net ionic equation with the smallest whole-number coefficients?


1. 5e-


2. 2e-


3. 7e-


4. 4e-


5. 1e-


(Part 2) What is the molar concentration of SO4^2- anions in the solution? Answer in units of M

Answers

The molar concentration of SO4^2- anions in the solution is about 3.867 M.

To answer your question, first we need to write the balanced net ionic equation:

Cr^2+(aq) + Cu^2+(aq) → Cr^3+(aq) + Cu(s)

Now, we need to determine the number of moles of Cr(NO3)2 and CuSO4:

Cr(NO3)2: 33.6 g / (130.87 g/mol) = 0.257 moles
CuSO4: 60.5 g / (159.61 g/mol) = 0.379 moles

From the balanced net ionic equation, we can see that 1 mole of Cr^2+ reacts with 1 mole of Cu^2+. Since we have more moles of Cu^2+ than Cr^2+, Cr^2+ is the limiting reagent.

Now, let's calculate the number of electrons transferred:

Since each Cr^2+ ion loses one electron, the number of electrons transferred is equal to the number of moles of Cr^2+ ions:
0.257 moles * 1e- = 0.257e-

Since we need the smallest whole-number coefficients, we'll multiply by the lowest common denominator (LCD) to make the number of electrons a whole number. The LCD for 0.257 is 7, so we'll multiply the entire equation by 7:

7Cr^2+(aq) + 7Cu^2+(aq) → 7Cr^3+(aq) + 7Cu(s)

Therefore, the number of electrons transferred is:
0.257e- * 7 = 1.799e- ≈ 2e-
So the correct answer is 2e-.

(Part 2) To find the molar concentration of SO4^2- anions in the solution, we need to use the moles of CuSO4 and the volume of the solution:

0.379 moles / 0.098 L = 3.867 M

The molar concentration of SO4^2- anions in the solution is approximately 3.867 M.

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Given the equation: 2C2H2 + 5O2 → 4CO2 + 2H2O How many grams of C2H2 are required to react completely with 2. 0 mole of O2?

Answers

20.8 grams of C2H2 are required to react completely with 2.0 moles of O2.

The balanced chemical equation is: [tex]2C2H2 + 5O2 → 4CO2 + 2H2O[/tex]

The stoichiometry of the balanced equation shows that 2 moles of C2H2 react with 5 moles of O2 to produce 4 moles of CO2 and 2 moles of H2O.

Therefore, the mole ratio of C2H2 to O2 is 2:5.

If 2.0 moles of O2 are completely reacted, then the required moles of C2H2 can be calculated as follows:

2.0 mol O2 x (2 mol C2H2/5 mol O2) = 0.8 mol C2H2

Now, we can use the molar mass of C2H2 to calculate the mass required:

0.8 mol C2H2 x 26.04 g/mol = 20.8 g C2H2

Therefore, 20.8 grams of C2H2 are required to react completely with 2.0 moles of O2.

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11. The latent heat of fusion of water is 334 J/g. The latent heat of
vaporization of water is 2257 J/g. The specific heat capacity of
water is 4.186 J/g °C How much heat is needed to evaporate 500
og of ice that starts at 0°C ? Hint: Sum of AQS...Q1: Solid to Liquid;
Q2 of Liquid water; Q3 Liquid to Gas

Answers

The amount heat needed to evaporate 500 g of ice that starts at 0 °C is 1504800 J

How do i determine the heat needed to evaporate the ice?

First, we shall determine the heat needed to melt the ice. Details below:

Mass of ice (m) = 500 gLatent heat of fusion (ΔHf) = 334 J/gHeat (H₁) =?

H₁ = m × ΔHf

H₁ = 500 × 334

H₁ = 167000 J

Next, we shall determine the heat required to change the water from 0 °C to 100°C. Details below:

Mass of water (M) = 500 gInitial temperature of water (T₁) = 0 °CFinal temperature of water (T₂) = 100 °CChange in temperature of water (ΔT) = 100 - 0 = 100°CSpecific heat capacity of water (C) = 4.186 J/gºC Heat (H₂) =?

H₂ = MCΔT

H₂ = 500 × 4.186 × 100

H₂ = 209300 J

Next, we shall determine the heat required to vaporize the water. Details below:

Mass of water (M) = 500 g Heat of Vaporization (ΔHv) = 2257 J/gHeat (H₃) =?

H₃ = m × ΔHv

H₃ = 500 × 2257

H₃ = 1128500 J

Finally, we shall determine the heat required to evaporate the ice. Details below:

Heat required to melt the ice (H₁) = 167000 JHeat required to change the steam from 0 °C to 100 °C(H₂) = 209300 JHeat required to vaporize the water (H₃) = 1128500 JTotal heat required (Q) =?

Q = H₁ + H₂ + H₃

Q = 167000 + 209300 + 1128500

Total heat required = 1504800 J

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Please help!!!
1.00 L of gas is collected in a sealed elastic container in outer space where the pressure is 1.54 x 104 mm Hg and the temperature is 88 K. What will the volume be if the container is moved to sea level ( 101.3 kPa) and room temperature ( 23 C)?

Answers

The volume of the gas in the container at sea level and room temperature would be approximately 0.00676 L.

PV = nRT

Where:

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

V = nRT / P

First, we need to calculate the number of moles of gas in the container:

n = PV / RT

Where:

P = 1.54 x[tex]10^{4}[/tex] mm Hg

V = 1.00 L

R = 8.31 J/mol*K (gas constant)

T = 88 K

Converting the pressure to kPa and the volume to m^3:

P = 1.54 x [tex]10^{4}[/tex] mm Hg * (101.3 kPa / 760 mm Hg) = 2054.59 Pa

V = 1.00 L * [tex]10^{-3}[/tex] [tex]m^{3}[/tex]/L) = 0.001 [tex]m^{3}[/tex]

n = (2054.59 Pa * 0.001 m^3) / (8.31 J/mol*K * 88 K) ≈ 0.000276 mol

P = 101.3 kPa

T = 23 + 273.15 K = 296.15 K

V = nRT / P

V = (0.000276 mol * 8.31 J/mol*K * 296.15 K) / 101.3 kPa

Converting the volume to liters:

V = 0.00676 L

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Now you are ready to explain what happened when Lee mixed sodium and hydrogen chloride. Be sure to use key


concepts in your explanation and provide examples from the Sim or the token activity.


Answer the following question: How did sodium and hydrogen chloride change into two different substances?


pls help

Answers

When Lee mixed sodium and hydrogen chloride, a chemical reaction occurred. Sodium has a single valence electron, which it donates to hydrogen chloride, forming Na⁺ and Cl⁻ ions.

These ions then combine to form solid sodium chloride (NaCl) and hydrogen gas (H₂). This reaction is an example of a redox reaction, where the sodium undergoes oxidation and the hydrogen chloride undergoes reduction.

In the simulation or token activity, the reaction can be represented by the following equation:

2 Na + 2 HCl → 2 NaCl + H₂

Thus, the sodium and hydrogen chloride changed into two different substances, solid sodium chloride and gaseous hydrogen, as a result of a chemical reaction involving the transfer of electrons.

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iz
Which sentence from the article shows humans' MAIN problem?
(A)
(B)
(C)
(D)
Solar radiation is the energy, both heat and light, that the sun gives off.
During the day, the sun shines through the atmosphere, causing Earth's surface to warm up.
This process is what keeps our Earth at an average global temperature of 14 degrees Celsius (58
degrees Fahrenheit).
The level of carbon dioxide in Earth's atmosphere has risen consistently for decades, trapping extra
heat near the surface of the Earth.

Answers

The sentence that shows humans' main problem is "The level of carbon dioxide in Earth's atmosphere has risen consistently for decades, trapping extra heat near the surface of the Earth," which is the last option.

The sentence that shows humans' main problem is "The level of carbon dioxide in Earth's atmosphere has risen consistently for decades, trapping extra heat near the surface of the Earth." This sentence indicates that the main problem is the increasing levels of carbon dioxide in the Earth's atmosphere caused by human activities. This increase in carbon dioxide is resulting in global warming, where heat is being trapped near the Earth's surface, leading to several negative effects, including rising sea levels, changes in weather patterns, and loss of biodiversity.

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the reaction of nitrogen gas with oxygen gas, , has a kp value of 0.50 at some temperature. if 0.100 atm of n2 and o2 are placed in a closed vessel and allowed to come to equilibrium, what is the approximate equilibrium partial pressure of no gas?

Answers

The approximate equilibrium partial pressure of NO gas is 0.005 atm.

The balanced chemical equation for the reaction is:

N2(g) + O2(g) ⇌ 2NO(g)

The equilibrium constant expression for this reaction is:

[tex]Kp = (PNO)2 / (PN2)(PO2)[/tex]

At equilibrium, let x be the partial pressure of NO gas. Then the partial pressures of N2 and O2 gas will both be (0.100 - x) atm. Substituting these values into the equilibrium constant expression and solving for x gives:

[tex]0.50 = x^2 / (0.100 - x)^2\\0.50(0.100 - x)^2 = x^2\\0.005 - 0.1x + 0.5x^2 = x^2\\0.5x^2 - 0.1x + 0.005 = 0[/tex]

Using the quadratic formula, we can solve for x:

[tex]x = [0.1 ± sqrt(0.1^2 - 4(0.5)(0.005))] / (2(0.5)) \\x = [0.1 ± 0.195] / 1 \\x = 0.295 or x = 0.005[/tex]

Since the initial partial pressures of N2 and O2 are both 0.100 atm, the equilibrium partial pressure of NO cannot be greater than 0.100 atm. Therefore, the only possible solution is:

x = 0.005 atm

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Determine the celsius temperature of 1.50 moles of ammonia contained in a 10.0-l vessel under a
pressure of 2.0 atm.
a
-1100
162
-50 c
с
0.0 c

Answers

The celsius temperature of 1.50 moles of ammonia contained in a 10.0-l vessel under a pressure of 2.0 atm can be determined using the ideal gas law.

The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. To calculate the temperature in Celsius, the Kelvin temperature is first determined by rearranging the equation and solving for T.

Then, the Kelvin temperature is converted to Celsius by subtracting 273.15 from the Kelvin temperature. In this case, the calculation would be T = (2.0 * 10.0) / (1.50 * 0.0821) = 1100.16 K. Subtracting 273.15 from 1100.16 K yields 827.01 °C, which is equal to 827.01 - 273.15 = -50.0 °C.

In conclusion, the celsius temperature of 1.50 moles of ammonia contained in a 10.0-l vessel under a pressure of 2.0 atm is -50.0 °C.

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How many grams of calcium oxide will be produced in a closed vessel containing 20. 0 kg of calcium and 20. 0 kg of oxygen gas if the reaction goes to completion?


2Ca(s)+0 (g) 2CaO(s)

Answers

A total of 28,000 grams of calcium oxide will be produced.

To find out how many grams of calcium oxide will be produced in a closed vessel containing 20.0 kg of calcium and 20.0 kg of oxygen gas, follow these steps:

1. Convert the given masses into moles using the molar mass of each element:
  - For calcium (Ca): 20,000 g / 40.08 g/mol ≈ 499 moles
  - For oxygen (O2): 20,000 g / 32 g/mol ≈ 625 moles

2. Determine the limiting reactant using the stoichiometry of the balanced equation:
  - The stoichiometric ratio of Ca to O2 is 2:1, so 625 moles of O2 would require 1,250 moles of Ca, but there are only 499 moles of Ca available. Therefore, calcium is the limiting reactant.

3. Calculate the moles of calcium oxide (CaO) produced using the stoichiometry of the balanced equation:
  - The ratio of Ca to CaO is 1:1, so 499 moles of Ca will produce 499 moles of CaO.

4. Convert the moles of calcium oxide back to grams using the molar mass:
  - The molar mass of CaO is 56.08 g/mol (40.08 g/mol for Ca + 16 g/mol for O). Therefore, 499 moles of CaO * 56.08 g/mol ≈ 28,000 grams of calcium oxide will be produced.

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Put these atoms in order from most positive overall charge to least positive overall charge.


Atom B: 24 protons, 19 electrons
Atom A: 14 protons, 16 electrons
Atom R: 26 protons, 24 electrons
Atom P: 8 protons, 11 electrons PLEASE HELP FAST

Answers

Order/Answer: 1. Atom B. 2. Atom R 3. Atom A. and then 4. Atom P. (Sorry this was rushed, I believe it is correct though.)

I need help on number 2

In this lab exercise we tried to predict what would be the limiting reagent in each beaker
based on observation of the amount (in mass) of reactant available. In determining the
limiting reagent in a chemical reaction, is it enough to just know the mass of each of the
reactant? Explain.

Answers

It is not enough to just know the mass of each reactant to determine the limiting reagent in a chemical reaction. The limiting reagent is the reactant that gets completely consumed during a chemical reaction, which limits the amount of product that can be formed.

To determine the limiting reagent, you need to compare the amount (in moles) of each reactant present, rather than just the mass. This is because different reactants have different molar masses, and therefore the same mass of two different reactants would have different numbers of moles.

Once you have determined the amount (in moles) of each reactant present, you can use stoichiometry to calculate how much product can be formed from each reactant. The reactant that produces the smallest amount of product is the limiting reagent.


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