Answer:
[tex]m_{CaBr_2}=115gCaBr_2[/tex]
Explanation:
Hello,
In this case, we remember molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, therefore, for the given concentration and volume, we should solve the moles of calcium bromide to subsequently compute the mass, as shown below:
[tex]M=\frac{n}{V}\\ \\n=M*V=1.15\frac{mol}{L} *500mL*\frac{1L}{1000mL}=0.575mol[/tex]
Next, since molar mass of calcium bromide is 200 g/mol, we can compute the mass:
[tex]m_{CaBr_2}=0.575mol*\frac{1mol}{200g} \\\\m_{CaBr_2}=115gCaBr_2[/tex]
Regards.
How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Answer:
Grams of mercury= 0.06 g of Hg
Note: The question is incomplete. The complete question is as follows:
A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Explanation:
Since one fluorescent light bulb contains 4 mg of mercury,
15 such bulbs will contain 15 * 4 mg of mercury = 60 mg
1 mg = 0.001 g
Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.
Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.
Total amount of mercury in 15 compact fluorescent light bulbs is 0.06 gram of mercury.
Compact fluorescent light bulbs and mercury:What information do we have?
Number of compact fluorescent light bulbs = 15 bulb
Amount of mercury in each bulb = 4 mg
Total amount of mercury = Number of compact fluorescent light bulbs × Amount of mercury in each bulb
Total amount of mercury = 15 × 4
Total amount of mercury = 60 mg
Total amount of mercury = 60 / 1000
Total amount of mercury = 0.06 gram of mercury
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An increase in the temperature of reactant causes an increase in the rate of reaction.
Which of the following is the best explanation?
a)the concentration of reactant increases
b) the activation energy decreases
c) the collision frequency increases
d) the fraction of collision with total kinetic energy larger than activation energy
increases
Answer:
d) The fraction of collision with total kinetic energy larger than activation energy increases.
Explanation:
Hello,
In this case, kinetic models explain how the rate of a chemical reaction is affected by several factors. In such a way, specifically for temperature, when it increases, the average velocity of the particles is also increased, for that reason, the collision frequency increases since the molecules are more likely to collide as they move faster and encounter to each other.
Nonetheless, it is the minor reason because the main reason is that the effective collisions increase when the temperature is increased, and they are related with the fraction of collision with total kinetic energy that turns out larger than the activation energy, therefore, answer is d).
Best regards.
Question 2
Bromine will react very fast (almost instantly) with which compound?
O 1-Pentene
Cyclohexane
Heptane
Benzene
Answer:
1-Pentene
Explanation:
If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.
For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.
Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.
For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.
How many moles of hydrogen atoms are there in 120 g of C6H12O6
Answer: 8.0 moles
Explanation:
0.6661 moles×12 H≈8.0 moles
120 g of C₆H₁₂O₆ contain 8 moles of Hydrogen atoms
To obtain the answer to the question given above, we'll begin by calculating the mass of 1 mole of C₆H₁₂O₆
1 mole of C₆H₁₂O₆ = (12×6) + (12×1) + (16×6)
= 72 + 12 + 96
= 180 gFrom the above,
We can see that 180 g of C₆H₁₂O₆ contains 12 moles of Hydrogen.
Finally, we shall determine the number of mole of Hydrogen atoms in 120 g of C₆H₁₂O₆. This can be obtained as follow:
180 g of C₆H₁₂O₆ contains 12 moles of Hydrogen.
Therefore,
120 g of C₆H₁₂O₆ will contain = [tex]\frac{120 * 12}{180}[/tex] = 8 moles of Hydrogen.
Thus, 120 g of C₆H₁₂O₆ contain 8 moles of Hydrogen atoms
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Reactions that undergo shifts in their equilibrium must be...
A.reversible reactions
B.chemical reactions
C.physical reactions
D.nuclear reactions
Answer:
A. Reversible reactions.
Explanation:
Reactions that undergo shifts in their equilibrium must be reversible reactions.
A reversible reaction is a chemical reaction that is capable of forming it's reactants back from the resulting formation of products. This simply means that, reversible reactions are chemical reactions that are in equilibrium because the forward and reverse path happens at the same rate.
For example, the reaction of hydrogen gas [tex]H_{2}[/tex] and iodine gas [tex]I_{2}[/tex] to form a chemical compound called hydrogen Iodide [tex]HI[/tex] is a reversible chemical reaction.
Forward reaction:
[tex]H_{2} + I_{2} ----> 2HI[/tex]
Reverse reaction:
[tex]2HI----> H_{2} + I_{2}[/tex]
A reversible reaction is denoted with a double arrow.
Also, reversible chemical reactions are controlled by the Le Chatelier's principle.
When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to effuse as the same amount of CO2 under the same conditions of temperature and pressure. What is the molar mass of the phosphorus chloride
Answer:
molar mass of the phosphorus chloride = 138.06 g/mol
Explanation:
mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.
mass = 3.51 g
lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec
From this we can say that
rate of effusion of CO2 = 3.51/1 = 3.51 g/s
rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s
From graham's equation of effusion,
[tex]\frac{Rc}{Rp}[/tex] = [tex]\sqrt{\frac{Mp\\}{Mc} }[/tex]
Rc = rate of effusion of CO2 = 3.51 g/s
Rp = rate of effusion of phosphorus chloride = 1.98 g/s
Mc = molar mass of CO2 = 44.01 g/mol
Mp = molar mass of the phosphorus chloride = ?
Imputing values into the equation, we have
[tex]\frac{3.51}{1.98}[/tex] = [tex]\sqrt{\frac{Mp\\}{44.01} }[/tex]
1.77 = [tex]\frac{\sqrt{Mp} }{6.64}[/tex]
11.75 = [tex]\sqrt{Mp}[/tex]
Mp = [tex]11.75^{2}[/tex]
Mp = molar mass of the phosphorus chloride = 138.06 g/mol
Which is a substance that could be found in air , water , or soil that is harmful to humans or animals?
Answer: Pollutant
Explanation:
Answer:
Explanation:
The substance that could be found in air, water or soil that is harmful to humans or animals is called pollutant.
Before running other simulations, try to think of solutions to this global energy dilemma. Write a paragraph in response to the following questions: Should we invest more heavily in renewables at a high up-front financial cost, and/or should we invest in finding more fossil fuels, and/or should we ignore the CO2 limit
Answer:
Yes, we should invest in renewable energies that decrease the levels of carbon dioxide in the atmosphere, that is, decrease the partial pressure of carbon dioxide in the environment since it is lethal for human life, in turn I consider that the exploitation of sources Oil was widely used but I think they should have an end, since they pollute oceans, oxygen, and fuels that are not necessary for today's technology since automobiles could be a base of electric or solar energy
Explanation:
Environmental pollution is a very serious current problem, since we should stop investing in oil sources and be able to keep those profits or that economy in another way or replace them with another resource, since if this greenhouse effect is not perceived as a current serious problem, we could run the risk of suffering natural catastrophes, systematic diseases, affections in the same life.
Given that a 15.00 g milk chocolate bar contains 9.500 g of sugar, calculate the percentage of sugar present
in 15.00 g of milk chocolate bar keeping in mind that the answer should have four significant figures (two
decimal places).
Answer:
63.33%
Explanation:
To find the percentage of sugar in the milk chocolate just divide 15 by 9.5. This will give you 0.63333 (recurring). To find the percentage, times it by 100, which will give you 63.33%.
Hope this helped :)
The percentage of sugar present in the milk would be 63.33 %.
What are significant figures?In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.
As given in the problem a 15.00 g milk chocolate bar contains 9.500 grams of sugar, and we have calculated the percentage of sugar present,
The percentage of sugar present in the milk = 9.500 ×100/15
= 63.33 %
Thus, the percentage of sugar present in the milk would be 63.33 %.
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A solution is made by mixing 10.2 grams of CaCl2 in 250 grams of water. What is the molality of the
solution?
Answer:
The molality of this solution is 0.368 molal
Explanation:
Step 1: Data given
Mass of CaCl2 = 10.2 grams
Molar mass of CaCl2 = 110.98 g/mol
Mass of water = 250 grams = 0.250 kg
Step 2: Calculate the number of moles CaCl2
Moles CaCl2 = mass CaCl2 / molar mass CaCl2
Moles CaCl2 = 10.2 grams / 110.98 g/mol
Moles CaCl2 = 0.0919 moles
Step 3: Calculate the molality of the solution
Molality solution = moles CaCl2 / mass water
Molality solution = 0.0919 moles / 0.250 kg
Molality solution = 0.368 mol / kg = 0.368 molal
The molality of this solution is 0.368 molal
Answer:
0.367 m
Explanation:
We have to start with the molality equation:
[tex]m=\frac{mol~of~solute}{Kg~of~solvent}[/tex]
If we want to calculate the molality we need the moles of the solute and the Kg of the solvent. In this case the solute is [tex]CaCl_2[/tex] and the solvent is [tex]H_2O[/tex].
Lets start with the the moles of solute. For this calculation we need the molar mass of [tex]CaCl_2[/tex] ([tex]111~g~CaCl_2[/tex]) and the mass, so:
[tex]10.2~g~CaCl_2\frac{1~mol~CaCl_2}{111~g~CaCl_2}[/tex]
[tex]0.0918~mol~CaCl_2[/tex]
Now, we can calculate the Kg of solvent, if we do the conversion from grams to Kg:
[tex]250~g~CaCl_2\frac{1~Kg}{1000~g}[/tex]
[tex]0.25~Kg[/tex]
With these values we can calculate the "molality":
[tex]m=\frac{0.0918~mol~CaCl_2}{0.25~Kg}[/tex]
[tex]m=0.3672[/tex]
I hope it helps!
If you are diving and playing in the waves near a beach and suddenly find that you are being
pulled out into the ocean at a relatively quick pace, what current are you most likely trapped in?
Answer: Wind current
Explanation:
The wind current is the main driving force along with the water currents when one reaches to the seashore. The wind currents above surface of water generates due to the evaporation of water.
The strong wind currents can pull any animal or human walking or running near the sea shore. These currents bring the animal or human towards the sea.
100 mL of a buffer that consists of 0.20 M NH3 and 0.20 M NH4Cl is titrated with 25 mL of 0.20 M HCl. Calculate the pH of the resulting solution given that the Kb for NH3 is 1.8 x 10-5.
Answer:
pH = 9.03
Explanation:
The equilibrium of the NH₄Cl / NH₃ buffer in water is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻
Initial moles of both NH₃ and NH₄⁺ are:
0.100L ₓ (0.20 mol / L) = 0.0200 moles
The NH₃ reacts with HCl producing NH₄⁺, thus:
NH₃ + HCl → NH₄⁺ + Cl⁻
That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺
Moles added of HCl were:
0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:
NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles
NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.
Using H-H equation for bases:
pOH = pKb + log [NH₄⁺] / [NH₃]
Where pKb is -log Kb = 4.745.
Replacing:
pOH = 4.745 + log 0.0250mol / 0.0150mol
pOH = 4.967
As pH = 14- pOH
pH = 9.03
43. Calculate the equilibrium constant at the temperature given. (a) O2 (g) + 2F2 (g) ⟶ 2F2 O(g) (T = 100 °C) (b) I2 (s) + Br2 (l) ⟶ 2IBr(g) (T = 0.0 °C) (c) 2LiOH(s) + CO2 (g) ⟶ Li2CO3 (s) + H2 O(g) (T = 575 °C) (d) N2 O3 (g) ⟶ NO(g) + NO2 (g) (T = −10.0 °C) (e) SnCl4 (l) ⟶ SnCl4 (g) (T = 200 °C)
Answer:
Explanation:
O₂(g) + 2F₂(g) ↔ 2F₂(g)
Stabdard ΔG values are
[tex]\Delta G_f[F_2O]=41.9kJ/mol =41900J/mol[/tex]
[tex]\Delta G_f[O_2]=0\\\\ \Delta G_f[F_2]=0[/tex]
[tex]\Delta G^0=\sum \Delta G^\circ (products)- \sum \Delta G ^\circ (reactants)[/tex]
[tex]\Delta G^\circ = [2 \times 41900]-0\\\\=83800J/mol[/tex]
Now,
[tex]\Delta G^\circ =-RTInK[/tex]
Given T = 100°C
= 100+ 273.15 = 373.15K
R = 8.314J/k / mol
so,
83800 = -8.314 * 373.15 * InK
InK = -27.0116
K = 1.858 * 10⁻¹²
Equilibrium constant = 1.858 * 10⁻¹²An electrochemical cell contains a standard hydrogen electrode and a cathode consisting of a metallic chromium electrode, Cr(s), in contact with a 1.00 M chromium solution, Cr3 (aq). The voltage produced by this cell was measured at 25oC. Which statements describe the results of this measurement, assuming the conditions are ideal? The cell voltage with the appropriate sign equals
I. the cell potential.
II. the electromotive force.
III. the standard cell potential.
IV. the standard reduction potential for Cr3+/Cr.
a. I only
b. l and Il
c. I, II and III
d. I, II, II, and IV
e. ll only
Answer:
c. I, II and III
Explanation:
The cell is as follows
Cr / Cr⁺²(1M) // H⁺ ( 1 M ) / H₂
Standard reduction potential of hydrogen cell is zero . Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell
Cell potential = Ecathode - Eanode
= 0 - ( - E)
= E
E is cell potential and also standard cell potential or emf of the cell .
Standard reduction potential that is for Cr3+/Cr. is - E .
Hence statement I , II , III are right . IV th statement is wrong because of sign
Option c is correct.
The branch of chemistry which deals with electricity is called electrochemistry.
The correct answer is C
The cell representation is as follows:-
[tex]Cr / Cr^{2+}(1M) // H^+ ( 1 M ) / H_2[/tex]
The standard reduction potential of hydrogen cells is zero. Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell because it will behave as a cathode.
The formula of the electric cell is as follows :-[tex]Cell potential = E_{cathode} - E_{anode[/tex]
After putting the value the cell potential will be:-
Cell potential = 0 - ( - E)
The cell potential will be= E
The standard reduction potential that is for [tex]Cr^{3+}/Cr = - E .[/tex]
Hence, the correct option is C that is I, II, and IV.
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An enzyme can catalyze a reaction with either of two substrates, S1 or S2. The Km for S1 was found to be 2.0 mM, and the Km, for S2 was found to be 20 mM. A student determined that the Vmax was the same for the two substrates. Unfortunately, he lost the page of his notebook and needed to know the value of Vmax. He carried out two reactions: one with 0.1 mM S1, the other with 0.1 mM S2. Unfortunately, he forgot to label which reaction tube contained which substrate. Determine the value of Vmax from the results he obtained:
Answer:
101
Explanation:
Provided that
[tex]S_1 = S_2 = same\ V_{max}[/tex]
And,
[tex]S_1\ k_M = 2.0mM\\S_2\ k_M = 20mM[/tex]
Now we expect the same
{S} (0.1mM)
This determines that [tex]S_1[/tex] generates a higher rate of product formation as compared to the [tex]S_2[/tex]
So we can easily calculate the [tex]V_{max}[/tex] for either of [tex]S_1[/tex] or [tex]S_2[/tex] as we know that Tube 1 is [tex]S_2[/tex] and tube 2 is [tex]S_1[/tex]
As we know that
[tex]V_0 = V_{max}\ {S} / (K_M + {S})[/tex]
As the rates do not include any kind of units so we do not consider the units for [tex]V_{max}[/tex]
Now the calculation is
[tex]0.5 = V_{max} (0.1\ mM) / (20\ mM + 0.1\ mM)[/tex]
[tex]V_{max} = 0.5 (20.1\ mM) / 0.1\ mM[/tex]
= 100.5
≈ 101
What is the molarity of a solution that has 4.4 moles BaCl2 in 1.5 Liters of solution?
Answer:
2.9 M
Explanation:
Step 1: Given data
Moles of barium chloride (solute): 4.4 moles
Volume of solution: 1.5 liters
Step 2: Calculate the molarity of barium chloride in the solution
The molarity is a way to quantitatively express the concentration of a solute in a solution. The molarity is equal to the moles of solute divided by the volume, in liters, of solution.
[tex]M=\frac{4.4mol}{1.5L} =2.9 M[/tex]
Oxygen in air can be removed using
Answer:Fractional distillation of air. About 78 per cent of the air is nitrogen and 21 per cent is oxygen. These two gases can be separated by fractional distillation of liquid air.
Explanation:
Is electricity an essential property of matter ?
Answer: I think so.. (sorry if i am wrong)
Explanation: Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. A physical property is a characteristic of matter that is not associated with a change in its chemical composition.
Given the reaction: 2Na(s) + 2H20(1) 2Na+(aq) + 2OH(aq) + H2(g)
This reaction goes to completion because one of the products formed is
1.
an insoluble base
2.
a soluble base
3
a precipitate
4.
a gas
A soluble base is formed when sodium reacts with water.
What happens when sodium reacts with water?
When sodium reacts with water, it produces strongly alkalic sodium hydroxide which is also called caustic soda and hydrogen gas. In this chemical reaction, energy is absorbed which means it is an exothermic reaction so we can conclude that a soluble base is formed when sodium reacts with water.
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Which phrase best describes heat?
A. the energy that an object has as a result of its temperature
B
the average translational kinetic energy of the particles in an object
C
the energy transferred between objects at different temperatures
D
the total amount of energy possessed by the particles in an object
Answer: A
Explanation: Heat is the temperature of an object.
Answer:
Option C
the energy transferred between objects at different temperature.
Calculate the molar mass of a gaseous substance if 0.125 g of the gas occupies 93.3 mL at STP.
30.2 g/mol
30.4g/mol
30.6 g/mol
30.0 g/mol
None of the above
Answer:
30.0g/mol
Explanation:
Step 1: Given data
Mass of the gas: 0.125 gPressure (P): 1 atm (standard pressure)Temperature (T): 273.15 K (standard temperature)Volume (V): 93.3 mLStep 2: Calculate the moles of the gas
We will use the ideal gas equation.
[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1atm \times 0.0933L}{\frac{0.0821atm.L}{mol.K} \times 273.15K} = 4.16 \times 10^{-3} mol[/tex]
Step 3: Calculate the molar mass of the gas
4.16 × 10⁻³ moles correspond to a mass of 0.125 g. The molar mass of the gas is:
[tex]\frac{0.125g}{4.16 \times 10^{-3} mol} =30.0g/mol[/tex]
The molar mass of the 0.125 g of the gas occupies 93.3 mL at STP is 30.0 g/mol.
Number of moles of Gas at STP,
[tex]\bold{n =\dfrac {PV}{RT}}[/tex]
where,
P - pressure
V- volume
R- gas constant
T - temperature
Put the values in the formula,
[tex]\bold{n =\dfrac {1 \times 0.0933} {0.082 \times 273.15 }}\\\\\bold{n =4.16 \timesw 10^-^3}[/tex]
The molar mass of the gas can be calculated using formula,
[tex]\bold {m = \dfrac {w}{n}}\\\\\bold {m = \dfrac {0.125} {4.16 \times 10^-^3}}\\\\\bold {m = 30g/mol}[/tex]
The molar mass of the 0.125 g of the gas occupies 93.3 mL at STP is 30.0 g/mol.
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Determine whether the following pairs of elements can form ionic compounds
Answer:
Oxygen and Magnesium.
Based on the difference in electronegativity, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, whereas lithium and calcium, sulfur and bromine, and manganese and chlorine cannot.
Ionic compounds are formed when there is a high electronegativity difference between two elements.
To determine whether the following pairs of elements can form ionic compounds, we need to look at their electronegativity difference:
1. Lithium and calcium: The electronegativity difference between Lithium and calcium is only 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.
2. Sulfur and bromine: The electronegativity difference between sulfur and bromine is 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.
3. Manganese and chlorine: The electronegativity difference between manganese and chlorine is 1.5, which is less than 1.7. Therefore, they cannot form an ionic compound.
4. Potassium and sulfur: The electronegativity difference between potassium and sulfur is 2.4, which is greater than 1.7. Therefore, they can form an ionic compound.
5. Chlorine and lithium: The electronegativity difference between chlorine and lithium is 2.8, which is greater than 1.7. Therefore, they can form an ionic compound.
6. Oxygen and magnesium: The electronegativity difference between oxygen and magnesium is 1.7, which is equal to 1.7. Therefore, they can form an ionic compound.
Therefore, based on the electronegativity difference between the elements, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, while lithium and calcium, sulfur and bromine, and manganese and chlorine cannot form ionic compounds.
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Sulfurous acid is a diprotic acid with the following acid-ionization constants: Ka1 = 1.4x10−2, Ka2 = 6.5x10−8 If you have a 1.0 L buffer containing 0.252 M NaHSO3 and 0.139 M Na2SO3, what is the pH of the solution after addition of 50.0 mL of 1.00 M NaOH? Enter your answer numerically to 4 decimal places.
Answer:
pH = 7.1581
Explanation:
The equilibrium of NaHSO₃ with Na₂SO₃ is:
HSO₃⁻ ⇄ SO₃²⁻ + H⁺
Where K of equilibrium is the Ka2: 6.5x10⁻⁸
HSO₃⁺ reacts with NaOH, thus:
HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺
As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:
HSO₃⁻: 0.252 moles
SO₃²⁻: 0.139 moles
Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:
0.0500L ₓ (1mol /L): 0.050 moles of NaOH.
Thus, final moles of both compounds are:
HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles
SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles
Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:
pH = pka + log [SO₃²⁻] / [HSO₃⁻]
Where pKa is - log Ka = 7.187
Replacing:
pH = 7.187 + log [0.189] / [0.202]
pH = 7.1581516 mL of a 3.82 M sodium sulfate (Na2S04) solution is diluted with 0.875 L of water. What is the new concentration in molarity?
Answer
Molarity = [tex]1.4mol/L[/tex]
Explanation:
Molarity provides the number of moles of solute per liter of solution (moles/Liter). It is a means by which concentration of solution is measured.
SEE THE ATTACHMENT BELOW FOR STEP BY STEP SOLUTION.
Answer:
1.42 M
Explanation:
In this case have a dilution problem, therefore we need to use the dilution equation:
[tex]C_1*V_1=C_2*V_2[/tex]
What values we have?
[tex]C_1=3.82M[/tex]
[tex]V_1=526mL(0.516L)[/tex]
[tex]C_2=?[/tex]
[tex]V_2=?[/tex]
Now, we can calculate [tex]V_2[/tex] if we add the volumes, so:
[tex]0.516~L+~0.875~L=1.391~L[/tex]
So, [tex]V_2=1.391~L[/tex]
We can plug the values in the equation:
[tex]3.82~M*0.516~L=C_2*1.391~L[/tex]
[tex]C_2=\frac{3.82~M*0.516~L}{1.391~L}[/tex]
[tex]C_2=1.42~M[/tex]
I hope it helps!
There is a pattern to the phases of the moon. The time between one full moon to another is about a;new b;waxing crescent c;1st quarter d;full
Answer: a. new
Explanation:
The time between one full moon to another is new moon. The lunation is the average time taken for the formation of new moon to the next. The new moon is the initial visible crescent of the Moon after being conjugated with the Sun. Per lunar cycle can be assigned with a lunation number for the purpose of identification.
PLEASE HELP I HAVE LIMITED TIME!!
How many particles are in 0.500 moles of
N.05?
Answer:
3.01×10²³ particles.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles. This implies that 1 mole of N2O5 also contains 6.02×10²³ particles.
Now, if 1 mole of N2O5 contains 6.02×10²³ particles ,
Then 0.5 mole of N2O5 will contain = 0.5 × 6.02×10²³ = 3.01×10²³ particles.
Therefore, 0.5 mole of N2O5 contains 3.01×10²³ particles.
Select the correct answer from each drop-down menu.
Describe an oxidation-reduction reaction.
In an oxidation-reduction reaction, oxidation is what happens when a reactant___ . Reduction occurs when a reactant ___ in the reaction.
Explanation:
First off, we have to understand what we means by oxidation and reduction.
Oxidation is the loss of electrons during a reaction by a molecule, atom or ion. Oxidation occurs when the oxidation state of a molecule, atom or ion is increased.
Reduction on the other hand is the opposite of oxidation. It is the gain of electrons during a reaction by a molecule, atom or ion. Reduction occurs when the oxidation state of a molecule, atom or ion decreases.
Back to the question;
oxidation is what happens when a reactant loses electrons.
Reduction occurs when a reactant gains electrons in the reaction.
1. the purpose of the aqueous solutions in a galvanic cell is to?
a. provide ions to be oxidized and reduced
b. provide a path for the free flow of electrons
c. maintain charge balance in the cell
d. correct any volume changes in the cell
2. given these half-reactions,
B2 + 2e- -> 2B- Ecell= 0.662 V
A+ Ie- -> A Ecell= -1.305V
what is the standard potential for the overall reaction. 2A+B2 -> 2AB
a. -1.97 V
b. -0.643 V
c. +3.272 V
d. +1.967 V
Answer:
1. The correct option is;
c. maintains charge balance in the cell
2. The correct option is;
c. +3.272 V
Explanation:
The aqueous solution in a galvanic cell is the electrolyte which is a ionic solution containing that permits the transfer of ions between the separated compartment of the galvanic cell such that the overall system is electrically neutral
Therefore, the aqueous solution maintains the charge balance in the cell
2. Here we have;
B₂ + 2e⁻ → 2B⁻ Ecell = 0.662 V
A⁺ + 1e⁻ → A Ecell = -1.305 V
Hence for the overall reaction, we have;
2A + B₂ → 2AB gives;
(0.662) - 2×(-1.305) = +3.272 V.
Interpret and describe an example of a chemical formula. Summarize the two parts a chemical formula must have.
Answer and Explanation:
A chemical formula is composed by chemical symbols (in letters, which indicate the chemical elements) and subscripts (numbers). For example, let see the chemical formula for carbon dioxide:
CO₂
The letters C and O are the chemical symbols for the elements carbon (C) and oxygen (O). The number 2 in subscript indicates that there are 2 atoms of O per molecule. The subscript 1 is generally not indicated in chemical formulae, so it is assumed that the number of atoms of C is 1.
Summarizing, the chemical formula CO₂ indicates us that the molecule if formed by 1 atom of the element carbon (C) and 2 atoms of the element oxygen (O).
The functional groups in an organic compound can frequently be deduced from its infrare d absorption spectrum. A compound, C3H6O2, exhibits intense absorption at 1740 cm-1, accompanied by a band at 1050 cm-1. No absorption above 3000 cm-1 is present .
What functional class(es) does the compound belong to?
List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly.
Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm-1.
The functional class(es) of this compound is(are) .(Enter letters from the table below, in any order, with no spaces or commas.)
a. alkane (List only if no other functional class applies.)
b. alkene
c. terminal alkyne
d. internal alkyne
e. arene
f. alcohol
g. ether
h. amine
i. aldehyde or ketone
j. carboxylic acid
k. ester
l. nitrile
Answer:
(b.) alkene
(i) aldehyde/ketone
(k.) ester
Explanation:
Peaks observed at:
1740 cm-1 indicates the presence of a carbonyl group: C=O aldehyde, ketone, esters.1050 cm-1 shows presence of carbo bonded to electronegative elements e.g. C-N or C-O3000cm-1 is usually indicative of alcohols or carboxylic acid -OH group, which rules out those classes.