Answer:
A precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined. The insoluble salt that falls out of solution is known as the precipitate, hence the reaction's name.
Explanation:
What is the molarity of a solution of 3 moles (mol) of FeBr3 in 1/2 L water?
Answer:
3mol/0.5liters or 6 M
Explanation:
The molarity of a solution of 3 moles (mol) of FeBr₃ in 1/2 L water is 6 Molar Solution.
What is Molarity ?Molarity (M) is the amount of a substance in a certain volume of solution.
Molarity is defined as the number of moles of a solute per liters of a solution.
Molarity is also known as the molar concentration of a solution.
Molarity is number of moles of solute in 1 litre of solution.
Therefore,
If 0.5 ltr of solution consisting of 3 moles of FeBr₃
Thus,
1 ltr of solution consisting of 6 moles of FeBr₃.
6 moles of solute in 1 ltr of solution means 6 Molar solution.
Hence, The molarity of a given solution of 3 moles (mol) of FeBr₃ in 1/2 L water is 6 Molar Solution.
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which statement best describle for formation of solution
Explanation:
small amount of solute is dissolved in a large amount of solvent. it would not make sense to have more solute and solvent. you can think of it like this, you can dissolve a teaspoon of salt in a litre water but you can't resolve 1 kg of salt in a teaspoon of water.
also, the solute is what is begin dissolved and the solvent is what dissolving the solute so the one day the solvent is dissolved in the solute does not make complete sense.
3. The exosphere is the outer layer of the
O D. thermosphere
OC. mesosphere
O A. ozone layer
O B. troposhpere
ang
AT
HHH
Answer:
D. exosphere is the outer layer of the thermosphere
A 2.95 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder that contains 12.13 mL H2O. What will be the final volume reading on the graduated cylinder?
Answer:
12.54mL
Explanation:
Step 1:
Data obtained from the question:
Mass of Zn = 2.95g
Density of Zn = 7.14 g/mL
Volume of water = 12.13mL
Final volume reading of cylinder =..?
Step 2:
Determination of the volume of Zn. The volume of Zn can be obtained as follow:
Density = Mass /volume
7.14 = 2.95/volume
Cross multiply
7.14 x Volume = 2.95
Divide both side by 7.14
Volume = 2.95/7.14
Volume of Zn = 0.41mL
Step 3:
Determination of the final volume reading of the cylinder.
The final reading of the cylinder will be the addition of the volume of water and the volume of Zn as shown below:
Final volume of cylinder = volume of water + volume of Zn.
Volume of water = 12.13mL
Volume of Zn = 0.41mL
Final volume =..?
Final volume of cylinder = 12.13 + 0.41
Final volume of cylinder = 12.54mL
Therefore, the final volume reading of the cylinder is 12.54mL
The final volume reading of the graduated cylinder is 12.54 mL.
A 2.95 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder. The volume of the piece of zinc is:
[tex]2.95 g \times \frac{1mL}{7.14 g} = 0.413 mL[/tex]
The final volume reading of the graduated cylinder will be equal to the sum of the volumes of the piece of zinc and the water.
[tex]V = V(H_2O)+V(Zn) = 12.13 mL + 0.413 mL = 12.54 mL[/tex]
The final volume reading of the graduated cylinder is 12.54 mL.
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write briefly how oxygen gas is prepared in the industry
by fractional distillation of liquefied air or the use of zeolites to remove carbon dioxide and nitrogen from air or the electrolysis of water
How sensitive to changes in water temperature are coral reefs? To find out, measure the growth of corals in aquariums where the water temperature is controlled at different levels. Growth is measured by weighing the coral before and after the experiment.
(a) What are the explanatory and response variables?
explanatory variable:
response variable:
(b) Are they quantitative or categorical?
The explanatory variable is:
The response variable is:
Answers:
(a) The explanatory variable is the change in the temperature .
The response variable is the coral survivance.
(b) The explanatory variable is quantitative.
The response variable is categorical.
Explanation:
A quantitative variable is those that can take a range of values. In this case, the temperature may vary across a range of values associated with the survival of coral reefs.
A categorical (also known as qualitative) variable is characterized to present finite states. In this case, the coral reefs can survive or not depending on their responses to changes in the temperature.
Response and explanatory variables are used in experimental procedure ; the explanatory and response variables are :
Measured variable - > Change in weight of coral (quantitative) Explanatory variable - > Temperature of water. (categorical)Explanatory variables are independent variables which causes changes to the measured variable.
Measured variable is the variable we intend to measure in an experiment. They are called dependent variables.
Quantitative variables are numerical values, weight is measured as a number. Hence, it is quantitative.
Categorical variables are labels and are used to group objects into classes. They are Non-numeric. Levels of temperature could be (high, medium, low). Hence, temperature is categorical in the experiment.
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Calculate the final temperature of 12.0 g of Argon (considered as an ideal gas) that is expanded reversibly and adiabatically from V1 = 1.00 dm3 at T1 = 273.15 K to V2 = 3.00 dm3 . CV,m(Ar) = (3/2) * R.
Answer:
Explanation:
For adiabatic change the expression is
= [tex]PV^\gamma=constant[/tex]
[tex]=(\frac{RT}{V})V^\gamma = constant[/tex]
[tex]=TV^{\gamma-1} = constant[/tex]
[tex]=T_1V_1^{\gamma-1} =[/tex][tex]=T_2V_2^{\gamma-1}[/tex]
for Argon γ = 1.67
273.15 x 1 = T₂ x [tex]3^{1.67-1}[/tex]
T₂ = 273.15 / [tex]3^{0.67}[/tex]
= 273.15/ 2.0877
= 130.83 K.
g 1. An ideal gas is confined to a container with adjustable volume. The number of moles, n, and temperature, T, are constant. By what factor will the volume change if pressure decreases by a factor of 8.5 Hint: pV
Answer:
The volume will increase by a factor of 8.5
Explanation:
In an ideal gas, the volume is inversely proportional to the pressure. This means that the volume will decrease when the pressure increases and vice versa.
Let the initial volume be v1, and the initial pressure p1.
the final pressure is decreased by a factor of 8.5, i.e
p2 = p1/8.5 = 0.12p1
using p1v1 = p2v2
v2 = (p1/p2)v1
v2 = (p1/0.12p1)v1
v2 = 8.5v1
This means that the final pressure increases by a factor of 8.5
Answer:
Increases by a factor of 8.5
Explanation:
Hello,
In this case, with the given data, we can apply the Boyle's law in order to understand the pressure-volume behavior as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, for the given pressure decrease we have:
[tex]P_2=\frac{P_1}{8.5}[/tex]
So the volume changes by:
[tex]V_2=\frac{P_1V_1}{\frac{P_1}{8.5} } \\\\V_2=8.5V_1[/tex]
Therefore, it increases by a factor of 8.5.
Regards.
QUESTION 5
The attraction between the nucleus of the atom and the valence electrons for any given atom is called what?
Answer:
Electrostatic Force of Attraction
Hope this helps ;)
Explanation:
PLEASE HELP I HAVE LIMITED TIME!!
How many particles of water (H20) are in a collection of snowflakes
with a mass of 0.005 g?
Answer:
Explanation:
Si tomamos en cuenta el peso molecular del agua, que es equivalente a:
1 Átomo de H₂O
O = 16 gr/mol
H = 1 gr/mol
H₂O = 18 gr/mol
Teóricamente sabemos que en 1 mol de H2O habrá 18 gr.
Para obtener los moles presentes en 1 mg de H₂O, (como 1 gr = 1000 mg), decimos:
1 mol H2O ………………………….. 18000 mg
X …………………………… 1 mg
X = 1 / 18000 = 5,56 X 10⁻⁵ moles de H20
Y para obtener la cantidad de moléculas presentes, de acuerdo a los moles, multiplicamos por el número de Avogadro (6,023 X 10²³ moléculas /mol)
Moléculas de H₂O = 5,56x 10⁻⁵ mol x 6,023 x 10²³
Moléculas de H₂O = 3,34488 x10¹⁹ moléculas de H₂O
En el copo de nieve habrá 3,34488 x 10¹⁹ moléculas de H₂O.
Espero que te sirva =)
5,43 X 10²² particles of water (H20) are in a collection of snowflakes with a mass of 0.005 g.
What do you mean by mole ?The term mole is defined as the amount of substance of a system which contains as many elementary entities.
One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.
The mole concept can be used to convert between mass and number of particles.
1 atom of H₂O
O = 16 g/mol
H = 1 gr/mol
H₂O = 18 gr/mol
1 mol H₂O ………………………….. 18000 mg
? X …………………………… 1 mg
X = 1 / 18000
= 5,43 X 10²² moles de H20
Thus, 5,43 X 10²² particles of water (H20) are in a collection of snowflakes with a mass of 0.005 g.
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One kilogram is the same as
Answer:
1000 grams
Explanation:
Kilo- is the SI prefix meaning 1000. So, 1 kilogram is the same as 1000 grams.
Suppose there is 0.63 g of HNO3 per 100 mL of a particular solution. What is the concentration of the HNO3 solution in moles per liter?
Answer:
There are 0.09996826 moles per liter of the solution.
Explanation:
Molar mass of HNO3: 63.02
Convert grams to moles
0.63 grams/ 63.02= 0.009996826
Convert mL to L and place under moles (mol/L)
100mL=0.1 L
0.009996826/0.1= 0.09996826 mol/L
The concentration of the HNO₃ solution is 10⁻¹ moles per liter.
What is Molar Concentration ?
Molar Concentration is known as Molarity. It is the number of moles of solute present in per liter of the solution.
Molar concentration in terms of Molecular weight
C = [tex]\frac{m}{V} \times \frac{1}{M.W}[/tex]
where,
C is the molar concentration in mol/L
m is the mass or weight of solute in grams
V is volume of solution in liters
M.W is the molecular weight in g/mol
Molar concentration of HNO₃ = [tex]\frac{\text{Mass of}\ HNO_3}{\text{Liter of solution}} \times \frac{1}{\text{Molecular weight of}\ HNO_3}[/tex]
Molecular weight of HNO₃ = Atomic weight of H + Atomic weight of N + 3 (Atomic weight of O)
= 1 + 14 + 3 (16)
= 15 + 48
= 63 g/mol
Convert mL into L
1 mL = 0.001 L
100 mL = 100 × 0.001
= 0.1 L
Now, put the value in above formula we get
Molar concentration of HNO₃ = [tex]\frac{\text{Mass of}\ HNO_3 / \text{Liter of solution}}{\text{Molar mass of}\ HNO_3}[/tex]
= [tex]\frac{0.63\ g / 0.1\ L}{63\ g/mol}[/tex]
= [tex]\frac{6.3}{63}[/tex]
= 0.1
= 10⁻¹ moles per liter
Thus, we can say that 10⁻¹ moles per liter is the concentration of the HNO₃ solution.
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Many computer chips are manufactured from silicon , which occurs in nature as SiO2. WhenSiO 2 is heated to melting , it reacts with solid carbon to form liquid silicon and carbon monoxide gas. In an industrial preparation of silicon, 155.8 kg of SiO2 reacts with 78.3 kg of carbon to produce 66.1 kg of silicon. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.
Answer:
In the given case, SiO2 is the limiting reactant, 71.72 Kg is the theoretical yield, and 90.8 percent is the experimental yield.
Explanation:
A reactant that completely gets used up in a chemical reaction, and it limits the concentration of product, which can be produced is termed as a limiting reactant. The maximum concentration of product that can get generated from the limiting reactant is termed as the theoretical yield.
Based on the given question, the balanced chemical reaction is:
SiO₂ + 2C ⇒ Si + 2CO
Now the mole ratio of SiO₂: C = 1: 2
Based on the given information, the mass of SiO₂ is 155.8 Kg. The no. of moles can be calculated by using the formula, n = weight/molecular mass.
The molecular mass of SiO₂ is 60.08 gram per mole. By putting the values we get,
n = 155.8 × 10³ grams / 60.08 grams/mol
n = 2.59 × 10³ moles
The weight of Carbon given is 78.3 Kg or 78.3 × 10³ grams, the molecular mass of C is 12 gram per mole. Similarly, the moles of C will be,
n = 78.3 × 10³ grams / 12 g/mol
n = 6.52 × 10³ moles
The experimental molar ratio of SiO₂: C is,
= 2.59 × 10³: 6.52 × 10³
= 2.59: 6.52 = 1: 2.5
Hence, it is clear that carbon in the given case is present in excess amount, therefore, SiO₂ is a limiting reactant.
With the help of the balanced equation, the molar ratio of SiO₂: Si is 1: 1. As SiO₂ is the limiting reactant, therefore, theoretical yield of the reaction will be determined by SiO₂.
As calculated, the moles of SiO₂ used is 2.59 × 10³
Hence, the moles of Si produced in the given reaction is 2.59 × 10³.
Now the mass of Si produced theoretically can be determined by using the formula, moles = weight/molar mass. The molar mass of Si is 28.08 gram per mole. Now putting the values we get,
Weight = 2.59 × 10³ moles × 28.08 gram per mole
Weight = 72.72 × 10³ grams or 72.72 Kg
The theoretical yield of Si is 72.72 Kg, however, the experimental yield is 66.1 Kg. Therefore, the percent yield for the reaction will be,
= Experimental yield / Theoretical yield × 100 %
= 66.1 Kg / 72.72 Kg × 100%
= 90.8 %
The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 760.0 torr and the vapor pressure of water under the experimental conditions is 23.8 torr, what is the pressure (torr) of the H 2 gas
Answer:
the pressure (torr) of theH₂ gas is 736.2 torr
Explanation:
Given that
The H₂ gas produced in a chemical reaction is collected through water in a eudiometer; during this process, the gas collected contains some droplets of water vapor along with these gas.
So; the total pressure in the eudiometer = Pressure in the H₂ gas - Pressure of the water vapor
Where;
[tex]P_{Totsl}[/tex] = total pressure in the eudiometer = 760.0 torr
[tex]P_{H_2}[/tex] = Pressure in the H₂ gas = ???
[tex]P_{H_2O}[/tex] = Pressure in the water vapor = 23.8 torr
Now:
[tex]P_{Totsl}[/tex] = [tex]P_{H_2}[/tex] + [tex]P_{H_2O}[/tex]
- [tex]P_{H_2}[/tex] = + [tex]P_{H_2O}[/tex] - [tex]P_{Totsl}[/tex]
[tex]P_{H_2}[/tex] = - [tex]P_{H_2O}[/tex] + [tex]P_{Totsl}[/tex]
[tex]P_{H_2}[/tex] = (- 23.8 + 760) torr
[tex]P_{H_2}[/tex] = 736.2 torr
Thus; the pressure (torr) of theH₂ gas is 736.2 torr
If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.
A eudiometer is a laboratory device that measures the change in volume of a gas mixture following a physical or chemical change. Usually, the gas is collected over water.
According to Dalton's law of partial pressures, the sum of the sum partial pressure of the hydrogen and the partial pressure of the water is equal to the total pressure in the eudiometer.
[tex]P = pH_2 + pH_2O\\\\pH_2 = P - pH_2O = 760.0 torr - 23.8 torr = 736.2 torr[/tex]
If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.
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Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 25 grams of C is formed in 8 minutes. How much (in grams) is formed in 16 minutes
Answer: 45. 78 g is formed in 16 minutes.
Explanation:
Let [tex]A_{o} =40 g[/tex] and [tex]B_{o} =50 g[/tex]
we know that [tex]\alpha =A_{o}\frac{M+N}{M}\, and \, \beta =B_{o}\frac{M+N}{N}[/tex]
According to the question to create [tex]x[/tex] part of the chemical C we will need 2 part of A and one part of B.
Therefore, M = 2 and N = 1.
Now, we can easily solve for the value of α and β.
[tex]\alpha =40\frac{2+1}{2}=60\\\\\, and \, \beta =50\frac{2+1}{1}=150[/tex]
Now, the differential equation must be: [tex]\frac{dX}{dt}=k\left ( \alpha -X \right )\left ( \beta -X \right )[/tex]
I separate the variable and solve the equation
[tex]\int \frac{dx}{\left ( 60-x \right )\left ( 150-x \right )}=\int k\, dt[/tex]
[tex]ln\frac{150-x}{60-x}=90kt+C_{1}[/tex]
By using [tex]X(0) =0[/tex],
[tex]\frac{150-x}{60-x}=Ce^{90k_{o}},C=\frac{5}{2}[/tex]
and using [tex]X (8)=25[/tex] and solving for the value of 'k'
[tex]\Rightarrow \frac{150-25}{60-25}=\frac{5}{2}e^{450k}\\\\\Rightarrow 3.6\times \frac{2}{5}=e^{450k}\\\\\Rightarrow 1.4=e^{450k}[/tex]
Taking 'ln' both side, we get
[tex]\Rightarrow k=7.4716\times 10^{-4}[/tex]
We obtain:[tex]X(t)=\frac{60Ce^{90kt}-150}{Ce^{90kt}-1}[/tex]
Now, for the 16 min
[tex]\Rightarrow X(16)=\frac{150e^{1.07559}-150}{2.5e^{1.0759}-1}[/tex]
[tex]\Rightarrow X(16)=\frac{439.7583-150}{7.3293-1}\\\\\Rightarrow X(16)=\frac{289.7583}{6.3293}[/tex]
[tex]\Rightarrow X(16)=45.78 g[/tex]
Eduardo hurts his knee during a basketball game. The trainer applies a cold pack, which gets cold after being squeezed. The next day, Eduardo’s friend Beth is going sledding. Beth buys some hand warmers and puts them into her gloves. All morning, her hands stay toasty warm.
Required:
a. How do you think these devices work?
b. Where do you think the "cold" and the heat comes from?
Answer:
How do you think these devices work?
Both devices follows the principle and endothermic and exothermic reactions.The cold pack is made up of two bags. The first bag contains water.And it is placed inside the second bag which contains NH4NO3 , and urea. Therefore, when the water bag is squeezed, the bag busted releasing the water which dissolves the surrounding solid compounds. These compounds absolved heat energy in an endothermic reaction.
The components of hand warmers are activated carbon, salt iron, water,,and cellulose.The Oxidation of iron when exposed to the air (oxygen)generated heat in an exothermic reaction to keep the hand warm.
Thus the cold results from the endothermic reaction as the solid compounds in cold packs absorbed water to break bonds to melts,and the hotness comes from formation of new bonds(exothermic) as oxidation of Fe to form new compounds with other chemical occurs.
Explanation:
Imagine you are volunteering in a 1st grade classroom. The students recently learned about the water cycle. One student raises her hand and asks, "I heard that elements like carbon can cycle too. How does carbon cycle? Is it like how water cycles?" How would you explain the cycling of carbon to this 1st grade class? Keep in mind that the students have likely not taken an advanced science course yet!
Answer:
Carbon cycle is also important for the survival of life on earth.
Explanation:
In carbon cycle, carbondioxide gas is present in the atmosphere. This carbondioxide gas is taken inside by the plant body through stomata and made food from it. When this food is eaten by animals, this carbon is transferred to their bodies. During breathing, carbondioxide gas is removed from animal body and the carbondioxide gas again goes to the atmosphere.
Sort The following iron compound by whether the cation is iron(II) or iron(III)
Answer:
Left, Left, Right, Left, Right, Right
Explanation:
Follow that order!
g + In a coffee-cup calorimeter, when 3.25 g of NaOH is dissolved in 50.00 g of water initially at 22.0 oC, the temperature of the solution increases to 24.8 oC. Calculate the q of the reaction in kJ/mol of NaOH. Assume that the specific heat of the solution is 4.184 J/g oC, and that there is no other heat transfer than that of the solution process itself.
Answer:
THE STANDARD HEAT OF SOLUTION OF SODIUM HYDROXIDE IN WATER IS -7.68 KJ PER MOLE.
Explanation:
Variables:
Mass of NaOH = 3.25 g
Mass of water = 50 g
Initial temperature of water = 22°C = 22 + 273 K = 295 K
Final temperature of the reaction mixture = 24.8 °C = 24.8 + 273 K = 297.8 K
Assuming that:
1. specific heat of water = 4.184 J/g °C
2. total mass of the reaction mixture = 50 g + 3.25 g = 53.25 g
3. the rise in temperature = (297.8 K - 295 K ) = 2.8 K
4. Molar mass of sodium hydroxide = ( 23 + 16 + 1) = 40 g/mol
5. number of mole of sodium hydroxide = mass / molar mass
n = 3.25 g / 40 g/mol
n = 0.08125 moles
The rise in temperature for the reaction mixture produces how much of heat:
Heat = mass * specific heat * change in temperature
Heat = 53.25 * 4.184 * 2.8
Heat = 623.8344 J of heat.
Equation of reaction:
NaOH + H2O -------> NaOH + H2O + Heat
This is not a reaction but a dissolution as sodium hydroxide is very soluble in water and this reaction is exothermic where heat is given off.
So since 3.25 g having 0.08125 moles produces 623.8344 J of heat, 1 mole of the sodium hydroxide used will produce:
0.08125 mole of sodium hydroxide = 623.8344 J of heat
1 mole of sodium hydroxide = ( 623.8344 / 0.08125 J of heat
= 7677.96 J of heat per mole of sodium hydroxide.
= 7.68 kJ of heat
So therefore, the standard heat of solution of sodium hydroxide in water is -7.68 kJ of heat since its an exothermic reaction.
Sodium Chloride (NaCl) has a molecular mass of 58 grams/mole. What is the molarity of a NaCl solution with a
concentration of 232 grams/liter?
2.0 molar
4.0 molar
0.5 molar
3.0 molar
Answer:
2nd option
Explanation:
Molarity is the number of moles of the solute (NaCl) in 1 litre of the solution (NaCl solution).
Given: concentration= 232g/ L
what we are trying to achieve is __mol/ L.
So in 1 litre, we have 232g of NaCl.
To convert mass to mole, we divide it by the Mr.
Given that the Mr is 58g/mol,
number of moles
= 232 ÷58
= 4
Thus, 1 litre has 4 moles of NaCl.
Therefore, the molarity is 4.0 mol/L.
if two atoms are bonded to a central atom with no lone pairs , how will they be arranged?
How many hydrogen atom are present in 1.53 g of water
Answer:
[tex]=1.02x10^{23} atoms\ H[/tex]
Explanation:
Hello,
In this case, for water, whose molar mass is 18 g/mol, we can find two moles of hydrogen in one mole of water, therefore, for us compute the atoms, we should also use the Avogadro's number as shown below:
[tex]=1.53gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} *\frac{6.022x10^{23}atoms\ H}{1molH} \\\\=1.02x10^{23} atoms\ H[/tex]
Regards.
A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128 J/g oF; water = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium
Answer: The final temperature of both the weight and the water at thermal equilibrium is [tex]50.26^{o}C[/tex].
Explanation:
The given data is as follows.
mass = 7.62 g, [tex]T_{2} = 10.8^{o}C[/tex]
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q = [tex]mC \times \Delta T[/tex]
= [tex]7.62 g \times 4.184 J/^{o}C \times (52.3 - T)[/tex]
Now, heat gained by lead will be calculated as follows.
q = [tex]mC \times \text{Temperature change of lead}[/tex]
= [tex]2.04 \times 0.128 \times (T - 11.0)[/tex]
According to the given situation,
Heat lost = Heat gained
[tex]7.62 g \times 4.184 J/^{o}C \times (52.3 - T)[/tex] = [tex]2.04 \times 0.128 \times (T - 11.0)[/tex]
T = [tex]50.26^{o}C[/tex]
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is [tex]50.26^{o}C[/tex].
The conversion of methyl isonitrile to acetonitrile in the gas phase at 250 °C CH3CN(g) is first order in CH3NC with a rate constant of 3.00×10^-3 s-1. If the initial concentration of CH3NC is 5.45×10^-2 M, the concentration of CH3NC will be 1.56×10^-2 M after_____________ s have passed.
Answer: The concentration of [tex]CH_3NC[/tex] will be [tex]1.56\times 10^{-2}M[/tex] after 416 seconds have passed.
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]3.00\times 10^{-3}s^{-1}[/tex]
t = age of sample = ?
a = let initial amount of the reactant = [tex]5.45\times 10^{-2}M[/tex]
a - x = amount left after decay process = [tex]1.56\times 10^{-2}M[/tex]
[tex]t=\frac{2.303}{3.00\times 10^{-3}}\log\frac{5.45\times 10^{-2}}{1.56\times 10^{-2}}[/tex]
[tex]t=416s[/tex]
The concentration of [tex]CH_3NC[/tex] will be [tex]1.56\times 10^{-2}M[/tex] after 416 seconds have passed.
describe how the relative density of a salt solution can be determined using the hydrometer
The hydrometer is an instrument used to determine the density of liquids, and looks like a large glass float filled at the bottom with grains of lead or mercury.
Because the mass of the hydrometer is known and thus the mass of the displaced liquid, the density can be determined using the volume.
There is a scale on the handle, the density can be read here. The deeper the hydrometer sinks, the lower its density.
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Why do heart diseas patient's eat oil instead of fat?
Answer:
lube
Explanation:
Consider an Al-4% Si alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at 578˚C, (d) the amounts and compositions of each phase at 576˚C, (e) the amounts and compositions of each microconstituent at 576˚C, and (f) the amounts and compositions of each phase at 25˚C.
Answer:
(a) Hypoeutectic
(b) Alpha solid, aluminium
(c) 70% α , 30% β
(d) 97.6% α, 2.4% β
(e) 97.6% α, 2.4% β
(f) 97% α, 3% β
Explanation:
(a) The eutectic composition for Al Si alloy is 11.7 wt% silicon, therefore, an Al-4% Si alloy is hypoeutectic
(b) For the hypoeutectic alloy, aluminium, Al, is expected to form first, such that the aluminium content is reduced till the point it gets to the eutectic proportion of 11.7 wt% silicon
(c) At 578°C we have
% α: Al (11 - 4)/(11 - 1) = 70% α
% L: Si 100 - 70 = 30% β
(d) At 576°C we have
α: 99.83% Si (99.83 - 4)/(99.83- 1.65) = 97.6% α
β: 1.65% Si (4 - 1.65)/(99.83- 1.65) = 2.4% β
(e) Primary α: 1.65% α (99.83 - 4)/(99.83 - 1.65) = 97.6% α
Eutectic 4% Si = 100 - 97.6 = 2.4% β
(f) At 25°C we have;
α%: (99.83 - 4)/(99.83 - 1) = 97% α
β%: 100 - 97 = 3% β.
What is mass per unit volume called?
Answer:
The correct answer is Density
Explanation:
Hope this helps you
Answer:density
Explanation:
which two compounds are structural isomers of each other
Answer:
Explanation:
Structural isomer is a type of isomer in which molecules with the same molecular formula have different bonding patterns and atomic organization.
For example Pentan-1-ol, pentan-2-ol, and pentan-3-ol are structural isomers that exhibit position isomerism.
Cyclohexane and hex-1-ene are examples of functional group structural isomers.
A salt contains a compound containing a metal and a nonmetal. This is an
compound. *
A.molecular
B.covalent
C.Tonic
D.None of the above
Answer:
Option C. Ionic
Explanation:
Ionic bond is formed between a metal and a non metal.
Ionic bond occurs when there is a transfer of electron(s) from a metallic atom to a non metallic atom.
