15.0 L of an ideal gas at 298 K and 3.36 atm are heated to 383 K with a new pressure of 5.60 atm. What is the new volume in liters

Answers

Answer 1

Answer:

Answer:

$V_2=11.6L$

Explanation:

Hello!

In this case, since we volume, pressure and temperature which are all changing, we can use the combined ideal gas law to write:

$(P_1V_1)/(T_1) =(P_2V_2)/(T_2)$

Thus, since the final volume V2 is required, by solving for it, we write:

$V_2=(P_1V_1T_2)/(T_1P_2)$

In such a way, we plug in the given data to obtain:

$V_2=(3.36atm*15.0L*383K)/(298K*5.60atm)\n\nV_2=11.6L$

Which means that the process compressed the gas.

Best regards.

Answer 2

Answer:

Final answer:

To find the new volume, we can use the combined gas law equation.

Explanation:

To solve this problem, you can use the combined gas law equation, which relates the initial and final conditions of pressure, volume, and temperature.

The combined gas law equation is: P1 * V1 / T1 = P2 * V2 / T2

Substituting the given values into the equation, we can solve for the new volume:

(3.36 atm * 15.0 L) / 298 K = (5.60 atm * V2) / 383 K

Simplifying and solving for V2, the new volume, we find V2 = 11.78 L.

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Assuming 54.81 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be produced if the reaction will only produce 66.93 % yield?

Answers

Answer:

181.39g of AlCl3 is produced

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3CuCl2•2H2O + 2Al → 2AlCl3 + 6H2O + 3Cu

Next, we shall determine the mass of Al that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:

Molar mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Summary:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Next, we shall determine the theoretical yield of AlCl3. This can be achieved as shown below:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Therefore, 54.81g of Al will react to produce = (54.81 x 267)/54 = 271.01g of AlCl3.

Therefore, the theoretical yield of AlCl3 is 271.01g.

Finally, we shall determine the actual yield of AlCl3 produced from the reaction.

This can be obtain as follow:

Percentage yield of AlCl3 = 66.93%

Theoretical yield of AlCl3 = 271.01g

Actual yield of AlCl3 =?

Percentage yield = Actual yield/Theoretical yield x 100

66.93% = Actual yield /271.01g

Actual yield = 66.93% x 271.01

Actual yield = 66.93/100 x 271.01g

Actual yield = 181.39g.

Therefore, 181.39g of AlCl3 is produced from the reaction.

Based on your Lewis structures for the postlab assignment, which molecules below have a three dimensional structure? (Select all that apply.)(A) CO2
(B) H2S
(C) SO3
(D) PCl3
(E) CH2Cl2
(F) NO2

Answers

Answer:

D and E

Explanation:

D. PCl3

E. CH2Cl2

A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 96.0kPa. The piston is now pulled up, expanding the gas, until the gas has a final volume of 45.0L. Calculate the final pressure of the gas. Be sure your answer has the correct number of significant digits.

Answers

Answer:

The final pressure of the gas is:- 21.3 kPa

Explanation:

Using Boyle's law

${P_1}* {V_1}={P_2}* {V_2}$

Given ,

V₁ = 10.0 L

V₂ = 45.0 L

P₁ = 96.0 kPa

P₂ = ?

Using above equation as:

${P_1}* {V_1}={P_2}* {V_2}$

${96.0\ kPa}* {10.0\ L}={P_2}* {45.0\ L}$

${P_2}=\frac {{96.0}* {10.0}}{45.0}\ kPa$

${P_2}=21.3\ kPa$

The final pressure of the gas is:- 21.3 kPa

Be sure to answer all parts.Calculate the percent composition by mass (to 4 significant figures) of all the elements in calcium
phosphate (Ca3(PO4)2), a major component of bone.
% Ca
%P
% 0

Answers

Answer:

38.7%

41.3%

20%

Explanation:

The percentage composition helps to know the what percent of the total mass of a compound is made up of each of the constituent elements or groups.

To solve this problem:

find the formula mass by adding the atomic masses of the atoms that makes up the compound.
place the mass contribution of the element or group to the formula mas and multiply by 100;

Compound:

Ca₃(PO₄)₂

Formula mass = 3(40) + 2[31 + 4(16)]

= 120 + 2(95)

= 120 + 190

= 310

%C = $(3(40))/(310)$ x 100 = 38.7%

%P = $(8(16))/(310)$ x 100 = 41.3%

%O = $(2(31))/(310)$ x 200 = 20%

I have a cup of hot coffee at 140 oC but I want to cool it to 110 oC. My cup holds about 0.3 kg of coffee. Fortunately, I have a bunch of aluminum cubes in the freezer that I can drop into my hot coffee to cool it down. If each aluminum cube has a mass of 1 g (not 1 kg!) and my freezer keeps its contents at a temperature of –10 oC, how many cubes do I have to drop into my coffee? The specific heat of water is around 4000 joules/kg/oC and aluminum is about 900 joules/kg/oC. (Pick the answer closest to the true value and ignore any thermal losses to surroundings.)

A. 200
B. 330
C. 400
D. 110
E. 88

Answers

Answer:

The correct answer is option B.

Explanation:

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)$

where,

$C_1$ = specific heat of metal = $900 J/kg^oC$

$C_2$ = specific heat of coffee= $4000 J/kg^oC$

$m_1$ = mass of metal = x

$m_2$ = mass of coffee = 0.3 kg

$T_f$ = final temperature of aluminum metal= $110^oC$

$T_1$ = initial temperature of aluminum metal = $-10^oC$

$T_2$ = initial temperature of coffee= $140^oC$

Now put all the given values in the above formula, we get

$x* 900 J/kg^oC* (110-(-10))^oC=-(0.3 kg* 4000 J/kg^oC* (110-140)^oC$

$x=0.333 kg$

Mass of aluminum cubes = 0.3333 kg = 333.3 g

If mass of 1 cube is 1 gram, then numbers of cubes in 333.3 grams will be:

$=(333.3 g)/(1 g)=333.3\approx 330$

330 cubes of aluminum cubes will be required.

What is an extensive property? *A property that changes if temperature changes
A property that will NOT change if temperature changes
A property that changes if the amount of substance changes
A property that does NOT change if the amount of substance changes
Help :( pls

Answers

Answer:

A property that changes if the amount of substance changes

Explanation:

An extensive property is a property that depends on the amount of matter in a sample.

Final answer:

An extensive property changes if the amount of substance changes. For instance, mass and volume are extensive properties as they would vary depending on the amount of substance.

Explanation:

An extensive property is a property that changes if the amount of substance changes. For example, mass and volume are extensive properties. If you have two separate samples of a substance, each with a different amount, their mass and volume would be different. On the other hand, the melting point or boiling point of the substance, which are examples of intensive properties, would not change regardless of the amount of substance.

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