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2 years ago

Given each of the following sets of values for an ideal gas, calculate the unknown quantity.

Chemistry

1 answer:

omeli [17]2 years ago

5 0

Answer:

a. 5.77L

b. 700mmHg

c. 395K

Explanation:

Using PV = nRT we can solve these problems where:

P is pressure of the gas in atm (1atm = 760mmHg)

V is volume in liters

n are moles of the gas

R is gas constant: 0.082atmL/molK

T is asbolute temperature in K

a. PV = nRT

V = nRT/P

P = 773mmHg*(1atm/760mmHg) = 1.017atm

T = 25°C+273 = 298K

V = 0.240mol*0.082atmL/molK*298K / 1.017atm

V = 5.77L

b. PV = nRT

P = nRT/V

P = 0.0947mol*0.082atmL/molK*309K/0.635L

P = 0.9216atm * (760mmHg/1atm) = 700mmHg

c. PV = nRT

PV/nR = T

P = 727mmHg * (1atm / 760mmHg) = 0.9566atm

0.9566atm*13.3L/0.393mol*0.082atmL/molK = T

T = 395K

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Explain why magnesium and aluminium react in a similar way?

postnew [5]

Because magnesium loses its electrons easily than aluminum

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2 years ago

How many grams of hydrogen are needed to react with 1.40 g of nitrogen to produce ammonia?

vazorg [7]

Answer:

Write a balanced chemical reaction:

N2 + 3H2 ==> 2NH3

Looking at the mole ratios in this balanced equation you can see it takes 3 moles H2 to make 2 moles NH3. So, next calculate the moles of NH3 represented by 1.80 g and then convert to moles of H2 needed:

moles of NH3 = 1.80 g x 1 mole/17 g = 0.106 moles NH3

Moles H2 needed = 0.106 moles NH3 x 3 moles H2/2 moles NH3 = 0.159 moles H2 needed

Grams H2 needed = 0.159 moles x 2 g/mole = 0.318 grams H2 needed

Explanation:

8 0

2 years ago

Ab2 has a molar solubility of 3.72×10−4 m. what is the value of the solubility product constant for ab2?

Gekata [30.6K]

Solubility product constants are values to describe the saturation of ionic compounds with low solubility. A saturated solution is when there is a dynamic equilibrium between the solute dissolved, the dissociated ions, the undissolved and the compound. It is calculated from the product of the ion concentration in the solution. For the generic salt, AB2, the dissociation would be as follows:

AB2 = A2+ + 2B-

So, the expression for the solubility product would be:

Ksp = [A2+] [B-]^2
Ksp = [x] [2x]^2 = 4x^3

where x = 3.72×10^−4 M

Ksp = 4( 3.72×10^−4 )^3
Ksp = 2.06x10^-10 M^3

The solubility product constant of AB2 would be Ksp = 2.06x10^-10 M^3.

4 0

2 years ago

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 1.74 g of butane is

xeze [42]

Answer : The mass of $CO_2$ produced will be, 5.3 grams.

Explanation : Given,

Mass of $C_4H_{10}$ = 1.74 g

Mass of $O_2$ = 11 g

Molar mass of $C_4H_{10}$ = 58 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $C_4H_{10}$ and $O_2$ .

$\text{Moles of }C_4H_{10}=\frac{\text{Mass of }C_4H_{10}}{\text{Molar mass of }C_4H_{10}}=\frac{1.74g}{58g/mole}=0.03moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{11g}{32g/mole}=0.34moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_4H_{10}$ react with 13 mole of $O_2$

So, 0.03 moles of $C_4H_{10}$ react with $\frac{13}{2}\times 0.03=0.195$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C_4H_{10}$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$ .

As, 2 moles of $C_4H_{10}$ react to give 8 moles of $CO_2$

So, 0.03 moles of $C_4H_{10}$ react to give $\frac{8}{2}\times 0.03=0.12$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$ .

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.12mole)\times (44g/mole)=5.3g$

Therefore, the mass of $CO_2$ produced will be, 5.3 grams.

5 0

3 years ago

A fuel tank has a width of 18 inches, a height of 11 inches and a depth of 5 inches. What is the volume of the tank in m3

cupoosta [38]

Answer:

0.016 m³

Explanation:

Step 1: Given data

Width of the tank (w): 18 in

Height of the tank (h): 11 in

Depth of the tank (d): 5 in

Step 2: Calculate the volume of the tank in in³

The fuel tank is a cuboid. We can calculate its volume using the following expression.

V = w × h × d

V = 18 in × 11 in × 5 in

V = 990 in³

Step 3: Convert 990 in³ to m³

We will use the relationship 1 m³ = 61023.7 in³.

990 in³ × (1 m³ / 61023.7 in³) = 0.016 m³

6 0

2 years ago