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3 years ago

What volume will 0.405 g of krypton gas occupy at STP?

Chemistry

2 answers:

Rufina [12.5K]3 years ago

8 0

Answer:

The answer to your question is V = 0.108 L or 108 ml

Explanation:

Data

Volume = ?

mass = 0.405 g

Temperature = 273°K

Pressure = 1 atm

Process

1.- Convert mass of Kr to moles

83.8 g of Kr -------------------- 1 mol

0.405 g ------------------- x

x = (0.405 x 1) / 83.8

x = 0.0048 moles

2.- Use the Ideal gas law to solve this problem

PV = nRT

- Solve for V

V = nRT / P

- Substitution

V = (0.0048)(0.082)(273) / 1

- Simplification

V = 0.108 / 1

- Result

V = 0.108 L

Ganezh [65]3 years ago

4 0

Answer:

The volume of the krypton gas is 108.2 mL

Explanation:

Step 1: Data given

Mass of krypton = 0.405 grams

STP = 1 atm , 273 K

Atomic mass krypton = 83.80 g/mol

Step 2: Calculate moles of krypton

Moles krypton = mass krypton /molar mass krypton

Moles krypton = 0.405 grams / 83.80 g/mol

Moles krypton = 0.00483 moles

Step 3: Calculate volume of krypton gas

p*V = n*R*T

V = (n*R*T)/p

⇒with n = the moles of krypton =0.00483 moles

⇒with R = the gas constant = 0.08206 L*atm/L*mol

⇒with T = the temperature = 273 K

⇒ with p = the pressure = 1 atm

⇒ with V = the volume of the krypton gas = ?

V = 0.00483 * 0.08206 * 273

V = 0.1082 L = 108.2 mL

The volume of the krypton gas is 108.2 mL

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Answer:

616,0 ng is the right answer.

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You should know that 1 mole = 1 .10^9 nanomoles

Get the rule of three.

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MrRissso [65]

Since the density of water is 1 g /mL, hence there is 100 g of H2O. So total mass is:

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=> The heat of reaction can be calculated using the formula:

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4 years ago

the pressure in a sealed plastic container is 108 kPa at 41 degrees Celsius. What is the pressure when the temperature drops to

Katarina [22]

<u>Answer:</u> The new pressure will be 101.46 kPa.

<u>Explanation:</u>

To calculate the new pressure, we use the equation given by Gay-Lussac Law. This law states that pressure is directly proportional to the temperature of the gas at constant volume.

The equation given by this law is:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are initial pressure and temperature.

$P_2\text{ and }T_2$ are final pressure and temperature.

We are given:

By using conversion factor: $T(K)=T(^oC)+273$

$P_1=108kPa\\T_1=41^oC=314K\\P_2=?kPa\\T_2=22^oC=295K$

Putting values in above equation, we get:

$\frac{108kPa}{314K}=\frac{P_2}{295K}\\\\P_2=101.46kPa$

Hence, the new pressure will be 101.46 kPa.

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