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Artemon [7]
3 years ago
13

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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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

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Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

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