Chemical formula for sodium and sulfur

A sample of 1.44 g of helium and an unweighed quantity of O₂ are mixed in a flask at room temperature. The partial pressure of h

skad [1K]

Answer:

Mass of O₂ present in the mixture = 14.7 g

Explanation:

We apply the mole fraction to solve this excersise:

Mole fractio of a gas is:

Moles of X gas / Total moles = Partial pressure of X gas / Total pressure

We determine total pressure:

42 Torr + 159 Torr = 201 Torr

Sum of mole fraction = 1

Mole fraction of O₂ = 159 Torr / 201 Torr → 0.791

1 - Mole fraction of O₂ = Mole fraction of He → 1 - 0.791 = 0.209

We convert the mass of He, in order to determine the moles

1.44 g / 4g/mol = 0.36 moles

Mole fraction He = Moles of He / Total moles

0.36 moles / Total moles = 0.209

0.209 / 0.36 moles = Total moles → 0.580 moles

Mole fraction of O₂ = 0.791 = Moles of O₂ / Total moles

0.791 . 0.580 moles = Moles of O₂ → 0.459 moles

We convert the moles of O₂ to mass → 0.459 mol . 32g / 1mol = 14.7 g

4 0

3 years ago

At a given temperature, 4.06 atm of H2 and 3.5 atm of Cl2 are mixed and allowed to come to equilibrium. The equilibrium pressure

Llana [10]

Answer: The value of $K_p$ for the given chemical reaction is 0.1415

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_p$ is written as:

$K_p=\frac{p_{C}^cp_{D}^d}{p_{A}^ap_{B}^b}$

For the given chemical equation:

$H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g)$

The expression for $K_p$ for the following equation is:

$K_p=\frac{(p_{HCl})^2}{(p_{H_2)}(p_{Cl_2})}$

We are given:

$p_{HCl}=1.418atm\\p_{H_2}=4.06atm\\p_{Cl_2}=3.5atm$

Putting values in above equation, we get:

$K_p=\frac{(1.418)^2}{(4.06)\times (3.5)}\\\\K_p=0.1415$

The value of $K_p$ for the given chemical reaction is 0.1415

5 0

3 years ago

A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What is the new v

guapka [62]

Ans: Final volume = 25.0 ml

Given:

Initial volume V1 = 50.0 ml

Initial pressure P1 = 20.0 atm

Final pressure P2 = 40.0 atm

To determine:

The final volume V2

Explanation:

Ideal gas equation: PV = nRT

under constant temperature, T and number of moles n we have:

PV = constant

or, P1V1 = P2V2

V2 = P1V1/P2 = 20*50/40 = 25 ml.

6 0

3 years ago

Ammonia gas has a molar mass of approximately 17 grams per mole. At 290K and 1.2 atm, a sample of ammonia has a volume of 3.7 L.

jeka94

Ammonia gas will have a mass of 3.17grams.

HOW TO CALCULATE MASS:

The mass of a substance can be calculated by multiplying the number of moles in the substance by its molar mass.

However, to calculate the number of moles present in the ammonia gas given in this question, we use the formula as follows:

PV = nRT

where;

P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant
T = temperature

n = PV ÷ RT

n = (1.2 × 3.7) ÷ (0.0821 × 290)

n = 4.44 ÷ 23.81

n = 0.186mol

Mass = 0.186 × 17

Mass of ammonia gas = 3.17g

Therefore, the mass of ammonia gas is 3.17grams.

Learn more at: brainly.com/question/21085277?referrer=searchResults

6 0

3 years ago

Percent yield for sodium bromide if 3.87 moles of iron 3 bromide yielded 10.4 moles of sodium bromide

Marizza181 [45]

Answer:

89.57 %

Explanation:

Don't have one just thought id help <3

6 0

3 years ago