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A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure

Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember "Dalton's Law of Partial Pressures". This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a proportional relationship between moles and pressure. In other words, more moles indicate more pressure and vice-versa.

$P_i=P_t_o_t_a_l*X_i$

Where:

$P_i$ =Partial pressure

$P_t_o_t_a_l$ =Total pressure

$X_i$ =mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

moles of hydrogen gas

The molar mass of hydrogen gas ( $H_2$ ) is 2 g/mol, so:

$6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2$

moles of oxygen gas

The molar mass of oxygen gas ( $O_2$ ) is 32 g/mol, so:

$64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2$

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

$P_H_2=\frac{3}{5}*P_t_o_t_a_l$

$P_O_2=\frac{2}{5}*P_t_o_t_a_l$

So, the answer would be 3/5P.

I hope it helps!

5 0

3 years ago

Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s

Nina [5.8K]

Answer:

D) Ca(OH)₂ will not precipitate because Q < Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂ is produced:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca(OH)₂ is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

8 0

3 years ago

Type your answer in decimal form. Do not round. 7 days = hours

djyliett [7]

168.0
Hope this helps u

5 0

2 years ago

Read 2 more answers

7g of gold is heated from 17 degrees C to 42 degrees C. How much energy (heat) was used? The specific heat capacity of gold is .

wel

Heat needed=mcθ
=7×0.031×(42-17)
=5.425cal

7 0

3 years ago

How many grams of water can be produced from 4.6 grams of Hydrogen and 7.3 grams of Oxygen?

12345 [234]

Answer:

The answer to your question is 8.21 g of H₂O

Explanation:

Data

mas of water = ?

mass of hydrogen = 4.6 g

mass of oxygen = 7.3 g

Balanced chemical reaction

2H₂ + O₂ ⇒ 2H₂O

Process

1.- Calculate the atomic mass of the reactants

Hydrogen = 4 x 1 = 4 g

Oxygen = 16 x 2 = 32 g

2.- Calculate the limiting reactant

Theoretical yield = H₂/O₂ = 4 / 32 = 0.125

Experimental yield = H₂/ O₂ = 4.6/7.3 = 0.630

From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.

3.- Calculate the mass of water

32 g of O₂ ---------------- 36 g of water

7.3 g of O₂ --------------- x

x = (7.3 x 36) / 32

x = 262.8 / 32

x = 8.21 g of H₂O

4 0

3 years ago