The dependent variable measured was ____

A sample of pure NH4HS is placed in a sealed 2.0-L container and heated to 550 K at which the equilibrium constant is 3.5 x 10-3

Zolol [24]

Answer:

The mass of $NH_{3}$ in the container is 2.074 gram

Explanation:

Given:

Volume of $NH_{4} HS$ $V = 2$ lit

Equilibrium constant $k _{eq} = 3.5 \times 10^{-3}$

The reaction in which $NH_{3}$ is produced

$NH_{4} HS$ ⇄ $NH_{3} + H_{2}S$

Here equal moles of $NH_{3}$ and $H_{2}S$ is formed.

From the formula of equilibrium constant,

$k_{eq} = (NH_{3})(H_{2}S )$

$x^{2} = 3.5 \times 10^{-3}$

$x = 0.061$ M

Above value shows,

$NH_{3} = 0.061$ $\frac{moles}{L}$

So in 2 L no. moles of $NH_{3}$ = $0.061 \times 2 = 0.122$ moles.

So mass of 0.122 mole of $NH_{3}$ is = $0.122 \times 17 = 2.074$ g

Therefore, the mass of $NH_{3}$ in the container is 2.074 gram

8 0

3 years ago

5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H

cricket20 [7]

We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

7 0

3 years ago

How can you explain what happened?

galina1969 [7]

Answer:

What do you mean what happened?

Explanation:

???

3 0

2 years ago

What are some insulators and conductors of electricity? Name 2 of each

hammer [34]