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3 years ago

How many moles of solute are contained in a 250.0ml solution with a concentration of 2.5m

1 answer:

5 0

Mole solute=Molar concentration(mol/L)*Volume solution(L)

250.0 mL=0.2500L
2.5M=2.5 mol/L
mole solute=2.5 mol/L*0.2500L= 0.625 ≈ 0.63 mol

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Explanation:

The law of included fragments states: Fragments found in a rock must be older than the rock itself.

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Stainless steel is an alloy of iron, chromium, nickel, and manganese metals. If a 2.000 g sample contains 10.0% nickel, what is

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200 g

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3 years ago

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3 years ago

Read 2 more answers

The decomposition of N2O4 into NO2 has Kp = 2. Some N2O4 is placed into an empty container, and the partial pressure of NO2 at e

Sunny_sXe [5.5K]

Answer: Option (B) is the correct answer.

Explanation:

Expression for the given decomposition reaction is as follows.

$N_{2}O_{4} \rightarrow 2NO_{2}$

Let us assume that x concentration of $N_{2}O_{4}$ is present at the initial stage. Therefore, according to the ICE table,

$N_{2}O_{4} \rightarrow 2NO_{2}$

Initial : x 0

Change : - 0.1 $2 \times 0.1$

Equilibrium : (x - 0.1) 0.2

Now, expression for $K_{p}$ of this reaction is as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

Putting the given values into the above formula as follows.

$K_{p} = \frac{P^{2}_{NO_{2}}}{P_{N_{2}O_{4}}}$

$2 = \frac{(0.2)^{2}}{(x - 0.1)}$

$2 \times (x - 0.1) = (0.2)^{2}$

x = 0.12

This means that $P_{N_{2}O_{4}}$ = x = 0.12 atm.

Thus, we can conclude that the initial pressure in the container prior to decomposition is 0.12 atm.

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3 years ago