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Ronch [10]
3 years ago
12

What’s the answer please

Chemistry
1 answer:
DaniilM [7]3 years ago
4 0

Answer:

5 × 10⁻³ mol

Explanation:

Step 1: Given data

  • Volume of HCl solution (V): 25 cm³
  • Molar concetration of the HCl solution (C): 0.2 mol/dm³

Step 2: Convert "V" to dm³

We will use the conversion factor 1 dm³ = 1000 cm³.

25 cm³ × 1 dm³/1000 cm³ = 0.025 dm³

Step 3: Calculate the moles of HCl (n)

We will use the definition of molarity.

C = n/V

n = C × V

n = 0.2 mol/dm³ × 0.025 dm³ = 5 × 10⁻³ mol

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8 0
3 years ago
At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 (
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Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

                   N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>         0.3411      1.661

<u>At eqllm:</u>     0.3411-x  1.661-3x      2x

Evaluating for 'x', we get:

\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141

Volume of the container = 2.50 L

The expression of K_c for the above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

We are given:

[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M

[N_2]=\frac{0.2001}{2.5}=0.08004M

[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M

Putting values in above expression, we get:

K_c=\frac{(0.1128)^2}{0.08004\times (0.4952)^3}\\\\K_c=1.31

Hence, the equilibrium constant for the above reaction is 1.31

5 0
3 years ago
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ki77a [65]

Answer:

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                    16 mmol

We have an excess of OH⁻, the ones from the NaOH and the ones that formed the salt NaHCOO, because this salt has this hydrolisis:

NaHCOO  →  Na⁺  +  HCOO⁻

HCOO⁻  +  H₂O  ⇄   HCOOH  +  OH⁻   Kb →  Kw / Ka = 5.55×10⁻¹¹

These contribution of OH⁻ to the solution is insignificant because the Kb is very small

So:  [OH⁻] =  16 mmol / 400 mL →  0.04 M

- log  [OH⁻]  = pOH →  1.39

pH = 14 - pOH → 12.61

6 0
3 years ago
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