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babymother [125]
2 years ago
13

Which volume, in cm3 , of 0.20 mol dm-3 naoh (aq) is needed to neutralize 0.050 mol of h2s (g)? h2s (g) + 2naoh (aq) → na2s (aq)

+ 2h2o (l)?
Chemistry
1 answer:
BigorU [14]2 years ago
6 0
Answer is: volume of sodium hydroxide is 500 cm³.
Chemical reaction: H₂S + 2NaOH → Na₂S + 2H₂O.
From chemical reaction: n(H₂S) : n(NaOH) = 1 : 2.
n(NaOH) = 2 ·0.050 mol.
n(NaOH) = 0.1 mol.
V(NaOH) = n(NaOH) ÷ c(NaOH).
V(NaOH) = 0.1 mol ÷ 0.2 mol/dm³.
V(NaOH) = 0.5 dm³.
V(NaOH) = 0.5 dm³ · 1000 cm³/dm³.
V(NaOH) = 500 cm³.
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A reaction vessel contains nh3, n2, and h2 at equilibrium at a certain temperature. the equilibrium concentrations are [n2] = 0.
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Glucose has a molecular mass of 180 daltons. to make a 2-molar (2 m) solution of glucose, __________.
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6 0
2 years ago
What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution
EastWind [94]

Answer:

The volume required for complete neutralize is 32.29 mL

Explanation:

The computation of the volume required for complete neutralize is shown below:

As we know that, the balanced equation is

Ba(OH)_2 + 2Hcl\rightarrow Bacl_2 + 2H_2O

Now

The number of moles of Ba(OH)_2 = n_1 = 1

And, the number of moles of Hcl = n_2 = 2

Therefore

The equation i.e. to be used to find out the volume is given below:

\frac{M_1V_1}{n_1} =  \frac{M_2V_2}{n_2}

V_2 = \frac{M_1V_1}{n_1} \times \frac{n_2}{M_2} \\\\ = \frac{0.2350 \times 23.55}{1} \times \frac{2}{0.3428} \\\\ = \frac{11.0685}{0.3428}

= 32.29 mL

Hence, the volume is 32.29mL

6 0
2 years ago
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