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KATRIN_1 [288]
2 years ago
15

0.20 L of HNO3 is titrated to equivalence using 0.12 L of 0.2 M NaOH. What is the concentration of the HNO3?

Chemistry
1 answer:
AleksandrR [38]2 years ago
4 0

M_{A}V_{A}=M_{B}V_{B}\\(0.20)(M_{A})=(0.12)(0.2)\\M_{A}=\frac{(0.12)(0.2)}{(0.20)}=\boxed{0.12 \text{ M}}

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Will the electronegativity of barium be larger or smaller than that of strontium? Explain. EN of Ba (0.89) is smaller than EN of Sr (0.95). Because of larger size and more inner core e- shielding, Barium will more readily share its valence electron in the 6s sublevel than Sr will share it valence e- in the 5s sublevel.

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How many grams of zinc metal will react completely with 7.8 liters of 1.6 M HCl? Show all of the work needed to solve this probl
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<u>Answer:</u>

For 1: The correct answer is 407.97 grams.

For 2: The correct answer is 76.72 grams.

<u>Explanation:</u>

  • <u>For 1:</u>

We are given the molarity of HCl, to find the moles of HCl, we use the formula:

Molarity=\frac{\text{Moles}{\text{Volume}}

We are given:

Molarity = 1.6 M

Volume = 7.8 L

Putting values in above equation, we get:

1.6mol/L=\frac{\text{Moles of HCl}}{7.8L}\\\\\text{Moles of HCl}=12.48mol

For the given reaction:

Zn(s)+2HCl(aq.)\rightarrow ZnCl_2(aq.)+H_2(g)

By Stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of zinc metal.

So, 12.48 moles of HCl react with = \frac{1}{2}\times 12.48=6.24mol of Zinc metal.

To calculate the mass of zinc metal, we use the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}    .....(1)

Molar mass of Zinc metal = 65.38 g/mol

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We are given that 298 grams of 83.1 % by mass of iron (II) nitrate solution are present.

So, mass of Iron (II) nitrate solution will be = \frac{83.1}{100}\times 298=247.638g

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Putting values in equation 1, we get:

\text{Moles of }Fe(NO_3)_2=\frac{247.638g}{180g/mol}=1.37mol

For the following reaction:

2Al(s)+3Fe(NO_3)_2(aq.)\rightarrow 3Fe(s)+2Al(NO_3)_3(aq.)

By Stoichiometry of the reaction:

3 moles of Fe(NO_3)_2 produces 3 moles of iron metal.

So, 1.37 moles of Fe(NO_3)_2 will produce = \frac{3}{3}\times 1.37=1.37mol of iron metal.

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1.37=\frac{\text{Mass of iron metal}}{56g/mol}\\\\\text{Mass of iron metal}=76.72g

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