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STALIN [3.7K]
3 years ago
12

Aqueous hydrochloric acid HCl will react with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid wat

er H2O. Suppose 20. g of hydrochloric acid is mixed with 7.56 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Chemistry
1 answer:
Arada [10]3 years ago
8 0

here's the answer to your question

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The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:
Nimfa-mama [501]

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of AgNO_{3} in 50.0 mL of 0.100 M of AgNO_{3}

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= \frac{0.100}{1000}\times 50.0 moles

= 0.00500 moles

According to balanced equation, 1 mol of AgNO_{3} reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of AgNO_{3} react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = (100.0\times 1.00) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= \frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}

= \frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}

= -67716 J/mol

[m = mass, c = specific heat capacity, \Delta T = change in temperature and negative sign is included as it is an exothermic reaction]

4 0
3 years ago
100 ml is drawn from 0.1 M solution of KCl and added to 900 ml of water. What is the
grandymaker [24]

Answer:

The new concentration will be 0.01 M.

Explanation:

To determine the new concentration we use the following formula.

concentration (1) × volume (1) = concentration (2) × volume (2)

concentration (1) = 0.1 M

volume (1) = 100 mL

concentration (2) = unknown

volume (2) = 100 mL + 900 mL = 1000 mL

concentration (2) = [concentration (1) × volume (1)] / volume (2)

concentration (2) = (0.1 × 100) / 1000 = 0.01 M

3 0
3 years ago
A procedure calls for the bromination of 12 mmol of trans‑cinnamic acid with 12 mmol of Br 2 in glacial acetic acid solvent. Sel
SashulF [63]

Answer: The yield of dibromide product will be approximately one‑half of the expected yield.

Explanation:

6 0
3 years ago
A sample of metal has a mass of 21.35 g, and a volume of 6.28 mL. What is the density of this metal?
Scilla [17]

Answer:

3.40g/mL

Explanation:

Density is a measure of mass over volume, so to get the density all we have to do is divide the mass by the volume.

21.35g ÷ 6.28 mL = 3.40g/mL

5 0
3 years ago
How many moles are there in 77.1 g of cl2?
natita [175]
There are 70.90g of Cl2 in 1 mol (because 35.45g Cl in one mol). So 77.1g/70.9g≈1.0874 mol.
8 0
3 years ago
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