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atroni [7]
2 years ago
13

When 50 ml (50 g) of 1.00 m hcl at 22oc is added to 50 ml (50 g) of 1.00 m naoh at 22oc in a coffee cup calorimeter, the tempera

ture increases to 28.87oc. how much heat is produced by the reaction between hcl and naoh? (the specific heat of the solution produced is 4.18 j/goc)?
Chemistry
1 answer:
vitfil [10]2 years ago
7 0

Answer:

\boxed{\text{2700 J}}

Explanation:

HCl + NaOH ⟶ NaCl + H₂O

There are two energy flows in this reaction.  

Heat of reaction + heat to warm water = 0  

           q₁             +              q₂                 = 0  

           q₁             +          mCΔT              = 0  

Data:

    m(HCl) = 50 g

m(NaOH) = 50 g

           T₁ = 22       °C

          T₂ = 28.87 °C

           C = 4.18 J·°C⁻¹g⁻¹

Calculations:

 m = 50 + 50 = 100 g

ΔT = 28.87 – 22 = 6.9 °C

 q₂ = 100 × 4.18 × 6.9 = 2900 J

q₁ + 2900 = 0

q₁ = -2900 J

The negative sign tells us that the reaction produced heat.

The reaction produced \boxed{\textbf{2900 J}}.

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What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?
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Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.
3 0
2 years ago
The statement that percent yield can never be greater than theoretical yield is another example of the ________.
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We can rephrase the statement with a little more specificity in order to understand the answer here.

The mass of the products can never be more than the The mass that is expected.
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Answer:

In order to take steps by humans for controlling the change in climate.

Explanation:

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Answer:

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Review Graham's Law => Effusion Rate ∝ 1/√formula mass.

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2 years ago
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