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Aliun [14]
3 years ago
14

The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its co

njugate, and the other half still remains unreacted. If 0.580 moles of a monoprotic weak acid (Ka = 4.5 _ 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?
Chemistry
1 answer:
kirill [66]3 years ago
5 0
The only thing you need to know to find  pH of the solution at the half-equivalence point is to remember that  pH is always equal to pKa, therefore you can easily get ehat you need - <span>4.35.</span>
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7 0
3 years ago
The temperature of a 95.4 g piece of Cu increases from 25.0 °C to 48.0 °C when the Cu absorbs 849 J of heat. What is the specifc
melisa1 [442]
<h3>Answer:</h3>

0.387 J/g°C

<h3>Explanation:</h3>
  • To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
  • Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
  • That is, Q = mcΔT

in our question we are given;

Mass of copper, m as 95.4 g

Initial temperature = 25 °C

Final temperature = 48 °C

Thus, change in temperature, ΔT = 23°C

Quantity of heat absorbed, Q as 849 J

We are required to calculate the specific heat capacity of copper

Rearranging the formula we get

c = Q ÷ mΔT

Therefore,

Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)

                                        = 0.3869 J/g°C

                                        = 0.387 J/g°C

Therefore, the specific heat capacity of copper is 0.387 J/g°C

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3 years ago
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Given the following values for the heats of formation, what is the number of moles of ethane (C2H6, MW 30.0) required to produce
baherus [9]

Answer:

0.641 moles of ethane

Explanation:

Based on the equation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

We can determine ΔH of reaction using Hess's law. For this equation:

<em>Hess's law: ΔH products - ΔH reactants</em>

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<em />

ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}

ΔH = -1559.7kJ/mol

That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:

1.00x10³kJ * (1mole ethane / 1559.7kJ) =

<h3>0.641 moles of ethane</h3>
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