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Rasek [7]
3 years ago
11

The heat of vaporization for liquid zinc is 1.76 kj/g. How much heat is needed to boil 11.2 g of liquid zinc already at its boil

ing point?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
6 0
<h3>Answer:</h3>

= 19.712 kJoules

<h3>Explanation:</h3>
  • Heat of vaporization refers to the amount of heat required to change a unit mass of a substance from liquid to gaseous state without change in temperature.

To calculate the amount of heat, we use,

Amount of heat = Mass × Heat of vaporization

Q = m×Lv

Given;

Mass of liquid Zinc = 11.2 g

Lv of liquid Zinc = 1.76 kJ/g

Therefore;

Q = 11.2 g × 1.76 kJ/g

  = 19.712 kJ

Thus, the amount of heat needed to boil 11.2 g of zinc is 19.712 kilo-joules.

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Question 10 OT 20
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8 0
3 years ago
On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

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[H^{+}] = [NO_2^{-}] = x at equilibrium

x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

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[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

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= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

6 0
3 years ago
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