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adelina 88 [10]
4 years ago
14

The specific heat capacity of gold is 0.128 J/g C. How much would be needed to warm 250.0 grams of gold from 25.0 C to 100.0 C.

Chemistry
1 answer:
madam [21]4 years ago
7 0

Answer:

3,200 joules

Explanation:

q = mcΔT = (250.0 g)(0.128 j/g°C)(100°C -25°C) = 3,200 joules

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The atomic mass is greater than the atomic number.

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Carbon has four valence electrons, and oxygen has six valence electrons. If carbon and oxygen bond covalently, which of the foll
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8 0
4 years ago
For the titration of 50. mL of 0.10 M ammonia with 0.10 M HCl, calculate the pH. For ammonia, NH3, Kb
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Answer:

11.12 → pH

Explanation:

This is a titration of a weak base and a strong acid.

In the first step we did not add any acid, so our solution is totally ammonia.

Equation of neutralization is:

NH₃ + HCl → NH₄Cl

Equilibrium for ammonia is:

NH₃ + H₂O ⇄  NH₄⁺  +  OH⁻      Kb = 1.8×10⁻⁵

Initially we have 50 mL . 0.10M = 5 mmoles of ammonia

Our molar concentration is 0.1 M

X amount has reacted.

In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.

Expression for Kb is : x² / (0.1 - x)  = 1.8×10⁻⁵

As Kb is so small, we can avoid the x to solve a quadratic equation.

1.8×10⁻⁵ = x² / 0.1

1.8×10⁻⁵  .  0.1 = x²

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√1.8×10⁻⁶ = x → 1.34×10⁻³

That's the value for [OH⁻] so:

1×10⁻¹⁴ = [OH⁻] . [H⁺]

1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²

- log [H⁺] = pH

- log 7.45×10⁻¹² = 11.12 → pH

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