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spin [16.1K]
3 years ago
11

What amount of heat is required to raise the temperature of 350 grams of copper to cause a 25°C change? The specific heat of cop

per is 0.39 J/g°C.. . A. 2500 joules. B. 2600 joules. C. 3400 joules. D. 3900 joules.
Chemistry
2 answers:
jarptica [38.1K]3 years ago
4 0
The heat required to raise the temperature to a specific temperature change of a sample is related to the specific heat capacity of the substance.  In this case, the heat can be calculated through mCpΔT = 350 g * 0.39 J/g C *25 C. This is equal to 3412. 5 Joules. Closest answer is C.
Gnesinka [82]3 years ago
4 0

Answer   c

Explanation: Because I said don't doubt the all knowing

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borishaifa [10]

Answer:

\large \boxed{\mathbf{1.36; 3.6 \times 10^{\mathbf{-9}}}\textbf{mol/L}}

Explanation:

The HF is about five million times as strong as phenol, so it will be by far the major contributor of hydronium ions. We can ignore the contribution from the phenol.

1 .Calculate the hydronium ion concentration

We can use an ICE table to organize the calculations.

                    HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹:       2.7                   0       0

C/mol·L⁻¹:      -x                   +x      +x

E/mol·L⁻¹:   2.7 - x                 x        x

K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}] \text{F}^{-}]} {\text{[HF]}} = 7.2 \times 10^{-4}\\\\\dfrac{x^{2}}{2.7 - x} = 7.2 \times 10^{-4}\\\\\text{Check for negligibility of }x\\\\\dfrac{2.7}{7.2 \times 10^{-4}} = 4000 > 400\\\\\therefore x \ll 2.7\\\dfrac{x^{2}}{2.7} = 7.2 \times 10^{-4}\\\\x^{2} = 2.7 \times 7.2 \times 10^{-4} = 1.94 \times 10^{-3}\\x = 0.0441\\\text{[H$_{3}$O$^{+}$]}= \text{x mol$\cdot$L$^{-1}$} = \text{0.0441 mol$\cdot$L$^{-1}$}

2. Calculate the pH

\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.0441} = \large \boxed{\mathbf{1.36}}

3. Calculate [C₆H₅O⁻]

C₆H₅OH + H₂O ⇌ C₆H₅O⁻ + H₃O⁺

     2.7                         x        0.0441

K_{\text{a}} = \dfrac{0.0441x} {2.7} = 1.6 \times 10^{-10}\\\\0.0441x = 1.6 \times 10^{-10}\\x = \dfrac{1.6 \times 10^{-10}}{0.0441} = \mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}\\\text{The concentration of phenoxide ion is $\large \boxed{\mathbf{3.6 \times 10^{\mathbf{-9}}}\textbf{ mol/L}}$}

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3 years ago
How do I calculate the boiling point of 25g of calcium chloride in 120g of water
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Answer:

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