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vladimir2022 [97]
1 year ago
8

What is the temperature change on a 75.0 gram sample of mercury if 480.0 cal of heat are added to it? The specific heat of mercu

ry = 0.033 cal/g C
Chemistry
1 answer:
Olegator [25]1 year ago
6 0

Answer:

\Delta T=194^oC

Explanation:

Hello,

In this case, in terms of the heat, mass, heat capacity and change in temperature, we can analyze thermal changes as:

Q=mCp\Delta T\\

In such a way, we compute the required change in temperature as shown below:

\Delta T=\frac{Q}{mCp}=\frac{480.0cal}{75.0g*0.033\frac{cal}{g^oC} }  \\\\\Delta T=194^oC

Such change in temperature is positive indicating an increase in the temperature as the involved heat is positive, in means that heat was added to increase the temperature.

Best regards.

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The half reaction is:  

V⁵⁺ (aq) + 4e⁻ → V¹⁺ (aq)


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Answer:

[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;

pH = 11.4; pOH = 2.6

Explanation:

The chemical equation is

\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \,\text{OH}$^{-}$

For simplicity, let's re-write this as

\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$

1. Calculate [OH]⁻

(a) Set up an ICE table.  

   B + H₂O ⇌ BH⁺ + OH⁻

0.310               0        0

   -x                  +x      +x

0.310-x               x        x

K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}

Check for negligibility:

\dfrac{0.310}{1.8 \times 10^{-5}} = 17 000 > 400\\\\x \ll 0.310

(b) Solve for [OH⁻]

\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}

2. Calculate the pOH

\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}

3. Calculate the pH

\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.6 = \mathbf{11.4}

4 Calculate [H₃O⁺]

\text{H$_{3}$O$^{+}$} = 10^{-\text{pH}} = 10^{-11.4} = \mathbf{4.2 \times 10^{-12}} \textbf{ mol/L}

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