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ElenaW [278]
3 years ago
6

Melinda filled two glasses equal size half-full with water. The water in one glass was 50 degrees Celsius. The water in the othe

r glass was 10 degrees Celsius. She poured one glass into the other, stirred the liquid, and measured the temperature of the full glass of water. What do you think the temperature of the full glass of water will be after the water is mixed?
Chemistry
1 answer:
Deffense [45]3 years ago
5 0

Answer:

30 °C

Explanation:

When you pour water from one glass to the other, the heat energy will flow from the hot water to the cold.

You have equal masses of water in each glass. The hot water should cool down as much as the cold water warms up.

The final temperature should be halfway between the starting temperatures in each glass.

(10 °C + 50°C)/2 = 60 °C/2 = 30°C

The temperature of the full glass of water should be 30 °C.

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If a buffer contains 1.05M B and 0.750M BH+ has the pH of 9.5. What would be the pH after 0.005mol of HCL is added to 0.5L of so
Leviafan [203]

Answer:

Final pH: 9.49.

Round to two decimal places as in the question: 9.5.

Explanation:

The conjugate of B is a cation that contains one more proton than B. The conjugate of B is an acid. As a result, B is a weak base.

What's the pKb of base B?

Consider the Henderson-Hasselbalch equation for buffers of a weak base and its conjugate acid ion.

\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}}.

\text{pOH} = \text{pK}_w - \text{pH}.

\text{pK}_w = 14.

\text{pOH} = 14 - 9.5 = 4.5

\displaystyle \text{pK}_b = \text{pOH} -\log{\frac{[\text{Salt}]}{[\text{Base}]}}\\\phantom{\text{pK}_b} = 4.5 - \log{\frac{0.750}{1.05}} \\\phantom{\text{pK}_b} =4.64613.

What's the new salt-to-base ratio?

The 0.005 mol of HCl will convert 0.005 mol of base B to its conjugate acid ion BH⁺.

Initial:

  • n(\text{B}) = c\cdot V = 1.05 \times 0.5 = 0.525\;\text{mol};
  • n(\text{BH}^{+}) = c\cdot V = 0.750 \times 0.5 = 0.375\;\text{mol}.

After adding the HCl:

  • n(\text{B}) = 0.525 - 0.005 = 0.520\;\text{mol};
  • n(\text{BH}^{+}) = 0.375+ 0.005 = 0.380\;\text{mol}.

Assume that the volume is still 0.5 L:

  • \displaystyle [\text{B}] = \frac{n}{V} = \frac{0.520}{0.5} = 1.04\;\text{mol}\cdot\text{dm}^{-3}.
  • \displaystyle [\text{BH}^{+}] = \frac{n}{V} = \frac{0.380}{0.5} = 0.760\;\text{mol}\cdot\text{dm}^{-3}.

What's will be the pH of the solution?

Apply the Henderson-Hasselbalch equation again:

\displaystyle \text{pOH} = \text{pK}_b + \log{\frac{[\text{Salt}]}{[\text{Base}]}} = 4.64613 + \log{\frac{0.760}{1.04}} = 4.50991

\text{pH} = \text{pK}_w - \text{pOH}= 14 - 4.50991 = 9.49.

The final pH is slightly smaller than the initial pH. That's expected due to the hydrochloric acid. However, the change is small due to the nature of buffer solutions: adding a small amount of acid or base won't significantly impact the pH of the solution.

3 0
3 years ago
A geologist visits an environment to make observations. The scientist predicts that the environment will undergo secondary succe
kow [346]

Answer:

b c e

Explanation:

i did it

6 0
2 years ago
Read 2 more answers
The mother of a child has a cleft chin while the father of a child has a smooth chin. The child also has a cleft chin
timurjin [86]
The mother has stronger genetics
5 0
3 years ago
A particular solution of NaOH has hydronium concentration of 4x10-9 M. what is the solution’s pH?
bearhunter [10]

Answer:

The pH of the solution is 8, 40-

Explanation:

The pH indicates the acidity or basicity of a substance. PH values between 0 and less than 7 indicate acidic solutions, 7 neutral and greater than 7 to 14 basic. It is calculated as

pH = -log (H30+)

pH= -log (4x10-9M)

<em>pH=8,40</em>

8 0
3 years ago
What is the volume of 3.00 M sulfuric aid that contain 9.809 g of H2SO4 (98.09g/mol)
Slav-nsk [51]

Given :

Molarity of sulfuric acid solution is 3.0 M.

Amount of sulfuric acid present in solution is 9.809 g.

To Find :

The volume of solution.

Solution :

We know, molarity is given by :

Molarity = \dfrac{number \ of \ moles \times 1000}{Volume\ ( ml )}\\\\M = \dfrac{w \times 1000}{M.M \times V}\\\\3 = \dfrac{9.809\times 1000}{98.09 \times V}\\\\V = \dfrac{1000}{10\times 3}\  ml\\\\V = 33.33 \ ml

Therefore, volume required is 33.33 ml .

5 0
3 years ago
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