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3 years ago

Explain the meaning of the expression “like dissolves like.”

1 answer:

8 0

"Like dissolves like" is an expression used by chemists to remember how some solvents work. It refers to "polar" and "non-polar" solvents and solutes. Basic example: Water is polar. Oil is non polar. Like dissolves like, that means polar dissolves polar, so water dissolves salt.

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Two descriptions about physical quantities are given below:

Mashcka [7]

Answer:

quantity A is mass and quantity B is wieght

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3 years ago

Do you think it's possible for a renewable resource such as water to become depleted or

Naya [18.7K]

Yes if we use the resource up before it can renew itself then it could be used up and be depleted or near depletion. The resource has to have time to come back and if we use it up to fast then the resource can not take the time it needs to come back.

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2 years ago

Tell how many moles of each element are in the following amounts of each compound

qwelly [4]

Take note of the subscript written for each element in the compound. To find the total number of moles, make sure to multiply the subscript with the number of moles of compound. The answer for each is written below:

a. 3*1 = 3 moles Nitrogen; 3*3 = 9 moles Hydrogen
b. 0.25*2 = 0.5 moles Hydrogen; 0.25*1 = 0.25 moles Oxygen
c. 5*2 = 10 moles Hydrogen; 5*1 =5 moles Sulfur; 5*4 = 20 moles Oxygen
d. 0.75*1 = 0.75 moles Calcium; 0.75*1*2 = 1.5 moles Nitrogen; 0.75*3*2 = 4.5 moles Oxygen

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3 years ago

when heating copper in the final step, the bright copper color changes to a dull brown. will the percent recovery be too high or

jasenka [17]

<span>When heating copper in the final step, the bright copper color changes to a dull brown. I think this process will make the percent recovery low because a change in color can mean some reacted. Hope this answers the question. Have a nice day.</span>

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3 years ago

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