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anygoal [31]
2 years ago
6

If 29.4 mL of ethanol is dissolved in water to make 359 mL of solution, what is the concentration expressed in volume/volume % o

f the solute?
Chemistry
1 answer:
Sveta_85 [38]2 years ago
6 0

Answer: 8.2\%

Explanation:- Volume percentage is the ratio of volume of solute to the volume of solution defined in terms of percentage.

{\text {volume percentage}}=\frac{\text {volume of solute}}{\text {volume of solution}}\times 100\%

Given: volume of solute = 29.4 ml

Volume of solution= 359 ml

{\text {volume percentage}=\frac{29.4}{359}\times 100\%=8.2\%


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Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces
Shalnov [3]

Answer:  The concentration of excess [OH^-] in solution is 0.017 M.

Explanation:

1. Molarity=\frac{moles}{\text {Volume in L}}

moles of HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles

1 mole of HCl give = 1 mole of H^+

Thus 0.019 moles of HCl give = 0.019 mole of H^+

2. moles of Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles

According to stoichiometry:

1 mole of Ba(OH)_2 gives = 2 moles of OH^-

Thus 0.012 moles of Ba(OH)_2 give = 2 \times 0.012=0.024 moles of OH^-

H^++OH^-\rightarrow H_2O

As 1 mole of H^+ neutralize 1 mole of OH^-

0.019 mole of H^+ will neutralize 0.019 mole of OH^-

Thus (0.024-0.019)= 0.005 moles of OH^- will be left.

[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M

Thus molarity of [OH^-] in solution is 0.017 M.

4 0
2 years ago
Read 2 more answers
25. In the investigation shown the changes in heat of the copper is greater than the change in the heat of the water. What error
forsale [732]

Answer:

Heat lost to the surroundings

Heat lost to the thermometer

Explanation:

All changes in heat, or energy, can be explained. Many of the reactions or changes we see in the world involve the conversion of energy. For example as we heat up a substance (eg. water), the amount of energy we put in should give us an exact temperature. However, this is a "perfect world" scenario, and does not occur in real life. Whenever heat is added to a substance like water, we always need to account for the energy that is going to be lost. For example, heat lost to evaporation or even the effect of measuring the temperature with a thermometer (the introduction of anything including a thermometer will affect the temperature).

3 0
2 years ago
You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
Write the equations that represent the first and second ionization steps for hydroselenic acid (H2Se) in water. (Use H3O+ instea
Otrada [13]

Answer:

The equations are

1) H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

2) HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

Explanation:

There are two ionization steps in the dissociation of hydroselenic acid.

In first dissociation the H₂Se loses one proton and forms hydrogen selenide ion as shown below:

H_{2}Se+H_{2}O--> H_{3}O^{+}+HSe^{-}

The next step is again removal of a proton from the base formed above.

HSe^{-} +H_{2}O--> H_{3}O^{+}+Se^{-2}

4 0
2 years ago
Determine the specific heat (in J/g C) for a 2.508 kilogram substance which increases its temperature from 4.051 C to 42.061 C w
just olya [345]

Answer: 0.036 J/g°C

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = 3.42 Kilojoules

[Convert 3.42 kilojoules to joules

If 1 kilojoule = 1000 joules

3.42 kilojoules = 3.42 x 1000 = 3420J]

Mass = 2.508Kg

[Convert 2.508 kg to grams

If 1 kg = 1000 grams

2.508kg = 2.508 x 1000 = 2508g]

C = ? (let unknown value be Z)

Φ = (Final temperature - Initial temperature)

= 42.061°C - 4.051°C

= 38.01°C

Apply the formula, Q = MCΦ

3420J = 2508g x Z x 38.01°C

3420J = 95329.08g•°C x Z

Z = (3420J / 95329.08g•°C)

Z = 0.03588 J/g°C

Round the value of Z to the nearest thousandth, hence Z = 0.036 J/g°C

Thus, the specific heat of the substance is 0.036 J/g°C

7 0
2 years ago
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