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Ksivusya [100]
3 years ago
9

What is it im confused

Chemistry
2 answers:
Oxana [17]3 years ago
6 0
It is h because it is closet to 35 and the original water was 30
-BARSIC- [3]3 years ago
3 0
It’s g cause it’s closest to 10 which would make 40
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A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol
katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of 0.80^oC and a freezing point depression constant K_f=7.82^oC.kg/mol . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is -17.6^oC

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m

OR

\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}} ......(1)

where,

Freezing point of pure solvent = 0.80^oC

Freezing point of solution = ?^oC

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

K_f = freezing point depression constant = 7.82^oC/m

m_{solute} = Given mass of solute (iron (III) chloride) = 81.1 g

M_{solute} = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

w_{solvent} = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC

Hence, the freezing point of the solution is -17.6^oC

6 0
3 years ago
10. Lead nitrate solution mixed with sodium sulfate solution forms lead sulfate as a
Leviafan [203]

Answer:

88%.

Explanation:

The percentage yield of lead sulfate in this experiment is 88% if 2.53 is divided by 2.85 and multiply by 100. The percentage yield can be calculated when the experimental yield is divided by theoretical yield and then multiply by 100. The percentage yield tells us about the actual yield that is gained in the end of experiment which is lower than theoretical yield.

8 0
3 years ago
How does a difference in 1 pH unit change in terms of H+ concentration? Ex: How does a pH of 3 differ from the pH of 4? Which on
yuradex [85]

Answer:

A difference of 1 would mean there is either x10 more or less depending on where you are starting.

Explanation:

The pH of 3 differs from 4 because the acidity of 3 is 10x higher than that of 4. This is due to how the scientific notation of 3 in relation to pH is 10x^-3 while that of 4 is 10^-4 so in terms of the stronger acid that would be the one with the pH of 3.

6 0
3 years ago
Select all of the BENEFITS of a PARALLEL circuit. If one bulb burns out the rest of the bulbs will stay lit. The bulbs do not ge
artcher [175]

Answer:

The first two options are correct

Explanation:

The first two options are part of the benefits of a parallel connection of bulbs in a circuit. Here, the voltage of each connecting bulb is the same as the voltage of the bulb in the circuit hence all the bulbs have the same voltage running through them. Thus, when one bulb is removed/burns out, it does not affect the remaining bulbs (those ones will remain lit). Also, the addition of bulb(s) does not cause the remaining bulbs in the circuit to get dimmer (since they will all have the same voltage).

4 0
3 years ago
Do CO2 and h2o have the same geometry
Svetach [21]
No, they do not. Carbon dioxide has a linear geometry because the lone pair and bond pair repulsion cancels out; however, water has a bent structure because only the oxygen atom possesses a lone pair which brings the bonding electron pairs closer.
7 0
3 years ago
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