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Ulleksa [173]
2 years ago
5

A dissolved solute:

Chemistry
1 answer:
Oduvanchick [21]2 years ago
4 0
It would be c because it has to settle out the solution
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How many grams of ethylene glycol (C2H6O2) must be added to 1.15 kg of water to produce a solution that freezes at -4.46°C?
jok3333 [9.3K]

Answer:

232.5 g C2H6O2

Explanation:

The equation you need to use here is ΔTf = i Kf m

Since pure water freezes at 0 C, your ΔTf is just 4.46 C

i = 1 (ethylene glycol is a weak electrolyte)

Kf = molal freezing constant, which for water is 1.86 C/m

m = molality = x mols C2H6O2 / 1.15 kg H2O (don't know the moles of ethylene glycol we're dissolving yet)

Than,

4.46 C = 1.86 C/m (x mol C2H6O2 / 1.15 kg H2O)

Solve for x, you should get x = 2.75 mol C2H6O2

3.75 mol C2H6O2 (62 g C2H6O2 / 1 mol C2H6O2) = 232.5 g C2H6O2

7 0
3 years ago
Use Lewis theory to determine the chemical formula for the compound formed between Sr and N.
mariarad [96]

Answer:

Sr_3N_2

Explanation:

Hello,

In this case, since the Lewis theory is based on the bonds formation between atoms via the valence electrons, we can verify the chemical formula of the compound formed by strontium and nitrogen by noticing that strontium has two valence electrons as it is in group IIA, for that reason, two nitrogens should be available for bonding. Therefore, since nitrogen is in group VA, it is said that three electrons are required to attain the octet (maximum amount bonded electrons), for that reason, three strontiums are should be available for bonding. In such a way, the formula should be:

Sr_3N_2

Regards.

8 0
3 years ago
I need help on this it’s due today!!!
stich3 [128]
Atmosphere. You can think of the atmosphere as the outer shell for earth.
6 0
2 years ago
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Listed in the Item Bank are key terms and expressions, each of which is associated with one of the columns. Some terms may displ
Elena-2011 [213]

I don't think it is possible.

4 0
3 years ago
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In a second trial of this experiment, the molarity of koh(aq) was determined to be 0.95 m. the actual molarity was 0.83 m. what
Cloud [144]
 The % error in the second trial  is calculated as follows

% error = actual molarity/  theoretical molarity  x100


= 0.83/0.95 x100 = 87.4% error of second trial
5 0
3 years ago
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