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sleet_krkn [62]
3 years ago
10

Question 10 of 25

Chemistry
2 answers:
disa [49]3 years ago
7 0

Answer:

d) 0

Explanation:

a p e x :)

Over [174]3 years ago
5 0

Answer: 0

Explanation: A P E X

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How many grams of ethylene glycol (c2h6o2 must be added to 1.25 kg of water to produce a solution that freezes at -5.88 ∘c?
e-lub [12.9K]
The freezing point depression is calculated through the equation,
                                    ΔT = (kf)  x m 
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
                                   5.88 = (1.86)(m)
m is equal to 3.16m

Recall that molality is calculated through the equation,
                                  molality = number of mols / kg of solvent
                                       number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.

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3 years ago
Can you help me with three please? We’re balancing electrons
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2 years ago
A mothball, composed of naphthalene (C10H8), has a mass of 1.52g. How many naphthalene molecules does it contain?
Anna35 [415]
The molar mass of naphthalene is calculated as follows,
                          10 x (12.01 g/ mol) + 8 x (1.01 g/mol) = 128.12 g/mol
Divide the given mass by the molar mass to determine the number of moles of naphthalene present in the ball.
                                  n =  1.52 g / 128.12 g/mol  = 0.01186 mole
Multiply the number of moles to number of moles per mole.
                                 (0.01186 mol) x (6.022 x10^23 molecules / mole) = 7.14 x 10^21 molecules

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