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lesya692 [45]
3 years ago
14

What is the density of water if you have 50.0 grams of water and a volume of 50.0 millimeters

Chemistry
1 answer:
Gnom [1K]3 years ago
8 0

Answer:

\boxed {\tt 1.0 \ g/mL}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass of the water is 50.0 grams.

The volume of the water is 50.0 milliliters.

m= 50.0\ g \\v=50.0 \ mL

Substitute the values into the formula.

d=\frac{50.0 \ g}{50.0 \ mL}

Divide.

d= 1.0  \ g/mL

The density of the water is 1.0 grams per milliliter. Also, remember that the density of pure water is always 1.0 g/mL or g/cm³

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Determine the change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizi
uysha [10]

Answer: The change in boiling point for 397.7 g of carbon disulfide (Kb = 2.34°C kg/mol) if 35.0 g of a nonvolatile, nonionizing compound is dissolved in it is 2.9^0C

Explanation:

Elevation in boiling point:

T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of pure carbon disulfide= 46.2^oC

k_b = boiling point constant  =2.34^0Ckg/mol

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

w_2 = mass of solute = 35.0 g

w_1 = mass of solvent (carbon disulphide) = 397.7 g

M_2 = molar mass of solute = 70.0 g/mol

Now put all the given values in the above formula, we get:

(T_b-46.2)^oC=1\times (2.34^oC/m)\times \frac{(35.0g)\times 1000}{70.0\times (397.7g)}

T_b=49.1^0C

Therefore, the change in boiling point is (49.1-46.2)^oC=2.9^0C

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Answer:it’s abc it’s just science you know

Explanation:

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Neporo4naja [7]

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