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1 year ago

A Certain gas weighs 60g. IF the relative molecular mass of the gas is 12 what is the volume of the gas?

Chemistry

1 answer:

GaryK [48]1 year ago

8 0

Hm I’m not sure could you support more detail

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A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

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Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

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3 years ago

When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200

GaryK [48]

Answer:

$2.023 m^3$ is the total displacement (volume) of the submarine.

Explanation:

Mass of water carried by submarine at 1000 ft depth = m = 2100 kg

The density of seawater at 1000 ft depth = d = $1033 kg/m^3$

Volume of the water displaced = V= ?

Total displacement of the submarine = Volume of the water displaced = V

$Density=\frac{Mass}{Volume}$

$V=\frac{m}{d}=\frac{2100 kg}{1033 kg/m^3}=2.023 m^3$

$2.023 m^3$ is the total displacement (volume) of the submarine.

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Answer:

Molecular Weight

Explanation:

Chromium(III) Carbonate Cr2(CO3)3 Molecular Weight -- EndMemo.

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A Certain gas weighs 60g. IF the relative molecular mass of the gas is 12 what is the volume of the gas?​

A Certain gas weighs 60g. IF the relative molecular mass of the gas is 12 what is the volume of the gas?