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4 years ago

In a 51.0-g aqueous solution of methanol ch4o the mole fraction of methanol is 0.270. what is the mass of each component?

1 answer:

4 0

Mole fraction is another way of expressing the concentration of a solution or mixture. It is equal to the moles of one component divided by the total moles.

mass CH4O = 0.270 (51) = 13.77 g methanol

water = 51 - 13.77 = 37.23 g water

Hope this answers the question. Have a nice day.

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calculate the mols of alt gas if the volume is 0.97 liters at a temperature of 12 C and the pressure is 152 Kpa’s

katrin2010 [14]

Answer:

0.062mol

Explanation:

Using ideal gas law as follows;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821Latm/molK)

T = temperature (K)

Based on the information provided;

P = 152 Kpa = 152/101 = 1.50atm

V = 0.97L

n = ?

T = 12°C = 12 + 273 = 285K

Using PV = nRT

n = PV/RT

n = (1.5 × 0.97) ÷ (0.0821 × 285)

n = 1.455 ÷ 23.39

n = 0.062mol

4 0

3 years ago

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fomenos

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3 years ago

Which observation does not indicate that a chemical reaction has occurred?

castortr0y [4]

Answer:

change in the total mass of substances

6 0

3 years ago

What part of the sun do we see from earth?

NISA [10]

Answer:

photosphere

Explanation:

photosphere

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8 0

4 years ago

Read 2 more answers

Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container

Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

Explanation:

Here Pressure in a container is given as

$P=\frac{1}{3} \rho$

Here

P is the pressure which is to be calculated

ρ is the density of the gas which is to be calculated as below

$\rho =\frac{mass}{Volume of container}$

Here

mass is to be calculated for 75 gas phase molecules as

$m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g$

Volume of container is 0.5 lts

So density is given as

$\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3$

is the mean squared velocity which is given as

$=RMS^2$

Here RMS is the Root Mean Square speed given as 605 m/s so

$=RMS^2\\=(605)^2\\=366025$

Substituting the values in the equation and solving

$P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa$

So the average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

6 0

4 years ago