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3 years ago

Which of the following actions will lead to a decrease in pressure?

Chemistry

2 answers:

Alex Ar [27]3 years ago

6 0

The action that would most likely lead to. Decrease in volume is decrease in temperature.

tekilochka [14]3 years ago

6 0

<u>Answer:</u> The correct answer is decreasing temperature.

<u>Explanation:</u>

The relation between pressure, temperature, volume and number of moles of gas is given by the equal known as ideal gas equation:

$PV=nRT$

or,

$P=\frac{nRT}{V}$

where,

P = Pressure of the gas

V = Volume of the gas

n = Number of moles of gas

R = Gas constant

T = Temperature of gas

From the above relation, it is visible that pressure is directly related to number of moles and temperature of the gas. So, if pressure decreases, the number of moles and temperature of the gas also decreases.

Also, pressure is inversely related to volume of the gas. So, if pressure gets decreased, the volume of the gas must increase.

And, pressure is directly related to number of collisions of the particles. Thus, if the pressure decreases, this means that number of collisions between the particles must also decrease.

Hence, the correct answer is decreasing temperature.

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Calculate the grams of solute in each of the following solution: 278 mL of 0.038 M Fe2(SO4)3

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Answer: 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Now put all the given values in the formula of molality, we get

$0.038M=\frac{n}{0.278L}$

$n=0.0105mol$

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Thus 4.22 grams of solute is there in 278 ml of 0.038 M $Fe_2(SO_4)_3$

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3 years ago

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3 years ago

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3 years ago

Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d = $4.50\times 10^1 m=45 m$

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

$V=\pi r^2 h$

$=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L$

$1 m^3=1000 l$

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

$m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg$

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

5 0

3 years ago