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aniked [119]
3 years ago
7

A cylinder contains 4.0 g of nitrogen gas. A piston compresses the gas to half its initial volume. Afterward, by what factor has

the mass density of the gas increased?
Chemistry
1 answer:
notka56 [123]3 years ago
8 0

Answer:

The mass density will be doubled

Explanation:

  • Density is given by dividing the mass of a substance by its volume.
  • An increase in mass causes an increase in density and vice versa, while a  decrease in volume causes an increase in density and volume.
  • Therefore, when the volume is halved, then the density will be doubled if the mass is kept constant.
  • This has no effect on the number of moles as the mass is constant.
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Explain how do you would count the sig figs in the following number, 0.00340670, and what is
loris [4]

Answer:

The zeros in front of the 3 show where the decimal is. Then start counting at the 3 to the right. There are 6 sig figs.

Explanation:

3 0
3 years ago
a 1.00litre vessel contains 0.215 mole of nitrogen gas and 0.0118 mole of hydrogen gas at 25°C. determine the partial pressure o
stepladder [879]

Answer:

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8 0
2 years ago
An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To
antoniya [11.8K]

Answer: The new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}         ........(1)

We are given:

P_{PCl_5}=217.0torr

P_{PCl_3}=13.2torr

P_{Cl_2}=13.2torr

Putting values in above equation, we get:

K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24

Now we have to calculate the new partial pressure of Cl_2.

P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}

217.0torr+13.2torr+P_{Cl_2}=263.0torr

P_{Cl_2}=32.8torr

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of Cl_2.

Now, the equilibrium is shifting to the reactant side. The equation follows:

                       PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initial:             13.2         32.8            217.0

At eqm:         13.2-x      32.8-x         217.0+x

Putting values in expression 1, we get:

1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38

Neglecting the 40.4 value of 'x'  because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of PCl_5 = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of PCl_3 = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of Cl_2 = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of PCl_5,PCl_3\text{ and }Cl_2 when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0
3 years ago
Which of the following is the car average speed?
kap26 [50]

Answer:

it would be 100 km/hr

Explanation:

if you divide each speed by the time you get 100 each time

4 0
3 years ago
1. A 5.05 g sample of quartz (SiO2) contains 2.36 g of silicon. What are the percentages of silicon
hammer [34]

Answer: B) 46.7% Si and 53.3% O

Explanation:

To calculate the mass percent of element in a given compound, we use the formula:

\text{Mass percent of element}=\frac{\text{Mass of element}}{\text{total mass}}\times 100\%

Mass of quartz  (SiO_2) = 5.05 g

Mass of silicon = 2.36 g

Mass of oxygen = Mass of quartz  (SiO_2) - mass of silicon = 5.05g - 2.36 g = 2.69 g

\text{Mass percent of silicon}=\frac{\text{Mass of silicon}}{\text{total mass of quartz}}\times 100=\frac{2.36}{5.05}\times 100=46.7\%

\text{Mass percent of oxygen}=\frac{\text{Mass of oxygen}}{\text{total mass of quartz}}\times 100=\frac{2.69}{5.05}\times 100=53.3\%

Thus the percentages of silicon and oxygen in quartz are B) 46.7% Si and 53.3% O

6 0
2 years ago
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