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Luba_88 [7]
3 years ago
6

Consider 10.0 g of helium gas (He) in a rigid steel container. If you add 10.0 g of neon gas (Ne) to this container, which of th

e following best describes what happens? (Assume the temperature is constant.)
a) The pressure in the container doubles.
b) The pressure in the container more than doubles.
c) The volume of the container doubles.
d) The volume of the container more than doubles.
e) The pressure in the container increases but does not double.
Chemistry
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer: (e) The pressure in the container increases but does not double.

Explanation:

To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant

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Frank has a sample of steel that weighs 80 grams. If the density of his sample of steel is 8 g/cm3 ,what is the samples volume?
kirill115 [55]

Answer: 10 cm3

Explanation:

Mass of sample of steel (m) = 80 grams

density of sample of steel (p) = 8 g/cm3 samples volume (v) = ?

Since density is obtained by dividing mass by volume, the density of the steel sample is expressed as:

density = mass / volume

p = m / v

make v the subject formula

v = m / p

v = 80 grams / 8 g/cm3

v = 10 cm3

Thus, the samples volume is 10 cm3

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shusha [124]

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Explanation:

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LHS          RHS

N = 1            1

H = 5          5

O = 1            1

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Name two factors that influence the viscosity of a lava flow
mr_godi [17]
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If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).​
11111nata11111 [884]

Answer:

{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 =  \frac{11}{m _{r}}  \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\  \\ { \bf{vapour \: density = 2 \times m _{r}}} \\  = 2 \times 14.85 \\  = 29.7 \: { \tt{g {dm}^{ - 3} }}

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