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3 years ago

[CM.01] Which states of matter appear only in the hydrosphere?

2 answers:

3 0

I will say liquid because the rest all consist of solids and atmosphere also contain gas, but only hydrosphere has liquid.

choli [55]3 years ago

3 0

Answer:

Liquid is the correct answer.

Explanation:

The hydrosphere contains the total amount of water on a planet. The water is on the surface, ground and in the atmosphere. Hydrosphere can be in liquid, gas and solid forms. On earth, the water exists in liquid form in lakes, oceans, and rivers. The frozen part of the water is in the form of an ice cap and icebergs. In hydrosphere, the water moves in a cycle. which leads to the transformation of water from one form to another, it is called the water cycle. The waterfalls in the form of rain and snow and are collected by lakes and ocean. It again evaporates into the atmosphere and the cycle starts again.

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Luba_88 [7]

Answer:

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7 0

3 years ago

Oxidation number of au in kaucl4

DanielleElmas [232]

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3 0

3 years ago

Read 2 more answers

Why is potassium placed in the periodic table where it is?

Virty [35]

Potassium is placed where it is based on its properties and it's reactivity. It's also placed there based on it's atomic number.

6 0

3 years ago

Hydrogen gas was collected by water displacement. what was pressure of the h2 collected if the temperature was 26°c?

Vedmedyk [2.9K]

The ideal gas law may be written as
$p= \frac{\rho R T}{M}$
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)

For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol

Therefore
$p= \frac{(0.09 \, kg/m^{3})*(8.314 \, J/(mol-K))*(299 \, K)}{1.008 \times 10^{-3} \, kg/mol} =2.2195 \times 10^{5} \, Pa$

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm

Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)

4 0

3 years ago

A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam

Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0

3 years ago