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vesna_86 [32]
2 years ago
5

When gasses are heated the average velocity (speed) of the gas particles decreases.

Chemistry
1 answer:
Kitty [74]2 years ago
6 0
I think that would be true!
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Most of the dissolved substance in sea water is: hydrogen chloride nitrogen sodium ANSWER NOW! I NEED HELP ASAP! WILL GIVE BRAIN
VMariaS [17]

Answer: Sodium chloride

Explanation:

Ocean water contains a number of substances. When a substance which has ionic bonds is dissolved in water it takes the form of ions.

The most common ions in ocean water are sodium and chloride. These are the ions formed when common salt, sodium chloride (NaCl) is dissolved in water.

Sodium chloride accounts for about 3% of ocean water by mass.

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3 years ago
Which of these common substances is a homogeneous mixture?
IgorC [24]

Answer:

whole milk ,table salt , maple syrup

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2 years ago
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Please put in order from LARGEST WAVELENGH TO SHORTEST YOU WILL GET BRAINLIEST AND 25 POINTS START FROM TOP TO BUTTON
sleet_krkn [62]
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3 years ago
For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁   -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0
3 years ago
10 pointss..******^^
Kisachek [45]

Answer:

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Explanation:

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7 0
2 years ago
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