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1 year ago

The 225 g FeCl2 is about 1.8 moles.

Chemistry

1 answer:

Lyrx [107]1 year ago

5 0

Answer:

$4 \frac{mol}{l}$

Explanation:

From the question, we have been asked to find the molarity of FeCl2 having a volume of 450 mL,

We have been provided with 225 g which is proportional to 1.8 moles.

We know that molarity of any solution should be in mol/L.

1 mole contained in 1 L means it has a molarity of 1 mol/L

Let's convert 450 mL to Litres which is,

$\frac{450}{1000} litres$

= 0.450 L

Thus,

1 mole is contained in 1L

x moles are contained in 0.450 L

Hence,

x mole/molarity = {1 mole x 1 L}/{0.450 L}

= 4 mol/L

Therefore 4 mol/L is the molar concentration.

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2 years ago

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2 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%

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