The big bang is through to have started by an?

The solubility of a sample will ____________ when the size of the sample increases.

devlian [24]

The solubility of a sample will decrease when the size of the sample increases.

When the Size of sample is bigger, it means there is more particles that make up that sample so the amount of solvent to dissolve the sample will also be in the greater amount. Thus, when the sample size increases the solubility decrease. We can also say that size of sample and solubility are inversely proportional to each other.

8 0

3 years ago

Which of the following gas samples would have the largest number of representative particles at STP? A. 0.520 L SO3 B. 0.10 L Xe

egoroff_w [7]

Answer: The correct option is D.

Explanation: To calculate the larges number of partivles of gas, we will use Avogadro's Law which says that volume of the gas is directly related to the number of moles of gas at constant pressure and temperature.

Mathematically,

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ ....(1)

The gas which has the largest number of moles, will have the largest number of particles.

We are provided to use STP conditions, which says that

1 mole of a gas occupies 22.4 L of volume.

Initial conditions:

$n_1=1mole$

$V_1=22.4L$

A. 0.520 L of $SO_3$

$V_2=0.520L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{0.520}{n_2}$

$n_2=0.0232moles$

B. 0.10 L of Xe

$V_2=0.10L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{0.10}{n_2}$

$n_2=0.0044moles$

C. 7.0 L of $N_2$

$V_2=7.0L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{7.0}{n_2}$

$n_2=0.3125moles$

D. 12.0 L of Ne

$V_2=12.0L$

Putting all the values, in equation 1, we get

$\frac{22.4}{1}=\frac{12.0}{n_2}$

$n_2=0.5357moles$

Hence, from the above calculations, we see that option D has the largest amount of moles and hence, will have the largest amount of particles.

6 0

4 years ago

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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

Zinc has a specific heat capacity of 0.390 J/goC. What is its molar heat capacity? Enter your answer numerically to three signif

ch4aika [34]

Answer:

The answer to your questions is Cm = 25.5 J/mol°C

Explanation:

Data

Heat capacity = 0.390 J/g°C

Molar heat capacity = ?

Process

1.- Look for the atomic number of Zinc

Z = 65.4 g/mol

2.- Convert heat capacity to molar heat capacity

(0.390 J/g°C)(65.4 g/mol)

- Simplify and result

Cm = 25.5 J/mol°C

3 0

3 years ago

2. Sasha used the volume control on her TV to make the sound louder or

kobusy [5.1K]

Answer:

amplitude

................

4 0

3 years ago

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