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2 years ago

12

A hydrogen bond is a dipole-dipole attraction between a partially positive hydrogen atom and the unshared electron pair of a str

ongly electronegative atom such as O, F, or N. Group of answer choices True False

1 answer:

podryga [215]2 years ago

6 0

True! The answer is true because I did this today and got that answer and it was right!

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Alex cut a pizza into 8 equal slices. He removed 2 of the slices of pizza. What is the measure of the angle made by the missing

Firlakuza [10]

Answer: 90 degrees.

Explanation:

4 0

2 years ago

As the water molecules re-freeze, what happens to the salt ions that are mixed in with the water?

Murljashka [212]

Answer:

Explanation:

Since salt is composed of two ions Na+ and Cl- it will attract to the opposite poles of the water and form a ion-dipole bond. This will cause the phenomenon freezing point depression where as the name states it will take a lower tempertature to freeze it. So instead of freezing at 0 degrees c it will freeze at a negative 2 or so on.

5 0

2 years ago

At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va

Bas_tet [7]

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

$P=X_{A}P_{A}+X_{B}P_{B}$

Here, $X_{A}$ is mole fraction of A, $X_{B}$ is mole fraction of B, $P_{A}$ is partial pressure of A and $P_{B}$ is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

$X_{A}+X_{B}=1$

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

$X_{octane}=1-X_{hexane}=1-0.58=0.42$

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

$P=X_{hexane}P_{hexane}+X_{octane}P_{octane}$

Putting the values,

$P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg$

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

4 0

3 years ago

A mixture of helium and methane gases, at a total pressure of 821 mm Hg, contains 0.723 grams of helium and 3.43 grams of methan

Nookie1986 [14]

<u>Answer:</u>

<u>For 1:</u> The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

<u>For 2:</u> The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

<u>Explanation:</u>

<u>For 1:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For helium:</u>

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol$

<u>For methane gas:</u>

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol$

To calculate the mole fraction , we use the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ......(3)

<u>For Helium gas:</u>

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{He}=\frac{0.181}{0.181+0.214}=0.458$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{He}=0.458$

Putting values in equation 3, we get:

$p_{He}=0.458\times 821mmHg=376mmHg$

<u>For Methane gas:</u>

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{CH_4}=0.542$

Putting values in equation 3, we get:

$p_{CH_4}=0.542\times 821mmHg=445mmHg$

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

<u>For 2:</u>

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

<u>For nitrogen gas:</u>

$363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392$

<u>For carbon dioxide gas:</u>

$564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608$

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

6 0

3 years ago

A group of students collects data to determine whether compounds were ionic or covalent. They placed each compound in a small bo

yan [13]

Answer:

the answer is B.

No, as ionic compounds are only conductive in an aqueous (water) solution

Explanation: i just know

8 0

3 years ago