Ask question

Ask question

All categories

3 years ago

10

PLEASE HELP!CHEMISTRY QUESTION

1 answer:

ozzi3 years ago

8 0

<u>Answer:</u> The partial pressure of nitrogen on Venus is 81 mmHg

<u>Explanation:</u>

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

$p_{A}=p_T\times \chi_{A}$

where,

$p_A$ = partial pressure of nitrogen gas = ?

$p_T$ = total pressure = 2700 mmHg

$\chi_A$ = mole fraction of nitrogen gas = 3.0 % = 0.03

Putting values in above equation, we get:

$p_{N_2}=2700mmHg\times 0.03\\\\p_{N_2}=81mmHg$

Hence, the partial pressure of nitrogen on Venus is 81 mmHg

You might be interested in

How does the oceanic crust compare to continental crust?

dimulka [17.4K]

I think the answer is number (4)

5 0

2 years ago

Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super

dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below: $X^{m+}Y^{n-}\rightarrow X_nY_m$

Thus, for all the given combinations, we obtain:

- Y⁻

$X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3$

- Y²⁻

$X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3$

- Y³⁻

$X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY$

Best regards!

8 0

2 years ago

Can anyone help please

Aleks [24]

Answer:

Here you go! Hopefully this helps your question.

1) Apples May Lower High Cholesterol and Blood Pressure.

2) Apples Can Support a Healthy Immune System.

3) Apples Are a Diabetes-Friendly Fruit.

Explanation:

None needed

4 0

1 year ago

Read 2 more answers

The atomic mass of an element is the weighted average of the(1) number of protons in the isotopes of that element(2) number of n

liberstina [14]

It is a weighted average of the atomic masses of the naturally occurring isotopes of the element.

4 0

2 years ago

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M, [B] = 1.35 M, and [C] = 0.300 M. The fo

dimaraw [331]

<u>Answer:</u> The value of $K_c$ for given reaction is 0.465

<u>Explanation:</u>

We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

$A+2B\rightarrow C$

<u>Initial:</u> 0.65 1.35 0.30

<u>At eqllm:</u> 0.65-x 1.35-2x 0.30+x

Evaluating the value of 'x'

$0.650-x=0.550\\\\x=0.100$

So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M

Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of $K_c$ for above equation follows:

$K_c=\frac{[C]}{[A][B]^2}$

Putting values in above equation, we get:

$K_c=\frac{0.400}{0.650\times (1.15)^2}\\\\K_c=0.465$

Hence, the value of $K_c$ for given reaction is 0.465

7 0

3 years ago