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3 years ago

Ions:

Chemistry

1 answer:

kozerog [31]3 years ago

5 0

Chlorine: gain, -1 charge.

iron: lose, +2 charge.

sulfur: 6, gain, -2 charge.

copper: lose, +1 charge.

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What is the hydronium ion concentration of a solution whose pH is 7.30

Assoli18 [71]

[ H₃O⁺] = 10 ^ - pH

[ H₃O⁺ ] = 10 ^ - 7.30

[ H₃O⁺ ] = 5.011 x 10⁻⁸ M

hope this helps!

6 0

3 years ago

Air is a mixture of gases that is about 78.0% n2 by volume. when air is at standard pressure and 25.0 ∘c, the n2 component will

ArbitrLikvidat [17]

Step first:

Since, 78.0 percent $N_{2}$ by volume is present in the air which implies there are 78 moles of $N_{2}$ present in 100 moles of air.

Now,

Mole fraction of $N_{2}$ = $\frac{Moles of N_{2}}{Total moles of gases in the mixture}$

Mole fraction = $\frac{78}{100} mol$

= 0.78

Partial pressure is equal to the multiplication of total pressure and mole fraction.

Partial pressure = $0.78\times 1 atm$ (as 1 atm is atmospheric pressure)

= 0.78 atm.

Step second:

Henry's law constant is calculated by:

$S_{g}=k\times P_{g}$

where,

$S_{g}$ is solubility of gas

$P_{g}$ is partial pressure of gas

$k$ is henry's law constant

Substitute the value of solubility and partial pressure to find the value of Henry's law constant in above formula:

$4.88\times 10^{-4} M=k\times 0.78 atm$

k = $\frac{4.88\times 10^{-4}M}{0.78 atm}$

= $6.25\times 10^{-4} M atm^{-1}$

Thus, Henry's Law constant is $6.25\times 10^{-4} mol L^{-1} atm^{-1}$ .

5 0

3 years ago

Read 2 more answers

Which of the following is a problem caused by acid rain?

marissa [1.9K]

Answer:

Explanation:

is a factor in global warming

3 0

3 years ago

How is sublimation and deposition are alike

Y_Kistochka [10]

They both skip the liquid phase obviously xdxdxd

4 0

3 years ago

In Part A, you found the number of moles of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxyge

n200080 [17]

Answer:

1.40 moles.

Explanation:

The balanced chemical equation for the reaction of Phosphorus and Oxygen is as follows -

$4P + 5O_2 = 2P_2O_5$

In Part A, the oxygen was taken in excess. So Phosphorus will be the limiting reagent.

Since, 2 moles of $P_2O_5$ is formed by 4 moles of $P$

So, for 1.8 moles of $P_2O_5$ amount of required moles of $P$ = $\frac{4}{2} \times 1.80 = 3.60$

In Part B, the phosphorus was taken in excess so oxygen will be the limiting reagent.

Since, 2 moles of $P_2O_5$ is formed by 5 moles of oxygen

So, for 1.40 moles of $P_2O_5$ moles of $O_2$ required = $\frac{5}{2} \times 1.40 = 3.50$

Thus as of now we have 3.60 moles of $P_2O_5$ and 3.50 moles of $O_2$ .

As in the reaction of formation of $P_2O_5$ , oxygen is the limiting reagent.

So the moles of $P_2O_5$ formed by the 3.50 moles of oxygen will be

$P_2O_5$ = $\frac{2}{5} \times 3.50$ = 1.40 moles.

5 0

3 years ago