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faltersainse [42]

3 years ago

12

ANYONE KNOW CHEMISTRY.............................,..

1 answer:

Alexus [3.1K]3 years ago

7 0

Yes

Explanation:

Of course uwu

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Balance the following reaction _ Al + _ Cl, → _ AICI,

Free_Kalibri [48]

Answer:

1 Al + 1 Cl -> 1 AlCl

Explanation:

It is already balanced.

3 0

3 years ago

A small amount wet of hydrogen gas can be prepared by the reaction of zinc with excess hydrochloric acid and trapping the gas pr

dybincka [34]

Answer : The partial pressure of the hydrogen is, 705.9 mmHg

Explanation :

According to the Dalton's law of partial pressure,

$P_T=P_{H_2}+P_{H_2O}$

where,

$P_T$ = total pressure of the gas = 729.7 mmHg

$P_{H_2}$ = partial pressure of the hydrogen gas = ?

$P_{H_2O}$ = partial pressure of the water = 23.8 mmHg (standard value)

Now put all the given values in the above expression, we get:

$729.7mmHg=P_{H_2}+23.8mmHg$

$P_{H_2}=705.9mmHg$

Therefore, the partial pressure of the hydrogen is, 705.9 mmHg

4 0

4 years ago

Ava, Inc., issued 7% bonds, dated January 1, with a face amount of $117,200 on January 1, 2016 for an issue price of 106.5. The

natta225 [31]

Complete Question

The complete question is shown on the first uploaded image

Answer:

10a The interest payment on the bond is = $ 8,204

10b The debit to cash for the bond proceeds is = $ 302,000

Explanation:

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8 0

3 years ago

What is reproduction

Zolol [24]

The action or process of making a copy of something.

3 0

3 years ago

Calculate the masses of oxygen and nitrogen that are dissolved in of aqueous solution in equilibrium with air at 25 °C and 760 T

shtirl [24]

Explanation:

Let us assume that the volume of given aqueous solution is 7.5 L.

Therefore, according to Henry's law, the relation between concentration and pressure is as follows.

C = $\frac{P}{K_{h}}$

where, pressure (P) = 760 torr = 1 atm

According to Henry's law, constants for gases in water at $25^{o}C$ are as follows.

$p(O_{2})$ = 0.21 atm = 0.21 bar

$p(N_{2})$ = 0.78 atm = 0.78 bar

$K_{h}$ for $O_{2}$ = $7.9 \times 10^{2} bar/mol$

$K_{h}$ for $N_{2}$ = $1.6 \times 10^{3} bar /mol$

Since, 21% oxygen is present in air so, its mass will be 0.21 g. Similarly, 78% nitrogen means the mass of nitrogen is 0.78 g.

Therefore, concebtrations will be calculated as follows.

$C(O_{2}) = \frac{0.21}{7.9 \times 10^{2}} = 2.66 \times 10^-4 mol/L$

$C(N_2) = \frac{0.78}{1.6 \times 10^3} = 4.875 \times 10^-4 mol/L$

Now, we will calculate the number of moles as follows.

$n(O_{2}) = 7.5 \times 2.66 \times 10^-4 = 1.995 \times 10^-3$ mol

$n(N_2) = 7.5 \times 4.875 \times 10^-4 = 3.66 \times 10^-3$ mol

As the molar mass of $O_2$ = 32 g/mol

Hence, mass of oxygen will be as follows.

Mass of $O_2 = 32 \times 1.995 \times 10^-3 \times 1000$

= 63.84 mg

As the molar mass of $N_{2}$ = 28

Mass of $N_{2} = 28 \times 3.66 \times 10^-3$ = 102.5 mg

Thus, we can conclude that mass of oxygen is 63.84 mg and nitrogen is 102.5 mg.

5 0

4 years ago

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