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andreev551 [17]

3 years ago

5

Calculate the number of molecules of Ammonia (NH3) in 7.75 moles of ammonia.

1 answer:

Phoenix [80]3 years ago

5 0

$4.67 \times 10^{24}$

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What, if anything is wrong with the model of the solar system shown below ? Give me a big or small answer

jekas [21]

Answer:

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5 0

3 years ago

If you have an altitude of 55 mi. (miles), which layer of the atmosphere are you in?

Kruka [31]

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8 0

2 years ago

Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

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

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

4 0

2 years ago

Which is the larger atom kr Or As

nasty-shy [4]

Answer:

I think As is larger

Explanation:

5 0

2 years ago

The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t

Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

$\Delta T_f=iK_f\times m$

where,

i = van't Hoff factor = ?

$\Delta T_f$ = depression in freezing point = 0.225°C

$K_f$ = Cryoscopic constant = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

$0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21$

Hence, the experimental van't Hoff factor is 1.21

7 0

3 years ago