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slamgirl [31]
3 years ago
15

0.03 mol of potassium nitrate, KNO3 dissolved in 1.2 dm3 distilled water. What is the solubility of potassium nitrate, KNO3 ?​

Chemistry
1 answer:
Rufina [12.5K]3 years ago
3 0

Answer:

solubility=0.25mol/dm^3

Explanation:

number of moles of solute=0.03mol

volume of solution in dm³=1.2dm³

solubility=?

as we know that

solubility=\frac{number of moles of solute}{volume of solution indm}

solubility=\frac{0.03mol}{1.2dm^3}

solubility=0.25mol/dm^3

i hope this will help you :)

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Answer:

pH = 1.853

Explanation:

For every mole of hydrochloric acid, one mole of hydronium ion is required. Thus, in order to neutralize 0.014 moles of HCL, 0.014 moles of hydronium is required.

[H_3O^+] = [HCl] = 0.014

pH  = -log [H^+] = -log [H_3O^+]

Substituting the available values in above equation, we can say that   the pH of the solution is equal to

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pH = 1.853

pH of a 0.014 M HCL solution = 1.853

8 0
3 years ago
What is another name for a coefficient? this is for chemistry
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Answer:

synergistic

Explanation:

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Formed when an amine is combined with a carboxyl group
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Have a nice day
7 0
3 years ago
4) What volume will the gas in the balloon at right occupy at 250k?<br><br> balloon: 4.3L 350K
swat32

Answer:

2.87 liter.

Explanation:

Given:

Initially volume of balloon = 4.3 liter

Initially temperature of balloon = 350 K

Question asked:

What volume will the gas in the balloon occupy at 250 K ?

Solution:

By using:

Pv =nRT

Assuming pressure as constant,

V∝ T

Now, let  K is the constant.

V = KT

Let initial volume of balloon , V_{1} = 4.3 liter

1000 liter = 1 meter cube

1 liter = \frac{1}{1000} m^{3} = 10^{-3} m^{3

4.3 liter = 4.3\times10^{-3}=4.3\times10^{-3} m^{3}

And initial temperature of balloon, T_{1} = 350 K

Let the final volume of balloon is V_{2}

And as given, final temperature of balloon, T_{2} is 250 K

Now, V_{1} = KT_{1}

4.3\times10^{-3}=K\times350\ (equation\ 1 )

V_{2} = KT_{2}

=K\times250\ (equation 2)

Dividing equation 1 and 2,

\frac{4.3\times10^{-3}}{V_{2} } =\frac{K\times350}{K\times250}

K cancelled by K.

By cross multiplication:

350V_{2} =4.3\times10^{-3} \times250\\V_{2} =\frac{ 4.3\times10^{-3} \times250\\}{350} \\          = \frac{1075\times10^{-3}}{350} \\          =2.87\times10^{-3}m^{3}

Now, convert it into liter with the help of calculation done above,

2.87\times10^{-3} \times1000\\2.87\times10^{-3} \times10^{3} \\2.87\ liter

Therefore, volume of the gas in the balloon at 250 K will be  2.87 liter.

4 0
3 years ago
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