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3 years ago

Calculate the percentage of water in the hydrate manganese (ii) nitrate tetrahydrate

Chemistry

2 answers:

IRISSAK [1]3 years ago

8 0

Answer:

Divide the mass of the water lost by the mass of hydrate and multiply by 100.

Explanation:

i think

Neporo4naja [7]3 years ago

7 0

Answer:

28.7%

Explanation:

First let us write the formula of manganese (ii) nitrate tetrahydrate. This is shown below:

Mn(NO3)2.4H2O

The molar mass of Mn(NO3)2.4H2O = 55 + 2[14 + (3x16)] + 4[(2x1) + 16]

= 55 + 2[14 + 48] + 4[2 + 16]

= 55 + 2[62] + 4[18]

= 55 + 124 + 72

= 251g/mol

Mass of water of crystallization in Mn(NO3)2.4H2O = 4H2O

= 4[(2x1) + 16]

= 4[2 + 16]

= 4[18]

= 72g

The percentage by mass of water of crystallization in manganese (ii) nitrate tetrahydrate Mn(NO3)2.4H2O, is given by:

Percentage of water of crystallization = Mass of water of crystallization / Molar Mass of Mn(NO3)2.4H2O x 100

= 72/251 x 100

= 28.7%

Therefore the percentage of water of crystallization in manganese (ii) nitrate tetrahydrate is 28.7%

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IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

(1) 53.1 mM HCl

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As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= $53.1\times 10^{-3}M$

pH : It is defined as the negative logarithm of hydrogen ion concentration.

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

$pH=-\log (53.1\times 10^{-3})$

$pH=1.27$

(2) 0.223 M KOH

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (0.223)$

$pOH=0.65$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

(3) 53.1 mM HCl + 0.223 M KOH

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

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Phosphorus-32 is radioactive and has a half life of 14.3 days. What percentage of a sample would be left after 12.6 days

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Answer:

There is 54.29 % sample left after 12.6 days

Explanation:

Step 1: Data given

Half life time = 14.3 days

Time left = 12.6 days

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X = 100 / 2^n

⇒ with X = The amount of sample after 12.6 days

⇒ with n = (time passed / half-life time) = (12.6/14.3)

X = 100 / 2^(12.6/14.3)

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There is 54.29 % sample left after 12.6 days

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Answer:

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Explanation:

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