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3 years ago

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Find the molecular formula of a compound that contains 42.56 g of palladium and 0.80 g of hydrogen. The molar mass of the compou

nd is 216.8 g/mol.

1 answer:

Crazy boy [7]3 years ago

7 0

Answer:

The answer to your question is Pd₂H₄

Explanation:

Data

mass of palladium = 42.56 g

mass of hydrogen = 0.8 g

Process

1.- Convert the grams of each substance to moles

106 g of Pd ----------------- 1 mol

42.56 g of Pd ------------- x

x = (42.56 x 1)/106

x = 0.402 moles

1 g of H --------------------- 1 mol

0.8 g of H ------------------ x

x = (0.8 x 1)/1

x = 0.8 moles

2.- Divide by the lowest number of moles

Palladium = 0.402/0.402 = 1

Hydrogen = 0.8/0.402 = 1.99 ≈ 2

3.- Write the empirical formula

PdH₂

4.- Calculate the mass of the empirical formula

PdH₂ = 106 + 2 = 108

5.- Divide the molar mass by the molar mass of the empirical formula

216.8/108 = 2

6.- Get the molecular formula

2(PdH₂) = Pd₂H₄

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3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

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Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the $OH^-$ concentration.

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

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

<u>(b) a solution of 0.10 M $NH_3$ </u>

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the $OH^-$ concentration.

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

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

<u>(d) a solution of 0.10 M NaOH</u>

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

<u>(e) a $1\times 10^{-4}M$ solution of $HNO_2$ </u>

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

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

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the $OH^-$ concentration.

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

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

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