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3 years ago

Calculate the mass percent of 0.485 g of H, which reacts with oxygen to form 2.32 g H2O

Chemistry

1 answer:

Nataly [62]3 years ago

4 0

Answer:

Mass % of hydrogen = 20.9 %

Explanation:

Given data:

Mass of hydrogen = 0.485 g

Mass of water = 2.32 g

Mass percent of hydrogen = ?

Solution:

Mass % of hydrogen = mass of hydrogen / mass of water × 100

Mass % of hydrogen = 0.485 g/ 2.32 g × 100

Mass % of hydrogen = 0.209 × 100

Mass % of hydrogen = 20.9 %

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What is the oh- in a solution with a poh of 5.71

Rudik [331]

Answer:- The hydroxide ion concentration of the solution is $1.95*10^-^6$ .

Solution:- The formula used to calculate pOH from hydroxide ion is:

$pOH=-log[OH^-]$

When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:

$[OH^-]=10^-^p^O^H$

pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:

$[OH^-]=10^-^5^.^7^1$

$[OH^-]$ = 0.00000195 or $1.95*10^-^6$

So, the hydroxide ion concentration of the solution is $1.95*10^-^6$ .

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sergejj [24]

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3 years ago

Which of these half-reactions represents reduction?

gogolik [260]

Answer: The half-reactions represents reduction are as follows.

$Cr_{2}O^{2-}_{7} \rightarrow Cr^{3+}$
$MnO^{-}_{4} \rightarrow Mn^{2+}$

Explanation:

A half-reaction where addition of electrons take place or a reaction where decrease in oxidation state of an element takes place is called reduction-half reaction.

For example, the oxidation state of Cr in $Cr_{2}O^{2-}_{7}$ is +6 which is getting converted into +3, that is, decrease in oxidation state is taking place as follows.

$Cr_{2}O^{2-}_{7} + 3 e^{-} \rightarrow Cr^{3+}$

Similarly, oxidation state of Mn in $MnO^{-}_{4}$ is +7 which is getting converted into +2, that is, decrease in oxidation state is taking place as follows.