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worty [1.4K]
3 years ago
11

If you have 110.0 grams of an unknown compound that contains 12.3 grams of hydrogen, what is the percent by mass of hydrogen in

the compound
Chemistry
2 answers:
Vladimir [108]3 years ago
8 0
All you have to do is a simple division "parts over the whole"

12.3 grams H/ 110 grams compound x 100= 11.2%
tankabanditka [31]3 years ago
3 0

Answer : The percent by mass of hydrogen in the compound is, 11.18 %

Explanation : Given,

Mass of unknown compound = 110.0 g

Mass of hydrogen = 12.3 g

Now we have to calculate the percent by mass of hydrogen in the compound.

\text{Mass percent of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Total mass of unknown compound}}\times 100

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

\text{Mass percent of hydrogen}=\frac{12.3g}{110g}\times 100

\text{Mass percent of hydrogen}=11.18\%

Therefore, the percent by mass of hydrogen in the compound is, 11.18 %

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In a certain experiment 10.0g of magnesium nitride is allowed to react with 5.00 g of water. Calculate the final mass of ammonia
valkas [14]

Answer:

The final mass of ammonia is 1.57 grams

The final mass of water is 0 grams

The final mass of magnesium nitride is 5.34 grams

Explanation:

Step 1: Data given

Mass of Mg3N2 = 10.0 grams

Molar mass of Mg3N2 = 100.95 g/mol

Mass of H2O = 5.00 grams

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3

Step 3: Calculate moles Mg3N2

Moles Mg3N2 = mass / molar mass

Moles Mg3N2 = 10.0 grams / 100.95 g/mol

Moles Mg3N2 = 0.0991 moles

Step 4: Calculate moles H20

Moles H2O = 5.00 grams / 18.02 g/mol

Moles H2O = 0.277 moles

Step 5: Calculate the limiting reactant

H2O is the limiting reactant. It will completely be consumed (0.277 moles).

Mg3N2 is in excess. There will react 0.277/6 = 0.0462 moles

There will remain 0.0991 - 0.0462 = 0.0529 moles

This is 0.0529 moles * 100.95 g/mol = 5.34 grams

Step 6: Calculate moles of NH3

For 1 mol Mg3N2 we need 6 moles H2O to produce 3 moles Mg(OH)2 and 2 moles NH3

For 0.277 moles H2O we'll have 0.277/3 = 0.0923 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.0923 moles * 17.03 g/mol

Mass NH3 = 1.57 grams

4 0
3 years ago
5.00 ml of commercial bleach was diluted to 100.0 ml. 25.0 ml of the diluted sample was titrated with 4.56 ml of 0.100 m s2o3 2-
aev [14]

The solution would be like this for this specific problem:

 

4 NaOCl + S2O3{2-} + 2 OH{-} → 2 SO4{2-} + H2O + 4 NaCl

<span>(0.00456 L) x (0.100 mol/L S2O3{2-}) x (4 mol NaOCl / 1 mol S2O3{2-}) x (100.0 mL / 25 mL) x </span><span>
<span>(74.4422 g NaClO/mol) = 0.54313 g </span></span>

<span>(5.00 mL) x (1.08 g/mL) = 5.40 g solution </span>

(0.54313 g) / (5.40 g) = 0.101 = 10.1%

 

So, the average percent by mass of NaClO in the commercial bleach is 10.1%.

3 0
3 years ago
4 main parts of a flower Include the function of each pls help me
ZanzabumX [31]

Answer:

  • Calyx or sepals
  • Corolla
  • Androecium
  • Gynoecium

6 0
2 years ago
Is each of the following a physical or chemical property? ● melting point ● ability to react ● color ● density ● ability to burn
ryzh [129]
These are definitely chemical properties
5 0
3 years ago
How many joules of heat are absorbed when 73 g water are heated from 30*C to 43*C? *
stepladder [879]

Answer:

3966.82 J

Explanation:

q=sm∆T

q=73×13×4.18

the specific heat for water is 4.18

6 0
3 years ago
Read 2 more answers
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