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grigory [225]
2 years ago
14

What mass of carbon is present in 0.500 mol of sucrose?

Chemistry
2 answers:
Vilka [71]2 years ago
5 0

Answer:

There are 72 g of carbon in 0.500 mol of sucrose.

Explanation:

The sucrose molecule has the formul C₁₂H₂₂O₁₁.

Each molecule of sucrose has 12 atoms of carbon.

Each carbon atom weights 12 a.u.

Each mole of sucrose weights 342.30 g.

Each mole of carbon weights 12 g.

12 carbons weight 144 g.

Therefore, the weight percentage of carbon in each mole of sucrose is:

%C = 144/342.30 x 100%

%C = 42.1%

0.500 mol of sucrose weights 342.30 g/2 = 171.15 g.

42.1% x 171.15 g = 72 g

AlexFokin [52]2 years ago
3 0
C12H22O11
12×12×0.5=72
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Ammonia can be produced in the laboratory by heating ammonium chloride
AnnyKZ [126]

Answer:

Mass = 2.89 g

Explanation:

Given data:

Mass of NH₄Cl = 8.939 g

Mass of Ca(OH)₂ = 7.48 g

Mass of ammonia produced = ?

Solution:

2NH₄Cl   +  Ca(OH)₂     →    CaCl₂ + 2NH₃ + 2H₂O

Number of moles of NH₄Cl:

Number of moles = mass/molar mass

Number of moles = 8.939 g / 53.5 g/mol

Number of moles = 0.17 mol

Number of moles of Ca(OH)₂ :

Number of moles = mass/molar mass

Number of moles = 7.48 g / 74.1 g/mol

Number of moles = 0.10 mol

Now we will compare the moles of ammonia with both reactant.

                      NH₄Cl          :          NH₃

                          2              :           2

                         0.17          :          0.17

                   Ca(OH)₂         :          NH₃

                        1                :           2

                    0.10              :          2/1×0.10 = 0.2 mol

Less number of moles of ammonia are produced by ammonium chloride it will act as limiting reactant.

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.17 mol × 17 g/mol

Mass = 2.89 g

6 0
2 years ago
In a titration, 25.0 mL of 0.500 M KHP is titrated to the equivalence point with NaOH. The final solution volume is 53.5 mL. Wha
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M1v1=m2v2 
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<span>m2=(25.0*0.500)/53.5
m2=12.5/53.5
m2=0.2336
by rounding off:
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so the answer is C: 0.234 M</span>
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