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maxonik [38]
3 years ago
8

What is the theoretical yield of so3 produced by the quantities described in part a? express your answer numerically in grams?

Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
8 0
Sulfur reacts with oxygen to yield SO3 as shown in the equation below;
 2S(g)+ 3O2(g) = 2SO3(g)
From part A 7.49 g of S were used.
The atomic mass of sulfur is 32.06 g/mol
Hence, the number of moles of sulfur used
 7.49 / 32.06 = 0.2336 moles
The mole ratio of S : SO3 is 1:1
Thus the mass of SO3 will be ( 1 mol of SO3= 80.06 g)
0.2336 moles × 80.06 = 18.7 g
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All the elements in group 1 react with water to form a hydroxide. What other substance is also
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3 0
3 years ago
What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?
8090 [49]
Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
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m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
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3 0
3 years ago
Pls help and show work!!!
kumpel [21]

Answer:

44.9g

Explanation:

You have to convert grams of CH4 to moles, use the mole-to-mole ratio of CH4 to water, and convert back to grams.

(20.0g CH4)(1 mol CH4/16.04g)(2 mol H2O/1 mol CH4)(18.01 g H2O/ 1 mol) = 44.9127 g

Hope this helps!

5 0
2 years ago
Read 2 more answers
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