C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate werereacted?
Answers
Answer:
Mass = 76.176 g
Explanation:
Given data:
Mass of lead(II) chloride produced = 62.9 g
Mass of lead(II) nitrate used = ?
Solution:
Chemical equation:
Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃
Number of moles of lead(II) chloride:
Number of moles = mass/molar mass
Number of moles = 62.9 g/ 278.1 g/mol
Number of moles = 0.23 mol
Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:
PbCl₂ : Pb(NO₃)₂
1 : 1
0.23 : 0.23
Mass of Pb(NO₃)₂:
Mass = number of moles × molar mass
Mass = 0.23 mol × 331.2 g/mol
Mass = 76.176 g
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ethanol oxygen gas
What elements will be present in the substances that are created when ethanol burns?
A.
nitrogen, sulfur, and sodium only
B.
carbon, hydrogen, and oxygen only
C.
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D.
carbon and hydrogen only
correct answer is B I answered my own question! lol
Answers
Answer: The correct answer is Option B.
Explanation:
Every balanced chemical equation follows law of conservation of mass. This law states that mass can neither be created nor be destroyed but it ca only be transformed from one form to another form.
In a chemical reaction, total number of individual atoms on the reactant side must be equal to the total number of individual atoms on the product side.
During a chemical reaction, the atoms on the reactant side must also be present on the product side.
For the given reactants, the chemical equation will follows:
As, the reactant side contained carbon, hydrogen and oxygen atoms only. So, the product will also contain the same atoms.
Hence, the correct answer is Option B.
Answers
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Answers
answer- release carbon dioxide when burned
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Answers
Answer:
1258 grams of AgN03
Explanation:
We calculate the weight of 1 mol of AgN03:
Weight 1 mol AgN03= Weight Ag + Weight N +( Weight 0)x3=108g+ 14g+16gx3=170 g/mol
1 mol----170 g AgN03
7,4mol---x= (7,4 mol x170 g AgN03)/1 mol=1258 g AgN03