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Andru [333]
3 years ago
7

C10H12O4S(s) + O2(g)  CO2(g) + SO2(g) + H2O(g)

Chemistry
1 answer:
ale4655 [162]3 years ago
8 0

Answer: The coefficient for O_2(g) is 12.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

C_{10}H_{12}O_4S(s)+12O_2(g)\rightarrow 10CO_2(g)+SO_2(g)+6H_2O

Thus in the reactants, there are 12 molecules of oxygen in balanced chemical equation. Thus the coefficient for O_2(g) is 12.

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Answer:

  • Question 19: the three are molecular compounds.
  • Question 20: CuSO₄.5H₂O

Explanation:

<em>Question 19.</em>

  • C₂H₄
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  • H₂O₂

All of them are the combination of two kinds of different atoms in fixed proportions.

  • C₂H₄: two carbon atoms per four hydrogen atoms
  • HF: one hydrogen atom per one fluorine atom
  • H₂O₂: two hydrogen atoms per two oxygent atoms

Thus, they all meet the definition of compund: a pure substance formed by  two or more different elements with a definite composition.

Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.

Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.

<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.

Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol

Water is H₂O. Its molar mass is 18.015g/mol

Calling x the number of water molecules in the hydrate, the percentage of water is:

       \dfrac{18.015x}{18.015x+159.609}=36\%\\ \\ \\ \dfrac{18.015x}{18.015x+159.609}=0.36

From which we can solve for x:

      18.015x=6.4854x+57.45924\\ \\ 11.5296x=57.45924\\ \\ x\approx4.98\approx5

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is:

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4 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

3 0
3 years ago
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