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avanturin [10]
3 years ago
15

How to turn form moles to grams​

Chemistry
2 answers:
professor190 [17]3 years ago
7 0

If you want to go from moles to grams you will need to multiply the mole value of the substance by its molar mass to get it.

miss Akunina [59]3 years ago
5 0
Turning things to gram so need to convert to the metric system
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Vitamin C contains the elements C, H, and O. It is known to contain 40.9% C and 4.58% H by mass. The molar mass of vitamin C has
soldier1979 [14.2K]

Answer:

C₆H₈O₆

Explanation:

First off, the<u> percent of oxygen by mass</u> of vitamin C is:

  • 100 - (40.9+4.58) = 54.52 %

<em>Assume we have one mol of vitamin C</em>. Then we would have <em>180 grams</em>, of which:

  • 180 * 40.9/100 = 73.62 grams are of Carbon
  • 180 * 4.58/100 = 8.224 grams are of Hydrogen
  • 180 * 54.52/100 = 98.136 grams are of Oxygen

Now we <u>convert each of those masses to moles</u>, using the <em>elements' respective atomic mass</em>:

  • C ⇒ 73.62 g ÷ 12 g/mol = 6.135 mol C ≅ 6 mol C
  • H ⇒ 8.224 g ÷ 1 g/mol = 8.224 mol H ≅ 8 mol H
  • O ⇒ 98.136 g ÷ 16 g/mol = 6.134 mol O ≅ 6 mol O

So the molecular formula for vitamin C is C₆H₈O₆

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3 years ago
A change in the internal energy of a system at constant pressure from a chemical reaction will result in an identical change in
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Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
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Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g)     ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

[]=\frac{P}{R.T}

<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>

<em>Part A  130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>

[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M

[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

<em>Part B  5.0atm SO₂, 3.0atm O₂, 30atm SO₃  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M

[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M

[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

<em>Part C Each reactant and product at a partial pressure of 1.0 atm.  Express your answer using four significant figures.</em>

<em />

[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M

Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0
3 years ago
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