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Advocard [28]
2 years ago
6

2 CH3OH + 3 O2 + 2 CO2 + 4H2O

Chemistry
1 answer:
geniusboy [140]2 years ago
7 0

Answer:

Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen. A mole of water molecules would be 2 moles of hydrogen atoms plus 1 mole of oxygen atoms. From the periodic table we see the atomic weight of hydrogen is 1.0079 and the atomic weight of oxygen is 15.9994

Explanation:

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6 0
3 years ago
Please answer these about Charles law
NNADVOKAT [17]

Answer:

1. V2.

2. 299K.

3. 451K

4. 0.25 x 451 = V2 x 299

Explanation:

1. The data obtained from the question include:

Initial volume (V1) = 0.25mL

Initial temperature (T1) = 26°C

Final temperature (T2) = 178°C

Final volume (V2) =.?

2. Conversion from celsius to Kelvin temperature.

T(K) = T (°C) + 273

Initial temperature (T1) = 26°C

Initial temperature (T1) = 26°C + 273 = 299K

3. Conversion from celsius to Kelvin temperature.

T(K) = T (°C) + 273

Final temperature (T2) = 178°C

Final temperature (T1) = 178°C + 273 = 451K

4. Initial volume (V1) = 0.25mL

Initial temperature (T1) = 299K

Final temperature (T2) = 451K

Final volume (V2) =.?

V1 x T2 = V2 x T1

0.25 x 451 = V2 x 299

6 0
4 years ago
Which type of data have elements in common but do not fit into rigid rows and columns in a table?.
Talja [164]

Answer: Semi-structured data is a type of data that has some consistent and definite characteristics. It does not confine into a rigid structure such as that needed for relational databases that which is why  it will not fit into row and columns in a table .

Explanation:

6 0
3 years ago
Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi
Fed [463]

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
g Determine the empirical formula for a compound that contains C, H and O. It contains 40.92% C, 4.58% H, and 54.50% O by mass.
Lady bird [3.3K]

Answer:

The empirical formula for the compound is C3H4O3

Explanation:

The following data were obtained from the question:

Carbon (C) = 40.92%

Hydrogen (H) = 4.58%

Oxygen (O) = 54.50%

The empirical formula for the compound can be obtained as follow:

C = 40.92%

H = 4.58%

O = 54.50%

Divide by their molar mass

C = 40.92/12 = 3.41

H = 4.58/1 = 4.58

O = 54.50/16 = 3.41

Divide by the smallest i.e 3.41

C = 3.41/3.41 = 1

H = 4.58/3.41 = 1.3

O = 3.41/3.41 = 1

Multiply through by 3 to express in whole number

C = 1 x 3 = 3

H = 1.3 x 3 = 4

O = 1 x 3 = 3

The empirical formula for the compound is C3H4O3

6 0
4 years ago
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