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lara [203]
4 years ago
7

WRITE INCOMPLETE COMBUSTION OF ETHANE

Chemistry
1 answer:
Rainbow [258]4 years ago
8 0
The equation for the incomplete combustion of ethane is as follows:
2C2H6(g)+5O2(g)—>4CO(g)+6 H2O(l)
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What is the mass of 7.00 moles of H2O2
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Answer:

0.206 mol H₂O₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

7.00 moles H₂O₂

<u>Step 2: Identify Conversions</u>

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O₂ - 2(1.01) + 2(16.00) = 34.02 g/mol

<u>Step 3: Convert</u>

<u />7.00 \ g \ H_2O_2(\frac{1 \ mol \ H_2O_2}{34.02 \ g \ H_2O_2} ) = 0.205761 mol H₂O₂

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.205761 mol H₂O₂ ≈ 0.206 mol H₂O₂

8 0
3 years ago
A halogen is example of a(n)
kenny6666 [7]

A halogen is example of nonmetal. The answer is letter A. When compounds containing halogens they are called salts thus the name “salt – former”. Halogen consists of Fluorine, Chlorine, Bromine, Iodine, Astatine.

6 0
3 years ago
In the absence of air a penny and a feather that are dropped from the same height at the same time will
erik [133]
Have the same acceleration/reach the ground at the same time.

6 0
3 years ago
Which measurement would not be possible to obtain from the triple beam balance shown? a. 558.6 g b. 463.455 g c. 2.4 g d. 100.0
kirza4 [7]

Answer: B. 463.455 g

Explanation:

8 0
3 years ago
A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

4 0
3 years ago
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