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Yuri [45]
3 years ago
14

the combustion of ethane (c 2 h 6 ) is represented by the equation: 2c 2 h 6 (g) 7o 2 (g) 4co 2 (g) 6h 2 o(l) in this reaction:

(a) the rate of consumption of ethane is seven times faster than the rate of consumption of oxygen. (b) the rate of formation of co 2 equals the rate of formation of water. (c) water is formed at a rate equal to two-thirds the rate of formation of co 2 . (d) the rate of consumption of oxygen equals the rate of consumption of water. (e) co 2 is formed twice as fast as ethane is consumed.
Chemistry
1 answer:
DedPeter [7]3 years ago
4 0

Answer:

(e) CO2 is formed twice as fast as ethane is consumed.

Explanation:

Since we have the equation of the reaction as;

2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l)

The rate of consumption of ethene =  -1/2 d[C2H6]/dt

Rate of production of CO2= 1/4 d[CO2]/dt

Hence, CO2 is formed twice as fast as ethane is consumed.

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3 years ago
How can you use the periodic table to find the number of protons in one atom of an element?
evablogger [386]
A is your answer.
On the periodic table the atomic number is the number of protons inside the nucleus.
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Alexeev081 [22]

Answer:

chloroplast organelle

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You are provided with a stock solution with a concentration of 1.0x10-5 M. You will be using this to make two standard solutions
artcher [175]

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

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<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>

C₁ . V₁ = C₂ . V₂

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<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>

C₁ . V₁ = C₂ . V₂

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If something has an actual yield of 2.5 grams and a theoretical yield of 7.5 grams, what is the percent yield?
aivan3 [116]
2.5/7.5 = 0. 3333 * 100% = 33.33%
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