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exis [7]
3 years ago
8

Please help I need this Now

Chemistry
1 answer:
frez [133]3 years ago
5 0

The above figure is example of system equilibrium because it is closed system.

In closed system there's no loss of product (or reactant) therefore equilibrium is established.

You might be interested in
Which of the following alkyl halides will react fastest with CH3OH in an SN1 mechanism?
yaroslaw [1]

Answer:

IV

Explanation:

The complete question is shown in the image attached.

Let us call to mind the fact  that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.

Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.

Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic  substitution reaction by SN1 mechanism.

5 0
3 years ago
Why is sugar considered an organic compound
Sliva [168]
Sugar is considered an organic compound,
Because it is made up a long chain of carbon atoms.
5 0
3 years ago
Chemically stable than they would
Sergio [31]

Answer:C

Explanation:

8 0
3 years ago
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Crazy boy [7]

1. The volume of ammonia consumed in the reaction is 23.2 L

2. The volume of oxygen consumed in the reaction is 29 L

<h3>Balanced equation</h3>

4NH₃(g) + 5O₂(g) --> 4NO(g) + 6H₂O(g)

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃ and 5 L of O₂

<h3>1. How to determine the volume of ammonia, NH₃ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 4 L of NH₃

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 4) / 6 = 23.2 L of NH₃

Thus, 23.2 L of NH₃ were consumed

<h3>2. How to determine the volume of oxygen, O₂ consumed</h3>

From the balanced equation above,

6 L of H₂O were produced from 5 L of O₂

Therefore,

34.8 L of H₂O will be produced from = (34.8 × 5) / 6 = 29 L of O₂

Thus, 29 L of O₂ were consumed

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

7 0
2 years ago
Determine the theoretical yield and the percent yield if 21.8 g of K2CO3 is produced from reacting 27.9 g KO2 with 57.0 g CO2. T
Alex_Xolod [135]

Answer:

26.9 g

81%

Explanation:

The equation of the reaction is;

4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)

Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles

4 moles of KO2 yields 2 moles of K2CO3

0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3

Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles

2 moles of CO2 yields 2 moles of K2CO3

1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3

Hence the limiting reactant is KO2

Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g

Percent yield = actual yield/theoretical yield × 100

Percent yield = 21.8/26.9 × 100

Percent yield = 81%

5 0
3 years ago
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