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2 years ago

Which statement is true about the concentration of the reactants and products in chemical equilibrium?

Chemistry

1 answer:

bulgar [2K]2 years ago

5 0

Answer:

Ans: 2

Explanation:

The concentration of reactants and the concentration of products are constant.

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Solve fasttt<br><br> C7H17+O2=CO2+H2O<br><br> Balancing Equations

sertanlavr [38]

Answer:

The answer to your question is

4C₇H₁₇ + 45 O₂ ⇒ 28 CO₂ + 34H₂O

Explanation:

Write the equation

C₇H₁₇ + O₂ ⇒ CO₂ + H₂O

Process

1.- Check if the equation is balanced

Reactants Element Products

7 C 1

17 H 2

2 O 3

As the number of reactants and products is different, we conclude that the reaction is unbalanced.

2.- Write a coefficient "7" to CO₂ and a coefficient of 17/2 to H₂O

C₇H₁₇ + O₂ ⇒ 7CO₂ + $\frac{17}{2}$ H₂O

Reactants Element Products

7 C 7

17 H 17

2 O 51/2

3.- Write a coefficient of 45/2 to the O₂, and multiply all the equation by 2.

4C₇H₁₇ + 45 O₂ ⇒ 28 CO₂ + 34H₂O

Reactants Element Products

28 C 28

68 H 68

90 O 90

5 0

3 years ago

4.8 moles of Kl are dissolved in 850 mL of water. What is the molarity of the solutions?

AURORKA [14]

Hey there!

Molarity = 4.8/850 x 1000 =5.647mol/lit or 5.65M

Hope this helps you dear :)
Have a good day :)

4 0

1 year ago

Chromuate found inside which of the following structures of a eukaryotic cell?

dimulka [17.4K]

Answer:

Chromuates are found in the chromosomes of the eukaryotic cells!!!!

Explanation:

I learned this in 7th grade!!!!

Mark Me Brainliest plzz

4 0

3 years ago

Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]

ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

rate = $k_{2} [SO_{3}][C_{6}H_{6}]$

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

rate = $k_{1}[H_{2}SO_{4}]^{2}$

Hence when we substitute equation 2 in equation one,

Rate comes out to be = $k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]$

6 0

3 years ago

Calculate the moles of calcium oxalate formed from 5 grams of potassium oxalate (molar mass 184.24 g/mol) the mole mole ratio fr

Nataly_w [17]

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

The molar mass of K₂C₂O₄ is 184.24 g/mol.
The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.

$5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4} \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4$

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Learn more: brainly.com/question/15288923

7 0

2 years ago