Chemistry, 26.07.2019 15:00 AdoNice

Let a represent the red spheres and b represent the blue spheres. write a balanced equation for the equilibrium reaction.

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I have attached the diagram related to this question.

5A₂ + 5B .................> 4A₂B + A₂ + B

Explanation:
1- For the reactants:
We can note that:
i- we have 5 moles of substance A which is diatomic. This means that we have 5A₂
ii- we have 5 moles of substance B which is monoatomic. This means that we have 5B
Based on the above, for the reactants side, we have:
5A₂ + 5B

2- For the products:
We can note that:
i- we have 4 moles of a substance formed from the combination of two moles of substance A with one mole of substance B. This means that we have 4A₂B
ii- we have one mole of substance A that is diatomic. This means that we have A₂
iii- we have one mole of substance B that is monoatomic. This means that we have B
Based on the above, for the products side, we have:
4A₂B + A₂ + B

3- Combining reactants and products parts:
Combining the reactants side (from part 1) with the products side (from part 2), we can find that the initial equation is as follows:
5A₂ + 5B .................> 4A₂B + A₂ + B

4- Balancing the equation:
Taking a look at the initial equation we reached in part 3, we will find that:
Number of moles of substance A in reactants = 2*5 = 10 moles
Number of moles of substance A in products = 2*4 + 2*! = 10 moles
SUBSTANCE A IS BALANCED
Number of moles of substance B in reactants = 5*1 = 5 moles
Number of moles of substance B in products = 4*1 + 1 = 5 moles
SUBSTANCE B IS BALANCED
Based on the above, the final balanced equation is the same as the initial equation reached in part 3:
5A₂ + 5B .................> 4A₂B + A₂ + B

Hope this helps :)
Let a represent the red spheres and b represent the blue spheres. write a balanced equation for the

Let a represent the red spheres and b represent the blue spheres. write a balanced equation for the

Answer from: Quest

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

Answer from: Quest

can you attach a picture of the lettuce? : )

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