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rjkz [21]
3 years ago
15

The absorbance of a garbanzo bean solution that had been diluted by a factor of three was 0.528. what was the concentration of t

he iron in the original solution (before dilution)?
Chemistry
1 answer:
Serjik [45]3 years ago
6 0
The Beer-Lambert law states that A = E*c*l where A is absorbance, E is the molar absorbance coeffecient, c is concentration and l is path length. Therefore the absorbance is directly proportional to concentration, and by increasing the concentration by a factor of 3, absorbance will increase by a factor of 3 giving   A = 1.584
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If 10.0 g water at 0.0°C is mixed with 20.0 g of water at 30.0°C what is the final temperature of the mixture?
Bond [772]

Answer:

1.67 gradius Celcius

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

7 0
3 years ago
What is the answer helpnme plz
kykrilka [37]
MgH2 + 2 H2O → Mg(OH)2 + 2 H2
8 0
2 years ago
Why was there so much disagreements among the ancient greeks about the nature of the atoms?
olga2289 [7]
It is because in that time science was not established. This is, the experimentation to test hypotheses, which is a fundamental part of the scientific method, was not applied.

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4 0
3 years ago
With the aid of a balanced chemical equation explain Chemical Reactions and the Types Of Chemical Reaction
e-lub [12.9K]

Answer:

Answers

1.)reactants: nitrogen and hydrogen; product: ammonia.

2.)reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water.

3.)N 2 + 3H 2 → 2NH 3

4.)Mg(OH) 2 + 2HNO 3 → Mg(NO 3) 2 + 2H 2O.

5.)2NaClO 3 → 2NaCl + 3O 2

6.)4Al + 3O 2 → 2Al 2O 3

7.)N 2(g) + 3H 2(g) → 2NH 3(g)

Explanation:

''.''

6 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
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