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nevsk [136]
3 years ago
13

What is the mass of an original 5.60-gram sample of iron-53 that remains unchanged after 25.53 minutes?

Chemistry
1 answer:
olga_2 [115]3 years ago
4 0
Answer is 0.7 g

Reason: Given: mass of Fe = 5.60 gWe know that, half life of Fe-53 = t1/2 (Fe-53) = 8.51 min.
Thus, after 25.53 mins, number of half life passed = 25.53/8.51= 3

Now, Amount left after first half life = 5.60g/2 = 2.80 g
Amount left after second half life = 2.80g/2 = 1.40 g
Amount left after third half-life = 1.40g/2 = 0.7g.
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Answer:

This question is incomplete

Explanation:

This question is incomplete but...

1) You can calculate the molarity of the NaOH for each trial by following the steps below.

The formula for Molarity (M) is

M = number of moles (n) ÷ volume (V)

where the unit of volume must be in Litres or dm³

The unit of molarity is mol/dm³ or mol/L or molar conc (M)

The final answer must have the unit of molarity

If the number of moles is not provided, look out for the mass of NaOH used and then calculate your number of moles (n) as

n = mass of NaOH used ÷ molar mass of NaOH

Where the atomic mass of sodium (Na) is 23, oxygen (O) is 16 and hydrogen (H) is 1. Hence, molar mass for NaOH is 23 + 16 + 1 = 40 g/mol

n = mass of NaOH used ÷ 40

2) Average Molarity will be (Trial 1 Molarity +Trial 2 Molarity) ÷ 2

Answer must be in mol/dm³ or mol/L or M

3) Label the volumentric flask containing the NaOH solution with the answer gotten from (2) above

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Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)


Rate Law: k[HI]^2


Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

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