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creativ13 [48]
3 years ago
12

You are asked to write observations about a 100g sample of Nitrogen-16 before it decays, but you're running late. In order to ma

ke observations, you require at least .5g of material. Can you make observations if you are 57 seconds late to the lab?
Chemistry
1 answer:
san4es73 [151]3 years ago
5 0

Answer:

You cannot make observations if you are 57 seconds late into the lab.

Explanation:

The atomic nucleus can split by decay into 2 or more particles as a result of the instability of its atomic nucleus due to the fact that radioactive elements possess an unstable atomic nucleus.

Now, the primary particles which are emitted by radioactive elements in order to make them decay are alpha, beta & gamma particles.

The half life equation is;

N_t = N₀(½)^(t/t_½)

Where:

t = duration of decay

t_½ = half-life

N₀ = number of radioactive atoms initially

N_t = number of radioactive atoms remaining after decay over time t

We are given;

t = 57 secs

N₀ = 100 g

Now, half life of Nitrogen-16 from online sources is 7.2 seconds. t_½ = 7.2

Thus;

N_t = 100(1/2)^(57/7.2)

N_t = 0.4139g

We are told that In order to make observations, you require at least .5g of material.

The value of N_t you got is less than 0.5g, therefore you cannot make observations if you are 57 seconds late.

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fenix001 [56]

1)The limiting reactant will be aluminum acetate

2) The mass of aluminum hydroxide formed will be 9.75 grams.

<h3>Stoichiometric problem</h3>

The equation of the reaction is as below:

Al(C_2H_3O_2)_3 + 3NaOH --- > 3C_2H_3NaO_2 + Al(OH)_3

The mole ratio of the 2 reactants is 1:3.

Mole of 100 mL. 1.25 mol/L Al(C_2H_3O_2)_3 = 0.1 x 1.25 = 0.125 mol

Mole of 300 mL, 2.30 mol/L NaOH = 0.3 x 2.3 = 0.69 mol

Thus, aluminum acetate is limited.

Mole ratio of  Al(C_2H_3O_2)_3 and Al(OH)_3 = 1:1

Equivalent mole of Al(OH)_3 = 0.125 mol

Mass of 0.125 mole Al(OH)_3 = 0.125 x 78 = 9.75 grams.

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

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