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aliya0001 [1]
2 years ago
15

Lawrencium-262 has a half-life of 4 hr. How much of a 40 mg sample remains after 12 hours?

Chemistry
1 answer:
CaHeK987 [17]2 years ago
3 0

Answer:

5 mg

Explanation:

If one half life is 4 hours, then 3 half lives is 12 hours.

This means that the sample will decay to 1/8 of its original amount.

So, the answer is 40(1/8) = 5 mg.

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Describe a polyhydroxyl alcohol.
Mademuasel [1]
With a name like poly hydroxyl alcohol, it suggests that the alcohol has more than one alcohol group. Thus any alcohol with more than one hydroxyl is a polyhydroxyl alcohol. An example that pops into mind is ethanediol (CH2OHCH2OH) this has 2 hydroxyl groups and is an alcohol, thus a polyhydroxyl alcohol
8 0
3 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
Sodium is a compound<br> O True<br> or<br> O False
MA_775_DIABLO [31]

Answer:

i think false.

Explanation:

tell me if this is wrong

6 0
2 years ago
Read 2 more answers
1. Any gentle wind is called a
liubo4ka [24]

It's called a breeze

5 0
2 years ago
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What would the rate law be for the following reaction:
lana [24]

Answer:

C. Rate = k[H2]^2[O2]

Explanation:

Rate law only cares about REACTANTS. Since, rate law can only be determined experimentally, I am assuming the given reaction mechanism is elementary reaction from which we can write the rate law.

Only H2 and O2 are part of rate law since they are reactants and also the coefficient in front of H2 goes as exponent on rate law to indicate the order of H2 in the reaction.

Rate= k [H2]^2 [O2]

6 0
3 years ago
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