In your own words, explain the rules for multiplication and division with significant figures

1 answer:

6 0

Your answer cannot be MORE precise than the least precise measurement. The following rule applies for multiplication and division: The LEAST number of significant figures in any number of the problem determines the number of significant figures in the answer.

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Why is water a pure substance.

Helen [10]

Explanation:

water is a pure substance because it only contains one type of molecule.

6 0

3 years ago

The zero order reaction 2N2O→2N2+O2 has the reaction constant k is 6.28×10−3 molL s. If the initial concentration of N2O is 0.96

Talja [164]

<u>Answer:</u> The concentration of $N_2O$ in three significant figures will be 0.899 mol/L.

<u>Explanation:</u>

For the given reaction:

$2N_2O\rightarrow 2N_2+O_2$

The above reaction follows zero order kinetics. The rate law equation for zero order follows:

$k=\frac{1}{t}([A_o]-[A])$

where,

k = rate constant for the reaction = $6.28\times 10^{-3}\text{ mol }L^{-1}s^{-1}$

t = time taken = 10 sec

$[A_o]$ = initial concentration of the reactant = 0.962 mol/L

[A] = concentration of reactant after some time = ?

Putting values in above equation, we get:

$6.28\times 10^{-3}=\frac{1}{10}(0.962-[A])$

$[A]=0.899mol/L$

Hence, the concentration of $N_2O$ in three significant figures will be 0.899 mol/L.

5 0

3 years ago

I really can’t do this don’t understand it

masya89 [10]

Answer:

7.96g, 33.79%

Explanation:

I'll try my best to explain the entire process behind this question ;)

From the question, you can write the reaction

$2H_2O(l)->2H_2(g)+O_2(g)$

Now, there are a few reasons it is like this. Oxygen is a diatomic element, meaning it doesn't and can't exist as just O. It exists as O₂. To balance, this, double the amount of water and hydrogen so there is an equal amount of each element on both sides of the reaction (4 H's, 2 O's on the reactant side, and 4 H's, 2 O's on the product side).

From this we can get a mole-to-mole ratio.

Onto the stoichiometry. Our goal in this is to convert from grams of water to grams of hydrogen, and we do so with a mole to mole ratio.

$71.0gH_2O*\frac{1molH_2O}{18.02g} *\frac{2molH_2}{2molH_2O}* \frac{2.02g}{1molH_2}\\\\ =7.96gH_2$

Basically, what I did was divide by water's molar mass to get moles of water, multiplied by the mole-to-mole ratio (2:2) to get moles of H2, and then multiplied by H2's molar mass to get what should be the amount of H2 produced by the reaction.

For percent yield, you can calculate it is such:

$\frac{Actual}{Theoretical}*100$