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wolverine [178]
4 years ago
15

Solve using significant figures 8.647 + 45.969

Chemistry
2 answers:
marishachu [46]4 years ago
7 0

Answer:

54.62

Explanation:

When you are adding numbers the significant figures for the result must be the smallest number of significant figures from the numbers you are adding.

In this case, you have 8.647 with 4 significant figures

And 45.969 with 5 significant figures

So, your result must have 4 significant figures.

Now, when you add the numbers you obtain:

8.647+45.969 = 54.616

But this result has 5 significant figures, so you need to round to the next integer, in this case, you should round to the number 2 because the number to follow the 1 is greater than 5.

Then you have that 54.616 rounded to the next integer is 54.62 with 4 significatn figures.

Murrr4er [49]4 years ago
5 0

Answer:

54.616

Explanation:

8.647 + 45.969

or rewrite for easier look:

45.969 +

 8.647 =

54.616

Hope this helped :3

 

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Consider the following reaction: 2 NO(g) + 5 H2(g) → 2 NH3(g) + 2 H2O(g) Which set of solution maps would be needed to calculate
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Answer : The correct option is, (I) gNO\rightarrow molNO\rightarrow molNH_3\rightarrow gNH_3

Solution : Given,

Mass of NO = 45.8 g

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Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of NO and O_2.

\text{ Moles of }NO=\frac{\text{ Mass of }NO}{\text{ Molar mass of }NO}=\frac{45.8g}{30g/mole}=1.53moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{12.4g}{2g/mole}=6.20moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2NO(g)+5H_2(g)\rightarrow 2NH_3(g)+2H_2O(g)

From the balanced reaction we conclude that

As, 2 mole of NO react with 5 mole of H_2

So, 1.53 moles of NO react with \frac{1.53}{2}\times 5=3.82 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and NO is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 2 mole of NO react to give 2 mole of NH_3

So, 1.53 mole of NO react to give 1.53 mole of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(1.53moles)\times (17g/mole)=26.0g

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Answer:

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Explanation:

Let's consider the complete neutralization of a diprotic acid H₂X with NaOH.

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