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4 years ago

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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The number of moles of b2o3 that will be formed is determined as 4 moles.

<h3>Limiting reagent</h3>

The limiting reagent is the reactant that will be completely used up.

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from the equation above;

4 b ------------> 2 b₂O₃

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3 : 2

b is the limiting reagent, thus, the amount of b2o3 to be formed is calculated as;

4 b ------------> 2 moles of b2o3

8 moles -------> ?

= (8 x 2)/4

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4 years ago

A 4 kg rock is rolling 10 m/s. Find its kinetic energy

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2 years ago

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the max

jolli1 [7]

<u>Answer:</u> The mass of aluminium chloride that can be formed are 46.3 g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

<u>For Aluminium:</u>

Given mass of aluminium = 32 g

Molar mass of aluminium = 26.98 g/mol

Putting values in above equation, we get:

$\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol$

<u>For Chlorine:</u>

Given mass of chlorine = 37 g

Molar mass of chlorine = 71 g/mol

Putting values in above equation, we get:

$\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol$

For the given chemical equation:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

By Stoichiometry of the reaction:

3 moles of chlorine gas is reacting with 2 moles of aluminium.

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium.

As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.

So, chlorine gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

3 moles of chlorine gas is producing 2 moles of aluminium chloride

So, 0.521 moles of chlorine gas will react with = $\frac{2}{3}\times 0.521=0.347moles$ of aluminium chloride.

Now, calculating the mass of aluminium chloride by using equation 1, we get:

Moles of aluminium chloride = 0.347 moles

Molar mass of aluminium chloride = 133.34 g/mol

Putting all the values in equation 1, we get:

$0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g$

Hence, the mass of aluminium chloride that can be formed are 46.3 g

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