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barxatty [35]
3 years ago
9

when the sum 4.9965 + 2.11 + 3.887 is calculated, to how many decimal places should the answer be reported?

Chemistry
1 answer:
jarptica [38.1K]3 years ago
3 0

Answer:

the correct answer is four decimal places

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Answer is : It is a good thermal and electrical conductor. -The main point to be noted is that aluminium is a highly reactive element and still it is used for making cooking utensils. The reason is that aluminium has a very high affinity for oxygen. So, it reacts with oxygen and forms a layer of aluminium oxide on its surface.

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Science quiz, i think its chem... grade 8
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Everything to the right of the arrow is a product.

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If the lab technician needs 30 liters of a 25% acid solution, how many liters of the 10% and the 30% acid solutions should she m
Mice21 [21]

Answer:

7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.

Explanation:

Let the volume of 10% acid solution used to make the mixture = x L

So, the volume of 30% acid solution used to make the mixture = y L

Total volume of the mixture = <u>x + y = 30 L .................. (1) </u>

For 10% acid solution:

C₁ = 10% , V₁ = x L

For 30% acid solution :

C₂ = 30% , V₂ = y L

For the resultant solution of sulfuric acid:

C₃ = 25% , V₃ = 30 L

Using  

C₁V₁ + C₂V₂ = C₃V₃

10×x + 30×y = 25×30

So,  

<u>x + 3y = 75 .................. (2) </u>

Solving 1 and 2 we get,

<u>x = 7.5 L </u>

<u>y = 22.5 L</u>

6 0
3 years ago
Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4
inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

M=\frac{n_{solute}}{V_{solution}}

So you can solve for the moles of the solute:

n_{solute}=M*V_{solution}

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

n_{solute}=1mol/L*1.5L=1.5mol

But this is just a supposition.

Regards.

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2 years ago
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