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Lemur [1.5K]
2 years ago
11

A volume of saline solution had a mass of 61.4 g at 4°C. An equal volume of water at the same

Chemistry
1 answer:
Maurinko [17]2 years ago
5 0
Multiple by 2 because it was give u right answer
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cricket20 [7]

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7 0
2 years ago
Read 2 more answers
It is safe to stand on the top step of a ladder provided it is braced properly. A) True B) False
Roman55 [17]

FALSE You are never allowed to stand on the top of the ladder even if all legs are on the ground

3 0
2 years ago
Number 2 I need help Scientific notation
Lelu [443]
  • (6.6 x 10^2) (9.6 x 10^6) = 6.336 × 10^6
4 0
3 years ago
Why is Potassium hydroxide considered a strong base?
zloy xaker [14]

Answer:

KOH is a strong base, and therefore dissociates completely into negative  and positive ions in water solution. pH 8.35 is a high pH (> 7), therefore this is a basic solution. 4.

Explanation:

7 0
2 years ago
If 11.7 g of aluminum reacts with 37.2 g of copper (II) sulfate according to the following reaction, how many grams of aluminum
zhannawk [14.2K]

Answer:

There is 26.59 grams of aluminium sulfate produced

Explanation:

<u>Step 1:</u> Data given

Mass of aluminium = 11.7 grams

Mass of copper (II) sulfate = 37.2 grams

Molar mass of Aluminium = 26.98 g/mol

Molar mass of CuSO4 = 159.61 g/mol

Molar mass of Al2(SO4)3 = 342.15 g/mol

<u />

<u>Step 2</u>: The balanced equation

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

<u>Step 3</u>: Calculate moles of Aluminium

Moles Al = mass Al / molar mass Al

Moles Al = 11.7 grams / 26.98 g/mol

Moles Al = 0.434 mol

<u>Step 4</u>: Calculate moles of CuSO4

Moles CuSO4 = 37.2 grams / 159.61 g/mol

Moles CuSO4 = 0.233 moles

<u>Step 5:</u> Calculate limiting reactant

For 2 moles of Al we need 3 moles of CuSO4

CuSO4 is the limiting reactant. It will be completely consumed (0.233 moles).

Al is in excess. There will be consumed 0.233 *(2/3) = 0.1553 moles

There will remain 0.434 - 0.1553 = 0.2787 moles

<u>Step 6: </u>Calculate moles of Al2(SO4)3

For 2 moles of Al we need 3 moles of CuSO4, to produce 1 mole of Al2(SO4)3 and  3 moles of Cu

For 0.233 moles CuSO4 we produce 0.233/3 = 0.0777 moles of Al2(SO4)3

<u>Step 7</u>: Calculate mass of Al2(SO4)3

Mass of Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3

Mass of Al2(SO4)3 = 0.0777 moles * 342.15g/mol

Mass of Al2(SO4)3 = 26.59 grams

There is 26.59 grams of aluminium sulfate produced

4 0
3 years ago
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