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3 years ago

PLEASE HELP this is chemistry 12 points

Chemistry

1 answer:

Inessa05 [86]3 years ago

8 0

Answer:

6.791

Explanation:

For proper significant figures with addition, you would use the significant figures of the number with lowest decimal place. 6.298 goes to the 10⁻³ place. 0.492712 goes to the 10⁻⁶ place. You will go out to the 10⁻³ place.

6.298 + 0.492712 = 6.790712 ≈ 6.791

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3 years ago

The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t

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Answer: The theoretical yield of iron(III) sulfate is 26.6 grams

Explanation:

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

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

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol$

The given chemical equation follows:

$2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4$

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = $\frac{1}{2}\times 0.133=0.0665moles$ of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

$0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g$

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

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3 years ago

How would you prepare 100 ml of 0.4 M MgSO4 from a stock solution of 2 M MgSO4?

miss Akunina [59]

OK, so to answer this question, you will simply use the molality equation which is as follows:
M1V1 = M2V2
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml

plug in the givens in the above equation:
2 x V1 = 0.4 x 100
therefore:
V1 = 20 ml

Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.

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3 years ago