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3 years ago

Phases of matter Soild * * * Liquid * * * Gas * *

Chemistry

1 answer:

Annette [7]3 years ago

3 0

Strawberry, bread, bananas/ lemonade, milk, water/ nitrogen, oxygen, carbon dioxide

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Studentka2010 [4]

First blank -Same
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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

slava [35]

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

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Resources found in lithosphere: gold and iron etc

Resources found in atmosphere: Water vapor, gases etc.

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2 years ago

What would be the most likely explanation if none of the tested paint samples had turned brown?

mariarad [96]

Answer:4/d

Explanation:

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2 years ago

A brick has a mass of 100 G and a volume of 25CM3 what is the density of the brick

kumpel [21]

Answer:

$\boxed {\boxed {\sf 4 \ g/cm^3}}$

Explanation:

We are asked to find the density of a brick. Density is a substance's mass per unit volume. The formula for calculating density is as follows.

$\rho= \frac{m}{v}$

The mass of the brick is 100 grams and the volume is 25 cubic centimeters.

m= 100 g
v= 25 cm³

Substitute the values into the formula.

$\rho= \frac{ 100 \ g}{25 \ cm^3}$

Divide.

$\rho=4 \ g/cm^3$

The density of the brick is 4 grams per cubic centimeter.

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3 years ago