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Bezzdna [24]
3 years ago
11

How do you know if a compound will dissociate or ionize in water?

Chemistry
1 answer:
Ivenika [448]3 years ago
4 0

when an ionic or covalent compound is dissolved in water they break apart into ions through process called dissociation ..the ions get attracted by the water molecules and hence they carry polar charges ..and if there is a conduction of electricity. .then we get to know that a compound is dissociated ..

<u>HOPE</u><u> </u><u>IT</u><u> </u><u>IS</u><u> </u><u>HELPFUL</u><u />

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What does carbon dioxide absorb the most heat energy
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What are metalloids?
goldenfox [79]

A metalloid is a type of chemical element which has a preponderance of properties in between, or that are a mixture of, those of metals and nonmetals.

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A box of jello has a mass of 250 g. How many boxes must be bought to have 1 Kg of jello? *
zhuklara [117]

Answer:

Number of boxes = 4

Explanation:

Given:

Mass of one box of jello = 250 grams

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Number of boxes in 1,000 grams = ?

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8 0
3 years ago
What is the mass, in grams, of 28.58 mL of acetone?
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7 0
3 years ago
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The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<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)

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

8 0
3 years ago
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