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3 years ago

The equation K2O represents the ionic bond between potassium and oxygen. What is the name of the compound?

Chemistry

2 answers:

eduard3 years ago

5 0

Ten name if this compound is Potassium Oxide

andriy [413]3 years ago

5 0

Answer: potassium oxide.

Explanation:

$K_2O$ is an ionic compound because potassium element is a metal and oxygen element is a non-metal. The bond formed between a metal and a non-metal is always ionic in nature.

The nomenclature of ionic compounds is given by:

1. Positive is written first.

2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.

Hence, the name of $K_2O$ is potassium oxide.

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3 years ago

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2 years ago

True or False: Both molecules and compounds are pure substances.

Solnce55 [7]

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Explanation:

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4 0

3 years ago

Read 2 more answers

Pls help me i dont know this

r-ruslan [8.4K]

Answer:

1. 12.6 moles

2. 8.95 moles

3. 2A + 5B → 3C

4. 48 moles

Explanation:

1. 2Fe + 3Cl₂ → 2FeCl₃

We assume the chlorine in excess. Ratio is 2:2

2 moles of Fe, can produce 2 moles of chloride

12.6 moles of Fe will produce 12.6 moles of chloride.

2. 2Fe + 3Cl₂ → 2FeCl₃

For the same reaction, first of all we need to convert the mass to moles:

500 g . 1mol / 55.85 g = 8.95 mol

As ratio is 2:2, the moles we have are the same, that the produced

4. The reaction for the combustion is:

2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (l)

We assume the oxygen in excess.

Ratio is 2:6, so 2 mol of ethane produce 6 moles of water

Therefore 16 moles of ethane may produce (16 .6) / 2 = 48 moles

3 0

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]

<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