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3 years ago

Rubidium has a mass number of 87 and an atomic number of 37. how many electrons does it have?

Chemistry

2 answers:

Elis [28]3 years ago

5 0

Rubidium has 37 elections. there are multiple other sources you could use too confirm this. :-)

Aleksandr [31]3 years ago

5 0

Rubidium's (Rb) Z = 37
Z ( atomic number ) means , No: of proton or no: of electrons.
.: Number of electrons = 37

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3 years ago

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lisov135 [29]

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4 years ago

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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]

<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

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3 years ago

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3 years ago