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3 years ago

What trends were seen in medeleevs periodic table

Chemistry

1 answer:

Paraphin [41]3 years ago

8 0

Answer:

groups are based on how many electrons to become stable

Explanation:

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Which of the following describes the location and energy of all the valence electrons

Nastasia [14]

Answer:

1s22s22p63s23p4

Explanation:

Sulfur is located in the p block and has 6 valence electrons (the 2 exponent on the 3s and the 4 exponent on the 3p add up to 6)

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2 years ago

Answer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.

fomenos

Answer:

verdadero/a

falso/b

verdadero/c

Explanation:

6 0

3 years ago

Sylvanite is a mineral that contains 28.0% gold by mass. How much sylvanite would you need to dig up to obtain 73.0 g of gold?

Scorpion4ik [409]

You would have to dig up 261 g of sylvanite.

Mass of sylvanite = 73.0 g Au × (100 g sylvanite/28.0 g Au) = 261 g sylvanite.

3 0

3 years ago

What is the name of a solution whose concentration of solute is equal to the maximum concentration that is predicted from the

Stels [109]

C because it is maximumly contracted

8 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]

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

8 0

3 years ago