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Levart [38]
3 years ago
15

How do you convert between the volume of a gas at STP and the number of moles of the gas?

Chemistry
1 answer:
Naya [18.7K]3 years ago
6 0
Standard temperature and pressure (STP) means a temperature of 0°c and a pressure of 1 atmosphere (atm). The molar gas volume is used to convert between the number of moles of a gas and the volume of the gas at STP. One mole of a gas occupies a volume of 22400 cm³ or 22.4 liters at STP according to the molar gas volume.
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6th grade help me pleaseeeee
Vesnalui [34]

Answer:

the answer would be A. cells

Explanation:

just trust me Im a 8th grader and I did that before

7 0
3 years ago
When 2.5 grams react, the will produce how many grams of reactant in a complete chemical reaction?
const2013 [10]

Answer:

2.5g

Explanation:

When the reaction goes into completion, they will produce 2.5g. This is complement the law of conservation of mass.

According to the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed but transformed from one form to another".

  • The mass of reactants and products in a chemical reaction must be the same.
  • There is no change in mass in moving from reactant to product
  • So, if we start with 2.5g of reactants, we must end with 2.5g of products.
5 0
2 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
How is a sodium ion symbol written? so so- na na-
vfiekz [6]
A sodium ion symbol is written as Na^+.
6 0
3 years ago
Read 2 more answers
Identify the correct equilibrium constant expression
SCORPION-xisa [38]

Answer: Keq= [CO2]^6[H2O]^6/ [O2]^6

Explanation:

5 0
2 years ago
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