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Serggg [28]
2 years ago
10

18. Does the diagram below represent an exothermic reaction, or an endothermic reaction? Why do you think this? (2 pts) *

Chemistry
1 answer:
EastWind [94]2 years ago
7 0

Answer:

exothermic reaction

Explanation:

:)

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If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0
Yuki888 [10]

Answer:

T₂ = 317.87 K

Explanation:

Given data:

Initial pressure = 15 atm

Final pressure = 16 atm

Initial temperature = 298 K

Final temperature = ?

Solution:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

15 atm / 298K = 16 atm/T₂

T₂ = 16atm × 298 K / 15 atm

T₂ = 4768 atm. K / 15 atm

T₂ = 317.87 K

4 0
3 years ago
Has any one done 2020 chemistry paper 1 aqa gcse
Mars2501 [29]

Answer:

what do you mean?

Explanation:

7 0
3 years ago
A compound that contains 63.15% carbon,5.30% hydrogen and the remaining percent oxygen
tamaranim1 [39]
Oxygen percentage is 31.55%
4 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
Which of the following statements is true about catalysts? and Why?
LUCKY_DIMON [66]
I believe the correct answer from the choices listed above is option D. Catalysts lower the activation energy of a chemical reaction. It <span>is a substance which speeds up a reaction, but is chemically unchanged at the end of the reaction. It provides another pathway for the reaction to occur.</span>
8 0
3 years ago
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