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pogonyaev
3 years ago
14

Wha do u do when u get in a fight wit yo best friend?

Chemistry
2 answers:
antiseptic1488 [7]3 years ago
8 0
Talk to them and listen to each other. if they aren’t ready to talk, give them space. once both of you are ready, you can make up and forgive each other. don’t bother them by asking a lot of questions and forcing them to talk to you. and if they’re doing that, tell them you need time to think. just be sure to talk to them, listen, and understand. tell each other both sides of the stories. of course, different situations can require different solutions. so resolve it when it’s time :)
Zinaida [17]3 years ago
5 0

Try to talk it out, it’s works for me when I get in a fight with my BFF.

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What is the mole to mole relationship between copper and zinc? Cu2+(aq)+->Cu(s)+Zn2+(aq)
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5 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
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