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andrew11 [14]
3 years ago
14

What is the mass of 0.572 moles of Al?

Chemistry
1 answer:
kozerog [31]3 years ago
5 0
Answer: 15.433704

Use the periodic table to check the atomic mass, this is the number of grams per mole → 1 mole of Aluminum is 26.982 g
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_AgNO3 + _<br> Cu → _<br> Cu(NO3)2 + -<br> Ag
GuDViN [60]

Answer:

2AgNO3 + Cu = Cu(NO3)2 + 2Ag

Explanation:

First you see which side has the most elements. Which is Cu(NO3)2 + Ag. But, both sides have the same elements? But, on the reactants side, there is 2 of NO3. On the products side there is only one.

Reactants:            Products:

Cu = 1                    Cu = 1

NO3 = 1                 NO3 = 2

Ag = 1                    Ag = 1

They are all equal, except for NO3. So on the reactants side, you add a two to make it even.

2AgNO3 + Cu = Cu(NO3)2 + Ag

Reactants:            Products:

Cu = 1                    Cu = 1

NO3 = 2                NO3 = 2

Ag = 2                   Ag = 1

Now, the NO3 is equal, But the Ag isn't. But, you can add a 2 on the <u>products</u> side so the whole equation becomes equal.

2AgNO3 + Cu = Cu(NO3)2 + 2Ag

Reactants:            Products:

Cu = 1                    Cu = 1

NO3 = 2                NO3 = 2

Ag = 2                   Ag = 2

6 0
3 years ago
Cecil writes the equation for the reaction of hydrogen and oxygen below.
Wittaler [7]

Answer:

C) to show that atoms are conserved in chemical reactions

Explanation:

When writing a chemical reaction, we should always consider the Mass Conservation Law, which basically states that; in an isolated system; the total mass should remain constant, this is, the total mass of the reactives should be equal to the total mass of the products

For this case, we should add the apporpiate coefficients in order to be in compliance with this law:

2H₂ + O₂ → 2H₂O

So, we can check the above statement:

For reactives (left side):

4H

2O

For product (right side):

4H

2O

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this way they can make sure that the experiment is correct.


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Answer:zigga

Explanation:

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I think that the answer is (D)... I hope this helped

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