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Nata [24]
3 years ago
7

F.ree points for all!

Chemistry
1 answer:
shepuryov [24]3 years ago
4 0

Answer:

here...I helped u! XD

jkjk...I'll got to the other questions too...dw! XD

Explanation:

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How do you balance a chemical equation
Dominik [7]
There should be mass balance and the charge balance between the reactants and the products

Mass balance : total no of individual atoms of each type should be balanced before and after the reaction

Charge balance : Overall charge of the reactants should be balanced with the overall charge of the products

You can balance,

1)by just looking at it

2)by Algebraic method given above or

3)by the redox method

You need to know how to get the oxidation numbers in order to use the oxidation method

6 0
3 years ago
What volume is occupied by 2.33 moles of an ideal gas at 0.953 atm
nekit [7.7K]

Answer:

WhAT

Explanation:

WTH

4 0
3 years ago
Read 2 more answers
9. Matter is anything that
balu736 [363]
A.has mass and takes up space
5 0
3 years ago
At a certain temperature the value of the equilibrium constant, Kc, is 1.27 for the reaction. 2As (s) + 3H2 (g) ⇌ 2AsH3 (g) What
Bess [88]

Answer:

0.887

Explanation:

Hello,

In this case, the law of mass action for the first reaction turns out:

Kc=\frac{[AsH_3]^2}{[As]^2[H_2] ^3}=1.27

Now, for the second reaction is:

Kc=\frac{[As][H_2] ^{3/2}}{[AsH_3]}

Therefore, by applying square root for the first reaction, one obtains:

\sqrt{Kc} =\sqrt{\frac{[AsH_3]^2}{[As]^2[H_2] ^3}} =\sqrt{1.27}

\frac{\sqrt{[AsH_3]^2} }{\sqrt{[As]^2} \sqrt{[H_2] ^3} } =\sqrt{1.27}

\sqrt{1.27}=\frac{[AsH_3]}{[As][H_2] ^{3/2}}

Finally, since Kc is asked for the inverse reaction, one modifies the previous equation as:

Kc'=\frac{1}{\sqrt{1.27} }=\frac{[As][H_2] ^{3/2}}{[AsH_3]}=0.887

Best regards.

8 0
4 years ago
A substance undergoes a change. Which of the following indicates that the change
LekaFEV [45]

Answer:

change of color

Explanation:

when a substance undergoes a change of color it is an indication that it was chemical

5 0
3 years ago
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