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Stels [109]
3 years ago
10

Plz help! The best answer will get brainliest! What is the net force acting on an object?

Chemistry
2 answers:
yulyashka [42]3 years ago
7 0

Answer:The net force is the sum of all the forces that act upon an object.

Explanation:

Jet001 [13]3 years ago
3 0

Answer:

Gravity

Explanation:

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How many moles of methane (CH4) are in 7.31 x 10^25 molecules
Ivan

molar mass of methane CH4

= C + 4 H  

= 12.0 + 4 x 1.008

= 12.0 +  4.032

= 16.042g/mol

7.31 x 10^25 molecules x <u>             1 mole  CH4    </u>  = 121.43 moles

                                       6.02 x 10^23 CH4 molecules

121.43 moles CH4 are present.

       


6 0
3 years ago
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What does the density of ocean water depend upon
OverLord2011 [107]

It depends on depth

Hope this helps :)

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3 years ago
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When a student combed his dry hair for a long time, his hair began to stand up. Which of these has MOST LIKELY happened? A) The
Alecsey [184]
B. 

Opposites attract, likes repel.
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3 years ago
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How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?
zaharov [31]

Answer:

m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2

Best regards!

7 0
3 years ago
When two objects push or pull on each other, which type of energy is being transferred?
frez [133]

Kinetic energy should be right

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3 years ago
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