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cestrela7 [59]
2 years ago
10

Law of conservation of energy A B C D

Chemistry
2 answers:
Rama09 [41]2 years ago
8 0
Choice A because the amount of Joules on both sides are equal to one another (balanced).
Vesna [10]2 years ago
8 0
Law of conservation of energy states that energy can neither be created nor destroyed, so whatever energy you put in, you get it back without any loss.
Therefore A is correct
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What two factors determine the thermal energy in a substance
Eduardwww [97]
The answer to that is mass and chemical
7 0
3 years ago
Match each interaction type to the correct amount of potential energy.
Zigmanuir [339]

The type of energy which is present between the repulsion interaction is the highest potential energy.

<h3>What is potential energy?</h3>

Potential energy is the amount of energy that is possess by any body with respect to the electric charge posses in that and other factors also.

  • When molecules have a high attraction force then they have the low potential energy in them.
  • When molecules have a great repulsion between them then they have the great potential energy in them.
  • And molecules have the middle amount of potential energy then they have the balanced interaction.

Hence, molecules in which repulsion interaction is present will have highest potential energy.

To know more about potential energy, visit the below link:

brainly.com/question/14427111

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3 0
2 years ago
Which branch of chemistry would be used to make steel?
Tpy6a [65]

Answer:

the five major branches of chemistry are organic, inorganic, analytical, physical, and biochemistry.

5 0
3 years ago
RANK THE FOLLOWING FROM WARMEST TO COOLEST... 500G ALUMINUM CUP, 500G GOLD CUP, 750G ALUMINUM CUP.
blagie [28]
The answer is:

1. 500g stainless steel cup 
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<span>3. 750g aluminum cup</span>
8 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

3 0
2 years ago
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