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1 year ago

Match each interaction type to the correct amount of potential energy.

Chemistry

1 answer:

Zigmanuir [339]1 year ago

3 0

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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Answer:

1. they reduce green house gases

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Explanation:

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2 years ago

How much CaCO3 would have to be decomposed to produce 247 g of CaO

vovangra [49]

441 g CaCO₃ would have to be decomposed to produce 247 g of CaO

<h3>Further explanation</h3>

Reaction

Decomposition of CaCO₃

CaCO₃ ⇒ CaO + CO₂

mass CaO = 247 g

mol of CaO(MW=56 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{247}{56}\\\\mol=4.41$

From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :

$\tt \dfrac{1}{1}\times 4.41=4.41$

mass CaCO₃(MW=100 g/mol) :

$\tt mass=mol\times MW\\\\mass=4.41\times 100\\\\mass=441~g$

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2 years ago

How many moles are in 5.025 grams of NaOH?

yKpoI14uk [10]

Answer:

The SI base unit for amount of substance is the mole. 1 grams NaOH is equal to 0.025001806380511 mole.

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3 years ago

Read 2 more answers

The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?

Katyanochek1 [597]

Answer:

$a=64.7mg$

Explanation:

Hello,

In this case, we need to remember that for the required time for a radioactive nuclide as radium-226 to decrease to one half its initial amount we are talking about its half-life. Furthermore, the amount of remaining radioactive material as a function of the half-lives is computed as follows:

$a=a_0(\frac{1}{2} )^{\frac{t}{t_{1/2}} }$