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Anna35 [415]
3 years ago
5

As someone falls toward the Earth, what happens to their GPE?

Chemistry
1 answer:
Dimas [21]3 years ago
3 0
<h3>Answer:</h3>

B. Decreases

<h3>Explanation:</h3>
  • The potential energy or stored energy possessed by an object relative to the earth's gravity is known as gravitational potential energy, GPE.
  • This energy depends on the position of the object from the earth's surface and its weight.
  • It is given by, GPE = mass × height × gravitational acceleration
  • Therefore, a heavier object has more gravitational potential compared to a lighter object at the same height.
  • Additionally, an object at a higher height from the surface of the earth will possess more gravitational potential energy compared to an object near the earth's surface.

In this case;

  • We would, therefore, conclude that when someone falls towards the earth his/her gravitational potential energy (GPE) decreases due to a decrease in height from the surface of the earth.
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the smallest particle of a chemical element that can exist.

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2 years ago
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You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from
GuDViN [60]

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.400mole/L\times 0.200L=0.08mol

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 200mL+200L=400mL

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 400mL=400g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 400 g

T_{final} = final temperature of water = 27.78^oC=273+25.10=300.78K

T_{initial} = initial temperature of metal = 25.10^oC=273+27.78=298.1K

Now put all the given values in the above formula, we get:

q=400g\times 4.18J/g^oC\times (300.78-298.1)K

q=4480.96J

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

\Delta H=\frac{-4480.96J}{0.08mole}=-56012J/mole=-56.012kJ/mol

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 0
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Answer:

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"Both atoms achieve a more stable configuration."

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I am pretty sure that the answer is B. extrusive rock. I hope that this helps you!! Good Luck!!
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