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Anna71 [15]
2 years ago
5

Which of the following best represents potential energy being converted to kinetic energy? (2 points) The chemical energy in a b

attery is used to power a cell phone. The movement of wind turns a windmill that generates electricity. A man pedals a bicycle to ride across town. A woman kicks a soccer ball and scores a goal.
Chemistry
2 answers:
ziro4ka [17]2 years ago
6 0
A women kicks a soccer bal and scores a goal
valentina_108 [34]2 years ago
3 0

Answer: Option (d) is the correct answer.

Explanation:

Potential energy is defined as the energy present in an object due to its position.

Kinetic energy is the energy obtained by an object due to its motion. For example, a woman kicks a soccer ball and scores a goal.

When a work is done by an object due to its position or motion then this is known as mechanical energy. For example, the movement of wind turns a windmill that generates electricity.

When chemical energy is utilized to generate electrical energy in an electrochemical cell then it is known as electric current. For example, the chemical energy in a battery is used to power a cell phone.

Thus, we can conclude that a woman kicks a soccer ball and scores a goal, best represents potential energy being converted to kinetic energy.

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If phosphorous has 5 valence electrons, how many does sulfur have?
VARVARA [1.3K]

Answer:

6

Explanation:

The atomic number for phosphorous is 15, meaning that it has 15 electrons (and protons). The first and second shells would be filled up with 2 and 8 electrons respectively, leaving 5 which goes on the third shell, which is also the valence shell, meaning phosphorous has 5 valence electrons.

Since the atomic number of sulfur is 16, the first and second shells are also filled up with 2 and 8 electrons respectively, leaving 6 to be on the third shell, the valence shell. Hence, sulfur has 6 valence electrons.

5 0
1 year ago
Please help!!! What is electrolysis?
Evgen [1.6K]

Explanation:

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5 0
2 years ago
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What is the molarity of the potassium hydroxide if 25.25 mL of KOH is required to neutralize 0.500 g of oxalic acid, H2C2O4? H2C
Greeley [361]

Answer:

0.444 mol/L

Explanation:

First step is to find the number of moles of oxalic acid.

n(oxalic acid) = \frac{0.5g}{90.03 g/mol} = 5.5537*10^{-3} mol\\

Now use the molar ratio to find how many moles of NaOH would be required to neutralize 5.5537*10^{-3} mol\\ of oxalic acid.

n(oxalic acid): n(potassium hydroxide)

         1           :            2                  (we get this from the balanced equation)

5.5537*10^{-3} mol\\ : x

x = 0.0111 mol

Now to calculate what concentration of KOH that would be in 25 mL of water:

c = \frac{number of moles}{volume} = \frac{0.0111}{0.025} = 0.444 mol/L

5 0
3 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
2 years ago
what would be the ph of an aqueous solution of sulphuric acid which is 5×10^_5 mol l^_1 in concentration​
guajiro [1.7K]

Answer:

the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature

6 0
2 years ago
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