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3 years ago

Any moving object has _______________ energy A.kinetic B.potential C.gravitational potential

Chemistry

2 answers:

nignag [31]3 years ago

7 0

Answer:

Kinetic

Explanation:

potential means it has the POTENTIAL to move, gravitational potential energy is when an object isn't moving but the it turns into gravitational kinetic energy since gravity causes it to move. since the object in question is already in motion it proves kinetic energy as the answer.

zysi [14]3 years ago

3 0

Any moving object has kinetic energy

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Which two particles in an atom have the same mass ? protons and electrons or protons and neutrons

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When a calcium atom forms an ion, it loses two electrons. What is the electrical charge of the calcium ion?. . A. -1 . B. -2 . C

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Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing

UkoKoshka [18]

Answer: 1.2 moles of carbon dioxide is produced for the given value of oxygen.

Explanation:

The chemical reaction for the combustion of acetylene follows the equation:

$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$

By stoichiometry of the reaction:

5 moles of oxygen produces 4 moles of carbon dioxide.

So, 1.5 moles of oxygen will produce = $\frac{4}{5}\times 1.5=1.2mol$ of carbon dioxide.

Hence, 1.2 moles of carbon dioxide is produced for the given value of oxygen.

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4 years ago

During studies of the reaction below,

Leni [432]

Answer: The percent yield of the nitrogen gas is 11.53 %.

Explanation:

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

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

For NO:

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

$\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol$

For $N_2O_4$ :

Given mass of $N_2O_4$ = 102.1 g

Molar mass of $N_2O_4$ = 92 g/mol

Putting values in equation 1, we get:

$\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol$

For the given chemical reactions:

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$ ......(2)

$N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)$ .......(3)

Calculating the experimental yield of nitrogen gas:

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of $N_2O_4$

So, 0.383 moles of NO will be produced from = $\frac{2}{6}\times 0.383=0.128mol$ of $N_2O_4$

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 0.128 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 0.128=0.384mol$ of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g$

Calculating the theoretical yield of nitrogen gas:

By Stoichiometry of the reaction 2:

1 mole of $N_2O_4$ produces 3 moles of nitrogen gas

So, 1.11 moles of $N_2O_4$ will produce = $\frac{3}{1}\times 1.11=3.33mol$ of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

$3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g$

To calculate the percentage yield of nitrogen gas, we use the equation:

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

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

$\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%$

Hence, the percent yield of the nitrogen gas is 11.53 %.

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