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katen-ka-za [31]
3 years ago
11

We know that the law of conservation of energy states that energy can not be created or destroyed. It only changes form. Conside

r this energy transformation diagram. Assume that the energy input in this case is chemical potential energy and the useful energy output is kinetic energy, specifically a small fan to keep you cool. The diagram also shows how to calculate the efficiency of the fan. If the law of conservation of energy holds true, explain why the efficiency of the fan is not 100%.
A) No machine is ever 100% efficient; some energy is not transformed.
B) Some of the chemical potential energy is destroyed during the energy transformation.
C) Some of the chemical potential energy is transformed into heat energy as well as kinetic energy.
D) Conservation of energy also implies that some of the chemical potential energy is conserved or not used to run the fan.
Chemistry
2 answers:
White raven [17]3 years ago
7 0

Answer:

C) Some of the chemical potential energy is transformed into heat energy as well as kinetic energy

Explanation:

Just did it on USA Test Prep

mario62 [17]3 years ago
4 0

I'm not sure but I think the answer is C or D.

I think A is not possible because where did the energy go if it's not transformed?

B is definitely not because energy cannot be destroyed.

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Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride
IrinaK [193]

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of hydrogen chloride produced in the reaction will be 4.4 mL

<u>Explanation:</u>

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

\frac{V_1}{n_1}=\frac{V_2}{n_2}

where,

V_1\text{ and }n_1 are the volume and number of moles of methane gas

V_2\text{ and }n_2 are the volume and number of moles of hydrogen chloride

We are given:

V_1=1.1mL\\n_1=1mol\\V_2=?L\\n_2=4mol

Putting values in above equation, we get:

\frac{1.1}{1}=\frac{V_2}{4}\\\\V_2=\frac{1.1\times 4}{1}=4.4mL

Hence, the volume of hydrogen chloride produced in the reaction will be 4.4 mL

5 0
3 years ago
The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
natta225 [31]

Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

μ = \dfrac{1 \times 35}{1 + 35}

μ = \dfrac{35}{36}

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ  = \\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \  kg

μ  = 1.6139 × 10⁻²⁷ kg

r_o = 127 \ pm = 127*10^{-12} \ m

The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

where ;

J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

E_3= 2.5665 \times 10^{-21} \ J

We know that :

1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

8 0
3 years ago
Ca(s)+2hno3(aq)→ca(no3)2(aq)+h2(g) identify the oxidizing agent.
aleksandr82 [10.1K]

To find for the oxidizing agent, first let us write the half reactions of this complete chemical reaction:

Ca = Ca2+ + 2e- <span>
2 H+ + 2e- = H2</span>

 

The oxidizing agent would be the substance of the element that is reduced. We know that an element is reduced when an electron is added to it. In this case, the element being reduced is H. Therefore the oxidizing agent is HNO3.

 

Answer:

<span>HNO3</span>

8 0
3 years ago
Read 2 more answers
What is the bond between Ca and Cl
gtnhenbr [62]

ionic bond is formed between ca and cl forming molecule cacl2 ca has 2 velancy and cl has one velancy (ca has 2 electrons in its outer most shell while cl has 1 electron vecancy in its outermost shell). So ca would make bond with 2 cl atoms

4 0
3 years ago
Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c
LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass  of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H =  4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:  

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%  

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0
3 years ago
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