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3 years ago

At 20% efficiency, how many kWh would actually be produced from the 5 gal of gasoline? How

1 answer:

3 0

Answer:c. At 20% efficiency, how many kWh would actually be produced from the 5 gal of gasoline? How many BTUs of heat would be released from burning the gasoline? (3 points) 1.3 x 10 J x 1 kWH / 3.6 x 10 J = 36.1 kWh 5.2 x 10 J x 1 BTU / 1055 J = 492891 BTU 36.1 kWh would actually be produced from 5 gallons of gasoline.

Explanation:

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A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the

DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × $10^{-46}$ KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × $10^{-18}$ ( $\frac{1}{n^{2}_{final} }$ - $\frac{1}{n^{2}_{initial} }$ )

(1/434 × $10^{-9}$ ) = -2.178 × $10^{-18}$ ( $\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }$ )

⇒ 434 × $10^{-9}$ = (1/-2.178 × $10^{-18}$ ) $\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }$

But, $n_{final}$ = 2

434 × $10^{-9}$ = (1/2.178 × $10^{-18}$ ) $\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }$

434 × $10^{-9}$ × 2.178 × $10^{-18}$ = $(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })$

⇒ $n_{initial}$ = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

= (hc) ÷ λ

= (6.626 × 10−34 × 3.0 × $10^{8}$ ) ÷ (434 × $10^{-9}$ )

= (1.9878 × $10^{-25}$ ) ÷ (434 × $10^{-9}$ )

= 4.58 × $10^{-19}$ J

= 4.58 × $10^{-22}$ KJ

But 1 mole = 6.02× $10^{23}$ , then;

energy in KJ/mole = (4.58 × $10^{-22}$ KJ) ÷ (6.02× $10^{23}$ )

= 7.61 × $10^{-46}$ KJ/mole

7 0

3 years ago

The whiye pigment TiO2 is prepared by the reaction of titanium tetrachloride, TiCl4, with water vapor in the gas phase:

omeli [17]

Answer:

$\boxed{\text{62.1 kJ}}$

Explanation:

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

$\Delta_{\text{r}}H^{\circ} = \sum \Delta_{\text{f}} H^{\circ} (\text{products}) - \sum\Delta_{\text{f}}H^{\circ} (\text{reactants})$

TiCl₄(g) + 2H₂O(g) ⟶ TiO₂(s) + 4HCl(g)

ΔH°f/kJ·mol⁻¹: -763.2 -241.828 -939.7 -92.307

$\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [-939.7 + 4(-92.307)] - [-763.2 + 2(-241.828)\\& = & [-939.7 - 369.228] - [-763.2 - 483.656]\\& = & -1308.928 + 1246.856\\& = & \mathbf{-62.1}\\\end{array}\\\text{The amount of heat evolved is } \boxed{\textbf{62.1 kJ}}$

5 0

4 years ago

Murcury is the only metal at room temperature. Its density is 13.6g/mL. How many grams of murcury will occupy a volume of 95.8mL

Hatshy [7]

Answer:

<h3>The answer is 1.30288 × 10³ g</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume = 95.8mL

density = 13.6g/mL

We have

mass = 13.6 × 95.8 = 1302.88

We have the final answer as

Hope this helps you

3 0

3 years ago

Which one is a true statement about endothermic reactions? A. I need you from an outside source is continuously being added. B.

tekilochka [14]

Correct answer: "A. Energy from an outside source is continuously being added."
An endothermic reaction is a reaction that is characterised by the system absorbing energy from its surroundings. That energy is usually in heat form. For example, when mixing water<span> with potassium chloride, this reaction will absorb heat and the container will feel cold - endothermic reaction.</span>

7 0

4 years ago

Read 2 more answers

A 2.0 L flexible container holds 5.0 moles of oxygen (O2) gas. An additional 15.0 moles of nitrogen gas (N2) is added to the con

frozen [14]

Answer 8.0 L.

2.0L / 5.0 moles = x / 20.0 => x = 20 / 5 * 2 = 8

4 0

3 years ago

Read 2 more answers