All categories

3 years ago

What happens when gasoline is used to power a vehicle?

1 answer:

3 0

The answer should be A. Because the energy in gasoline is called chemical. When burned it is heat, Then to power a vehicle, it is mechanical energy. But I don't know whether the question wants to mean that the energy in the gasoline will not convert totally to the heat, so it will lose. But if think like this, when heat energy transform to mechanical, it will lose again. So I think the answer is A.

You might be interested in

A sample of hydrogen gas will behave most like an ideal gas under the conditions of

elena55 [62]

Answer:

The Answer is gonna be C) high pressure and low temperature

8 0

3 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

Which is not a waste product of cellular respiration heat oxygen carbon dioxide or water?

horsena [70]

Carbon Dioxide is used in celular respiration. All other choices are waste products

7 0

3 years ago

Calculate the energy required to heat 322.0g of ethanol from −2.2°C to 19.6°C . Assume the specific heat capacity of ethanol und

just olya [345]

Answer:

There is 17.1 kJ energy required

Explanation:

Step 1: Data given

Mass of ethanol = 322.0 grams

Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K

Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K

Specific heat capacity = 2.44 J/g*K

Step 2: Calculate energy

Q = m*c*ΔT

⇒ m = the mass of ethanol= 322 grams

⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K

⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K

Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ

There is 17.1 kJ energy required

3 0

4 years ago

Why isn't every point exactly on the 'best fit line'? Give a reason and explain

vampirchik [111]

Answer:

thats all help me too:)...............

5 0

3 years ago