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noname [10]
3 years ago
15

The purpose of a car engine is to transform the chemical energy of gasoline into kinetic energy of the car in motion. Gasoline

is burned in the engine to create that movement. However, gasoline engines are typically only about 20% efficient. What happens to the rest of the energy released from the burning gasoline?
Chemistry
1 answer:
WARRIOR [948]3 years ago
5 0

Answer:

heat

Explanation:

most of the gasoline's energy is released as heat energy rather than kinetic energy on the pistons. this is why a car engine must constantly be cooled by the water to air heat exchanger we call a radiator

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goblinko [34]
C. Digesting a sandwich


Is your answer

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3 0
3 years ago
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What is the IMA of the lever pictured?<br><br><br><br> 0.18<br> 0.20<br> 0.90<br> 5
IceJOKER [234]
<span>The ideal mechanical advantage represents the number of times the input force is multiplied under ideal conditions, that is with no friction. Actual mechanical advantage on the other hand stands for the number of times the input force is multiplied. 
Hence; IMA (ideal mechanical advantage)=Le/Lr
The Lr =0.3 +1.2 = 1.5 and Le= 0.3
       = 0.3/1.5 = 1/5;
therefore the correct answer is 0.2</span>
6 0
3 years ago
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Please help me with number 43
Lana71 [14]

Answer:

The reason why atomic mass is usually not a whole number is because it is a weighted average of the mass numbers of isotopes

Explanation:

8 0
3 years ago
How many sulfur atoms are generated when 9.42 moles of H2S react according to the following equation: 2H2S+SO2→3S+2H2O
8_murik_8 [283]

Answer:

A) 8.51 × 10²⁴  

Explanation:

1. Gather all the information

            2H₂S + SO₂ ⟶ 3S + 2H₂O

n/mol:   9.42

2. Calculate the moles of S atoms

The molar ratio is 3 mol S:2 mol H₂S

\text{Moles of S} = \text{9.42 mol H$_{2}$S} \times \dfrac{\text{3 mol S }}{\text{2 mol H$_{2}$S }} = \text{14.13 mol S}

3. Calculate the atoms of S

\text{Atoms of S } = \text{14.13 mol S} \times \dfrac{6.022 \times 10^{23}\text{ S atoms}}{\text{1 mol S}} = \mathbf{8.51 \times 10^{24}}\textbf{ S atoms}

 

6 0
3 years ago
Read 2 more answers
If 20.0 g of NaOH is added to 0.750 L of 1.00 M Cd(NO₃)₂, how many grams of Cd(OH)₂ will be formed in the following precipitatio
bulgar [2K]

Answer:

m_{Cd(OH)_2}=36.6 gCd(OH)_2

Explanation:

Hello.

In this case, for the given chemical reaction, in order to compute the grams of cadmium hydroxide that would be yielded, we must first identify the limiting reactant by computing the yielded moles of that same product, by 20.0 grams of NaOH (molar mass = 40 g/mol) and by 0.750 L of the 1.00-M solution of cadmium nitrate as shown below considering the 1:2:1 mole ratios respectively:

n_{Cd(OH)_2}^{by\ NaOH}=20.0gNaOH*\frac{1molNaOH}{40gNaOH} *\frac{1molCd(OH)_2}{2molNaOH} =0.25molCd(OH)_2\\\\n_{Cd(OH)_2}^{by\ Cd(NO_3)_2}=0.750L*1.00\frac{molCd(NO_3)_2}{L}*\frac{1molCd(OH)_2}{1molCd(NO_3)_2}  =0.75molCd(OH)_2

Thus, since 20.0 grams of NaOH yielded less of moles of cadmium hydroxide, NaOH is the limiting reactant, therefore the mass of cadmium hydroxide (molar mass = 146.4 g/mol) is:

m_{Cd(OH)_2}=0.25molCd(OH)_2*\frac{146.4gCd(OH)_2}{1molCd(OH)_2} \\\\m_{Cd(OH)_2}=36.6 gCd(OH)_2

Best regards.

4 0
3 years ago
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