What are the 4 properties of hydrogen bonds?

What source of thermal energy does an internal combustion engine use?

Dafna1 [17]

Answer:

D

Explanation:

The source of the thermal energy in a combustion engine is the spark from the spark plug that ignites the fuel and air mixture converting chemical energy into thermal energy.

5 0

3 years ago

Read 2 more answers

Please help

professor190 [17]

Answer:

I don't know if this is right but try it. The amount of water vapor in the air is called absolute humidity. The amount of water vapor in the air as compared with the amount of water that the air could hold is called relative humidity. This amount of space in air that can hold water changes depending on the temperature and pressure.

6 0

3 years ago

Which chemical formula represents both an element and a molecule?

victus00 [196]

Answer:

Explanation:

element - a basic substance that can't be simplified (hydrogen, oxygen, gold, etc...) molecule - two or more atoms that are chemically joined together (H2, O2, H2O, C6H12O6, etc...) compound - a substance that contains more than one element (H2O, C6H12O6, etc...)

5 0

2 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago

Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.

Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

$\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%$

$\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%$

$\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%$

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

3 0

3 years ago