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3 years ago

Where do Alkanes come from

Chemistry

1 answer:

bagirrra123 [75]3 years ago

3 0

Answer:

Alkanes naturally occur in crude oil and are a major component of many fuels and solvents derived from petroleum

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PLEASE HURRY THIS IS TIMED!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

AURORKA [14]

Answer:

if i remember correctly i beleive its A 1.8 x 10^24

but im not for sure also i think you forgot the 24

Explanation:

5 0

3 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]

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

Explanation:

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

3.65 gram of hcl is dissolved in 180 gram of water. Find the total number of molecules of hydrogen

Morgarella [4.7K]

Answer:

$Molec_{\ H_{tot}}=1.206x10^{25}molec$

Explanation:

Hello.

In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

$molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH} =6.03x10^{22}molec$

Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:

$molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH} =1.20x10^{25}molec$

Thus, the total number of molecules turns out:

$Molec_{\ H_{tot}}=6.03x10^{22}+1.20x10^{25}\\\\Molec_{\ H_{tot}}=1.206x10^{25}molec$

Regards.

6 0

3 years ago

How do the differences in the polyatomic ions po3 3- and po4 3- help you determine whether each ends in -ite or -ate?

Tcecarenko [31]

There is a rule in naming polyatomic anions containing oxygen. Those names ending in –ate are given to the most common oxyanion of an element. While those names ending in –ite are for those that contains fewer O atom for the same charge. Since P $PO_{3} ^{3-}$ has fewer O atoms, its name ends in –ite. Thus, P $PO_{3} ^{3-}$ is phosphite while P $PO_{4} ^{3-}$ is phosphate.