Ask question

Ask question

All categories

3 years ago

10

Natural gas and petroleum are mostly alkanes t/f

2 answers:

dolphi86 [110]3 years ago

5 0

Been a year since i was in chem but my money is on true

Blizzard [7]3 years ago

3 0

Answer: True

Explanation:

Petroleum and natural gas are both fossil fuels. They contain a mixture of hydrocarbons but they commonly have alkanes as their composition. The petroleum and natural gas produce after the degradation and decomposition of the organic matter generated after the dead and decaying plant and animal matter under the heap of sediments for long years. The natural gas specifically have methane, butane and propane. Therefore, on the basis of the above description, this can be said that natural gas and petroleum are mostly alkanes.

You might be interested in

Ñspent fuelî that can no longer be used to create energy.<br> nuclear waste ?

svetoff [14.1K]

Spent fuel that can no longer be used to create energy is waste.

6 0

2 years ago

Give three properties of a Solid

NARA [144]

In a solid, molecules are packed together, and it keeps its shape. Liquids take the shape of the container. Gases spread out to fill the container. Solid is one of the three main states of matter, along with liquid and gas.

Hope that helps!

8 0

2 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

What is the therm energy of an object?

Elan Coil [88]

Answer:

B

Explanation:

B. the total kinetic and potential energies of its particles

3 0

2 years ago

How many grams of lithium hypochlorite (LiClO) are there in 0.594 moles?

Flauer [41]

Answer : The mass of lithium hypochlorite are, 34.7 grams.

Explanation : Given,

Moles of $LiClO$ = 0.594 g

Molar mass of $LiClO$ = 58.4 g/mol

Expression used :

$\text{ Mass of }LiClO=\text{ Moles of }LiClO\times \text{ Molar mass of }LiClO$

Now put all the given values in this expression, we get:

$\text{ Mass of }LiClO=(0.594moles)\times (58.4g/mole)$

$\text{ Mass of }LiClO=34.7g$

Therefore, the mass of lithium hypochlorite are, 34.7 grams.

3 0

3 years ago