What is an internal cost of driving a domestic car?

Chemistry

1 answer:

Nataly_w [17]2 years ago

3 0

Answer:

The correct option is: Cost of materials used in manufacture

Explanation:

There are two types of costs of an object: the internal costs and the external costs.

The internal cost of an object is the direct monetized cost. It refers to the cost involved in the <u>production or manufacturing of a given objec</u>t. Example: labor, <u>material required</u>, equipment, energy, and overhead expenses.

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Name the type of organic compound from each description of the functional group: (b) group that is not polar and has a double bo

krek1111 [17]

Some of the crucial functional groups in organic compounds are the hydroxyl, methyl, carbonyl, carboxyl, amino, phosphate, and sulfhydryl groups. The formation of molecules including DNA, proteins, carbohydrates, and lipids depends heavily on these groups.

<h3>What is Function Group ?</h3>

In organic chemistry, a functional group is a collection of atoms within molecules that interact to cause predictable reactions. Functional groups include the hydroxyl, ketone, amine, and ether groups, for instance.

(b) Most double bonds, such those found in alkenes, are formed when two carbon atoms are present. In a carbonyl group, for instance, where a carbon atom and an oxygen atom are together, there are several double bonds connecting the two separate components. Sulfoxides (S=O), imines (C=N), and azo compounds (N=N) also include common double bonds.

To know more about Functional Group please click here : brainly.com/question/493841

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8 0

1 year ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$