_Sn+ _ P4 _Sn3 P4<br>balancing equations

How many carbon atoms are there in .500 mol of CO2?

AURORKA [14]

Answer: There are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

Explanation:

According to the mole concept, there are $6.022 \times 10^{23}$ atoms present in 1 mole of a substance.

In a molecule of $CO_{2}$ there is only one carbon atom present. Therefore, number of carbon atoms present in 0.500 mol of $CO_{2}$ are as follows.

$1 \times 0.500 \times 6.022 \times 10^{23}\\= 3.011 \times 10^{23}$

Thus, we can conclude that there are $3.011 \times 10^{23}$ atoms present in 0.500 mol of $CO_{2}$ .

6 0

3 years ago

In a chemical reaction, the difference between the potential energy of the products of the potential energy of the reactants is

natka813 [3]

In a chemical reaction, the difference between the potential energy of the products and the potential energy of the reactants is equal to the heat of the reaction<span>. This is, the net energy released or absorbed (change) during a chemical reaction is the sum of the potential energy of the products less the sum of the potential energy of the reactants.</span>

3 0

4 years ago

For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)

aev [14]

Answer:

ΔS° = -268.13 J/K

Explanation:

Let's consider the following balanced equation.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:

ΔS° = ∑np.Sp° - ∑nr.Sr°

where,

ni are the moles of reactants and products

Si are the standard molar entropies of reactants and products

ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]

ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]

ΔS° = -268.13 J/K

7 0

3 years ago

Compound a is an alkene that was treated with ozone (followed by dms) to yield only 4-heptanone. Draw the major product that is

olga nikolaevna [1]

Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.

Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.

The reaction is as shown in the image.

The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid $H_3O^{+}$ is shown in the image.