How can two atoms still bond to become a molecule if there is an anti-bonding orbital?

The first law of thermodynamics is a restatement of the

klasskru [66]

the law of thermodyanamic is the restatement of the law of conservation of energy

4 0

3 years ago

What is the new concentration of a solution of CaSO3 if 10.0 mL of a 2.0 M CaSO3 solution is diluted to 100 ml?

kifflom [539]

Answer: The new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

Explanation:

Given: $V_{1}$ = 10.0 mL, $M_{1}$ = 2.0 M

$V_{2}$ = 100 mL, $M_{2}$ = ?

Formula used to calculate the new concentration is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\10.0 mL \times 2.0 M = M_{2} \times 100 mL\\M_{2} = 0.2 M$

Thus, we can conclude that the new concentration of a solution of $CaSO_{3}$ is 0.2 M 10.0 mL of a 2.0 M $CaSO_{3}$ solution is diluted to 100 mL.

5 0

3 years ago

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

laiz [17]

Answerit is a

Explanation:

8 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.

Adding all three together:

$\text{A} \rightarrow 2\text{C}+\text{E}$ ( $\bf{\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3 }$ )

Thus, the first option is the correct answer.

Supplementary:

To learn more about Hess's Law, do check out: brainly.com/question/26491956

4 0

2 years ago

Add groups to complete polyvinyl alcohol, formed by the polymerization of the following monomer. If a carbon atom contains only

e-lub [12.9K]

Answer:

The resulting structure is:

R-CH-CH2 (OH)-CH-CH2(OH)-CH-CH2(OH)-CH-...

Explanation:

It can be said that vinyl alcohol cannot be polymerized by addition polymerization, since a very unstable radical would be formed, which is not possible. Generally, it is formed by polymerization and subsequent hydrolysis of the vinyl polyacetate. According to the exercise, the structure that results with its four units can be seen that the hydroxyl groups are found in the methylene carbons

8 0

4 years ago