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3.04 Quiz: Chemical Reactions 1 to 10 question

vaieri [72.5K]

Answer:

1. Chemical reactions

2.Substances

3. Properties

4. Precipitate

5. Light

6. Temperature

7. Color

8. Gas

Explanation:

During chemical reactions, new substances that possess new properties are formed. During the process of chemical reactions, evidences are found. Some of the evidences are formation of precipitate or light gel, production of gases and changes in color and temperature.

Chemical reactions start with the substances or compounds which are known as reactants. The reactants react to form new substances known as the products which possess new properties.

I only could help with 1 - 8 sorry.

5 0

3 years ago

Janet runs 6.4 miles in one hour. Is it speed or velocity or acceleration

Murrr4er [49]

Answer:

acceleration

Explanation:

6 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.