1. Which change of state is accompanied by the removal of heat energy

Draw the product formed when cyclohexene is reacted with H2 in the presence of Pt. Note: If adding hydrogen atoms to a carbon at

tigry1 [53]

Answer:

It has been drawn and uploaded as an attachment. Please download it to see the structure.

Explanation:

The product formed as a result of the reaction of cyclohexene with H2 in presence of Pt (platinum) can be described as catalytic hydrogenation. Catalytic hydrogenation is defined as the process of hydrogen addition in the presence of a catalyst, which in this case is platinum.

Note that Cyclohexene (alkene) is a hydrocarbon molecule represented by the chemical formula, C6H10 .

It consists of a double bond. During the hydrogenation reaction, the alkene undergoes an addition reaction to give alkane which is a saturated hydrocarbon as the product.

The first step in order to derive the product is to draw the chemical structure of cyclohexene and identify the double bond present in it.

The final product can be derived by replacing the double bond with the single bond and satisfying all the valences of the carbon atom. The final product structure has been drawn and uploaded as an attachment. Please download it to see the structure.

Ans:

The structure of the cyclohexane thus, formed has been shown as follows with all the hydrogen atoms:

3 0

2 years ago

For the following systems at equilibrium C: CaCO3(s) ⇌ CaO(s)+CO2(g) ΔH=+178 kJ/mol D: PCl3(g)+Cl2(g) ⇌ PCl5(g) ΔH=−88 kJ/mol cl

Rama09 [41]

Explanation:

C: $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$ ΔH=+178 kJ/mol

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

$A+\text{heat}\rightleftharpoons B$

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a reactant and on increasing a reactant at equilibrium, shifts the reaction in the forward direction.

Increase temperature → increase in heat → forward direction

Decrease temperature → decease in heat → backward direction

System C - Increase temperature : Reaction will move forward

System C - Decrease temperature : Reaction will move backward

D: $PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$ ΔH=−88 kJ/mol

The total enthalpy of the reaction comes out to be negative .

The temperature of the surrounding will increase.

For an exothermic reaction, heat is released during a chemical reaction and is written on the product side.

$A\rightleftharpoons B+\text{ heat}$

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a product and on increasing a product at equilibrium, shifts the reaction in the backward direction.

Increase temperature → increase in heat → backward direction

Decrease temperature → decease in heat → forward direction

System D - Increase temperature : Reaction will move backward

System D - Decrease temperature : Reaction will move forward

7 0

2 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0

2 years ago

Read 2 more answers

________ properties depend on the amount of material present.

Kamila [148]

Answer: Extensive!

Explanation:

6 0

2 years ago

4. Suppose you want to deposit silver on an iron spoon using silver nitrate as

Levart [38]

The spoon should be connected to the negative terminal of the battery. The other electrode should be made of silver.

8 0

2 years ago

1. Which change of state is accompanied by the removal of heat energy​

1. Which change of state is accompanied by the removal of heat energy