All categories

2 years ago

2H2O2 → 2H2O+O2+196.6 kj.

Chemistry

1 answer:

Mrrafil [7]2 years ago

5 0

2mol of Hydrogen peroxide releases 196.6KJ heat

1mol releases

196.6/2
98.3

5mol releases

5(98.3)
491.5KJ

You might be interested in

The illustration above depicts a spring being compressed and then released. Changes in both potential and kinetic energy occur

Anestetic [448]

Answer:

the potential energy is increasing through steps A & B, then decreases at C.

Explanation:

5 0

3 years ago

Read 2 more answers

A solution is made by adding 360 g of table salt into 1 L of water. The solubility of the salt is 36 g/100 mL water. Which term

Oksanka [162]

To express the given concentration of the solution to the same unit as the solubility of the salt, use dimensional analysis.
(360 g/ 1 L ) x (1 L / 10 (100 mL)) = 36 g / 100 mL
This value is equal to the value of the solubility of the salt which means that the solution is SATURATED.

7 0

4 years ago

Read 2 more answers

Forensic biologist Manning is looking at a cell sample under the microscope. The cell she sees is oval, with a long tail or flag

Gre4nikov [31]

Answer:

Sperm Cell

Explanation:

8 0

3 years ago

EZ POINTS!!!<br>what are 3 physical changes of wood.

VARVARA [1.3K]

Answer: Cutting it, boiling it , freezing it

4 0

3 years ago

Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

4 0

4 years ago