All categories

2 years ago

What’s properties my change during a chemical reaction

Chemistry

1 answer:

sweet-ann [11.9K]2 years ago

7 0

Answer:

The signs of chemical change include changes in color, temperature, the production of light, changes in odor, and the formation of gases.

You might be interested in

Which of the samples pictured is most likely a nonmetal?

max2010maxim [7]

Answer:

you need to be more specific

try adding a picture of the samples

Explanation:

7 0

3 years ago

The chemical breakdown of enormous quantities of organic material buried in the sedimentary rocks has produced gas.

Lelu [443]

The chemical breakdown of enormous quantities of organic material buried in the sedimentary rocks has produced ethane gas.

3 0

3 years ago

Where would you find the element with least mass?

dexar [7]

C. The lowest mass will be the element in period 1, group 1 (this element is hydrogen)

8 0

3 years ago

Hey conner i miss you soooooo much and does anyone have tik tok

kherson [118]

Answer:

I have tik tok

Explanation:

6 0

3 years ago

Read 2 more answers

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

3 years ago