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3 years ago

What are the eleven elements that are gas freezing?

Chemistry

2 answers:

Leona [35]3 years ago

8 0

Answer:

hydrogen, helium, oxygen, nitrogen, argon and neon, fluorine, chlorine, krypton, xenon and radon

Explanation:

kenny6666 [7]3 years ago

4 0

hydrogen, helium, oxygen, nitrogen, argon, neon, fluorine, chlorine, krypton, xenon and radon i believe >__< not in any order ahaha

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The study of the composition of matter is?

NemiM [27]

Chemistry. More specifically, analytical chemistry.

6 0

3 years ago

If 1495 J of heat is needed to raise the temperature of a 319 g sample of a metal from 55.0°C to 66.0°C, what is the specific he

Olegator [25]

The specific heat capacity of the metal is calculated using the following formula
Q(heat)= MC delta T
Q= 1495 j
c= specific heat capacity =?
M(mass)=319 g
delta T = change in temperature = 66-55 =11 c
by making c the subject of the formula

C=Q /M delta T
c= 1495 j/ 319 g x 11 c =0.426 j/g/c

3 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:

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