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Y_Kistochka [10]
3 years ago
10

Scattered. The particles. are not

Chemistry
1 answer:
exis [7]3 years ago
5 0

the change in the direction of motion of a particle because of a collision with another particle. As defined in physics, a collision can occur between particles that repel one another, such as two positive (or negative) ions, and need not involve direct physical contact of the particles

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How many molecules of water of water are in a 821.3 g sample
svetlana [45]

Answer: 2.744 x 1025 molecules

6 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
A sample of an unknown gas effuses in 14.4 min. An equal volume of H2 in the same apparatus under the same conditions effuses in
Elena-2011 [213]

Answer:- molar mass of the unknown gas is 71.5 gram per mol.

Solution:- From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.

When we compare the effusion rates of two gases then the formula for Graham's law is:

\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}

In this formula, V stands for volume and M stands for molar mass

Rate is volume effused per unit time. Since, the volumes are same, the formula could be written as:

\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}

let's say in formula, subscript 1 is for hydrogen gas and 2 is for the unknown gas.

Molar mass of hydrogen is 2.02 grams per mol and the time taken to effuse it is 2.42 min. The time taken to effuse the unknown gas is 14.4 min and we are asked to calculate it's molar mass. let's plug in the values in the formula:

\frac{14.4}{2.42}=\sqrt{\frac{M_2}{2.02}}

5.95=\sqrt{\frac{M_2}{2.02}}

doing squares to both sides:

35.4=\frac{M_2}{2.02}

M_2=35.4*2.02

M_2=71.5

So, the molar mass of the unknown gas is 71.5 grams per mol.



4 0
3 years ago
The temperature in Jersey City, NJ in mid-Spring is 60*F. The temperature in Miami, FL is 90*F on the same day. Miami is about 1
timurjin [86]

Answer:

Temperature gradient = 30

90F - 60F = 30F

The temperature gradient is 30F.

If I am right, let me know.

7 0
3 years ago
Read 2 more answers
Cyanogen is a gas which contains 46.2% C and 53.8% N by mass. At a temperature of 25°C and a pressure of 750 mm Hg, 1.50 g of cy
sergiy2304 [10]

Answer:

Molecular formula of cyanogen is C₂N₂

Explanation:

We apply the ideal gases law to find out the mole of cyanogen

P . V =  n. R. T

Firstly let's convert the pressure in atm, for R

750 mmHg = 0.986 atm

25°C + 273 = 298K

0.986 atm . 0.714L = n . 0.082 L.atm/mol.K .298K

(0.986 atm . 0.714L) / (0.082 L.atm/mol.K .298K) = n

0.0288 mol = n

Molar mass of cyanogen = mass / mol

1.50 g /0.0288 mol = 52.02 g/m

Let's apply the percent, to know the quantity of atoms

100 g of compound contain 46.2 g of C and 53.8 g of N

52.02 g of compound contain:

(52.02 . 46.2) / 100 = 24 g  → 2 atoms of C

(52.02 . 53.8) / 100 = .28 g  →  2 atoms of N

3 0
3 years ago
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