Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
![R\propto \sqrt{\frac{1}{M}}](https://tex.z-dn.net/?f=R%5Cpropto%20%5Csqrt%7B%5Cfrac%7B1%7D%7BM%7D%7D)
or,
..........(1)
where,
= rate of effusion of nitrogen gas = ![79mL/s](https://tex.z-dn.net/?f=79mL%2Fs)
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
![(\frac{79mL/s}{R_2})=\sqrt{\frac{64g/mole}{28g/mole}}](https://tex.z-dn.net/?f=%28%5Cfrac%7B79mL%2Fs%7D%7BR_2%7D%29%3D%5Csqrt%7B%5Cfrac%7B64g%2Fmole%7D%7B28g%2Fmole%7D%7D)
![R_2=52mL/s](https://tex.z-dn.net/?f=R_2%3D52mL%2Fs)
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.
The answer is False. The formation of the moon did not cause the Paleozoic extinction. The moon formed during the Precambrian era.
First, we convert the moles of each substance into the concentration using the volume of the reactor.
[SO₃] = 0.425/1.5 = 0.283 M
[SO₂] = 0.208 / 1.5 = 0.139 M
[O₂] = 0.208/1.5 = 0.139 M
The equilibrium constant is calculated by:
Kc = [SO₃]² / [O₂][SO₂]²
Kc = (0.283)²/(0.139)(0.139)²
Kc = 29.8 = 2.98 x 10¹
The answer is C
Polar will always have the higher boiling point because they have strong van der waal forces
Answer:
<u>ATGGCCTA</u>
Explanation:
For this we have to keep in mind that we have a <u>specific relationship between the nitrogen bases</u>:
-) <u>When we have a T (thymine) we will have a bond with A (adenine) and viceversa</u>.
-) <u>When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa</u>.
Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:
TACCGGAT
<u>ATGGCCTA</u>
<u></u>
I hope it helps!