2.1653 g
Explanation:
The molar mass of Rubidium is;
85.468 g/mol
Therefore the moles of Rubidium that reacted with oxygen is;
1.98 / 85.468
= 0.0232 moles
If every two moles of Rubidium reacts with one mole of oxygen then the amount of oxygen consumed in the chemical reaction is;
0.5 * 0.0232
= 0.0116 moles
The molar mass of an oxygen atom is 16 g/mole. Then the amount of O in grams consumed is;
0.0116 * 16
=0.1853 g
The final weight of the Rubidium II Oxide is;
1.98 + 0.1853
= 2.1653 g
Answer:
3.01 L
Explanation:
V
1
: 2.75L
T
1
:
18
∘
C
V
2
: ?
T
2
:
45
∘
C
If you know your gas laws, you have to utilize a certain gas law called Charles' Law:
V
1
T
1
=
V
2
T
2
V
1
is the initial volume,
T
1
is initial temperature,
V
2
is final volume,
T
2
is final temperature.
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:
2.75
291
=
V
2
318
By cross multiplying, we have...
291
V
2
= 874.5
Dividing both sides by 291 to isolate
V
2
, we get...
V
2
= 3.005...
In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.
Therefore,
V
2
= 3.01 L
Answer: There will be a shift toward the products.
The effect of increasing volume will be similar to decreasing pressure, it will shift the equilibrium into the side with higher molecule count. In the reaction above, the product molecule count is 1 (one PCl5) while the product is 2 (one PCl3 + one Cl2). So, increasing volume will increase the rate of product formation.
