Answer:

Explanation:
We are given the compound: Cs₃PO₄
According to the formula, 1 mole of cesium phosphate contains 3 moles of cesium, 1 mole of phosphate, and 4 moles of oxygen.
Therefore, there are 3 moles of cesium for 1 mole of cesium phosphate.

We want to calculate the moles of cesium in 3 moles of cesium phosphate, so we multiply the ratio by 3.


3 moles of cesium phosphate contains <u>9 moles of cesium.</u>
Answer:
True
Explanation:
lithium atoms lose one electron each while chlorine atoms gain one electron each
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
Answer:
1. Yes
2.The solubility of X is 34.55g/L
Explanation:
Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm
or 1 Litre of water.
From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved
= 0.019kg*1000 = 19g.
if 19g is required to saturate 550mL at 30°C,
then
will saturate 1L
= 34.545g will saturate 1Litre
The solubility thus is 34.55g/L