Answer:
Percent loss of water = 25%
Explanation:
Given data:
Mass of hydrated salt = 15.6 g
Mass of anhydrous salt = 11.7 g
Percentage of water lost = ?
Solution:
First of all we will calculate the mass of water in hydrated salt.
Mass of water = Mass of hydrated salt - Mass of anhydrous salt
Mass of water = 15.6 g - 11.7 g
Mass of water = 3.9 g
Now we will calculate the percentage.
Percent loss of water = mass of water / total mass × 100
Percent loss of water = 3.9 g/ 15.6 g × 100
Percent loss of water = 25%
(~26grams/mole) and Avogadros # (6.022x10^23) 84.3grams x 1mole/26grams x 6.022x10^23 molecules/mole = 1.95x10^24 molecules of C2H2
Answer:
The molecular formula of the compound is
.
Explanation:
Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:
Carbon


Hydrogen


Oxygen


Now, the number of moles (
), measured in moles, of each element are calculated by the following expression:

Where:
- Mass of the element, measured in grams.
- Molar mass of the element, measured in grams per mol.
Carbon (
,
)


Hydrogen (
,
)


Oxygen (
,
)


For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:

Answer : The pH of 0.289 M solution of lithium acetate at
is 9.1
Explanation :
First we have to calculate the value of
.
As we know that,

where,
= dissociation constant of an acid = 
= dissociation constant of a base = ?
= dissociation constant of water = 
Now put all the given values in the above expression, we get the dissociation constant of a base.


Now we have to calculate the concentration of hydroxide ion.
Formula used :
![[OH^-]=(K_b\times C)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%28K_b%5Ctimes%20C%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
where,
C is the concentration of solution.
Now put all the given values in this formula, we get:
![[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%285.5%5Ctimes%2010%5E%7B-10%7D%5Ctimes%200.289%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
![[OH^-]=1.3\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.3%5Ctimes%2010%5E%7B-5%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)


Now we have to calculate the pH.

Therefore, the pH of 0.289 M solution of lithium acetate at
is 9.1