Answer:
Mass = 3732.62 g
Explanation:
Given data:
Number of formula units = 8.53×10²⁴
Mass of sample = ?
Solution:
1 mole of any substance contain 6.022×10²³ formula units.
8.53×10²⁴ formula units ×1 mol / 6.022×10²³ formula units
1.42×10¹ mol
14.2 mol
Mass of magnesium phosphate:
Mass = number of moles × molar mass
Mass = 14.2 mol × 262.86 g/mol
Mass = 3732.62 g
Answer:
about 37 years
Explanation:
If 100% of the estimated reserves are used at the current rate, it will take ...
(1.234·10^12 bbl)/(3.3·10^10 bbl/yr) = 37.4 yr
It will take about 37 years to deplete the reserves.
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Your calculator can perform arithmetic using scientific notation.
The molar mass of the compound:
If the solution has an osmotic pressure of 8.44 torr, then the molar mass of the unknown non-electrolyte is 223.14 g.
What is osmosis?
- Osmosis is defined as the flow of solvent molecules through semi-permeable membrane.
- Osmotic pressure is the pressure applied to stop the flow of solvent molecules.
- It is a colligative property that means osmotic pressure depends on the number of solute particles .
Therefore,
π
( for electrolytes)
Where, π= Osmotic pressure
i = Van 't Hoff factor
n= moles
R= Gaseous constant = 62.363577 L torr 
T= Temperature
V= Volume of solution
Given:
T= 298K
V= 150 mL= 0.150 L
Given mass of unknown electrolyte= 15.2 mg = 15.2 x 
g
Osmotic pressure= 8.44 torr
Molar mass= ?
For non-electrolytes:
πV = n RT
πV=
RT
Calculations:
Putting the given values in the formula:
8.44 x 0.150 =15.2 x / M x 62.36 x 298
1.266 = 282.5/M
M = 282.5/1.266
M = 223.14 g
Therefore,
The molar mass of the unknown non-electrolyte is 223.14g.
Learn more about Osmotic pressure here,
brainly.com/question/13680877
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When Ag₂S dissolves, it dissociates as follows;
Ag₂S ---> 2Ag⁺ + S²⁻
First we need to calculate molar solubility which gives the number of moles dissolved in 1 L of solution.
If molar solubility of Ag₂S is y, then molar solubility of Ag²⁺ and S²⁻ is 2y and y respectively.
ksp gives the solubility constant
ksp = [Ag⁺]²[S²⁻]
ksp = [2y]²[y]
4y³ = 8.00 x 10⁻⁵¹
y³ = 2 x 10⁻⁵¹
y = 1.26 x 10⁻¹⁷ mol/L
molar mass = 247.8 g/mol
solubility of Ag₂S = 1.26 x 10⁻¹⁷ mol/L x 247.8 g/mol = 3.12 x 10⁻¹⁵ g/L
Solubility of Ag₂S = 3.12 x 10⁻¹⁵ g/L
Conjugate base of Propanoic acid (
is propanoate where -COOH group gets converted to -CO
. The structure of conjugate base of Propanoic acid is shown in the diagram.
The 
above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.
= 
+ log![\frac{[Conjugate base]}{[acid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BConjugate%20base%5D%7D%7B%5Bacid%5D%7D)
=4.9+log
=5.85
As 90% conjugate base is present, so propanoic acid present 10%.
