Answer:
0.5188 M or 0.5188 mol/L
Explanation:
Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.
***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".
The formula for molarity is:
n = moles (unit mol)
V = volume (unit L)
<u>Find the molar mass (M) of potassium hydroxide.</u>
<u>Calculate the moles of potassium hydroxide.</u>
Carry one insignificant figure (shown in brackets).
<u>Convert the volume of water to litres.</u>
Here, carrying an insignificant figure doesn't change the value.
<u>Calculate the concentration.</u>
<= Keep an insignificant figure for rounding
<= Rounded up
<= You use the unit "M" instead of "mol/L"
The concentration of this standard solution is 0.5188 M.
Answer:
Oceans, Fossil fuels, atmosphere
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Answer:
V₂ = 2.96 L
Explanation:
Given data:
Initial volume = 2.00 L
Initial temperature = 250°C
Final volume = ?
Final temperature = 500°C
Solution:
First of all we will convert the temperature into kelvin.
250+273 = 523 k
500+273= 773 k
According to Charles's law,
V∝ T
V = KT
V₁/T₁ = V₂/T₂
V₂ = T₂V₁/T₁
V₂ = 2 L × 773 K / 523 k
V₂ = 1546 L.K / 523 k
V₂ = 2.96 L
