Answer:
Mg²⁵ = 10.00%
Mg²⁶ = 45.04%
Mg²⁴ = 44.96%
Explanation:
Given data:
Atomic mass of Mg²⁶ = 25.983
Atomic mass of Mg²⁵ = 24.986
Atomic mass of Mg²⁴ = 23.985
Abundance of Mg²⁵ = 10.00%
Abundance of Mg²⁶ = ?
Abundance of Mg²⁴ = ?
Solution:
Average atomic weight of Mg = 25.983 + 24.986+ 23.985 / 3
Average atomic weight of Mg = 74.954/3
Average atomic weight of Mg = 24.985 amu
Abundance of
Mg²⁵ = 10.00
Mg²⁶ = x
Mg²⁴ = 100- 10 - x = 90 -x
Formula:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
24.985 = (0.1×24.986)+(90-x×23.985) + ( x ×25.983 ) /100
24.985 = 249.86 + 2158.65 - 23.985x + 25.983x / 100
24.985 = 2408.51 + 1.998 x / 100
2498.5 = 2408.51 + 1.998 x
1.998 x = 2498.5 - 2408.51
1.998 x = 89.99
x = 89.99 /1.998
x = 45.04
Now we put the value of x:
Mg²⁵ = 10.00
Mg²⁶ = x (45.04)
Mg²⁴ = 90 -x (90 - 45.04 = 44.96)
Answer:
The ΔH of the reaction is + 12.45 KJ/mol
Explanation:
Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)
heat capacity of water = 4.18 Jk-1 Mol-1
Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)
Molar mass of NaHCO3 = 84 g/mol
Mole of NaHCO3 = 14.5 / 84 = 0.173 mol
Step 1 : Calculate the heat energy (Q) lost by the water.
Q = M x C x ΔT
Q = -100 x 4.18 x (-5.14)
Q = 2148.5 joules
Q = 2.1485 K J
Step 2: Calculating the ΔH of the reaction?
ΔH = Q / number of moles of NaHCO3
ΔH = 2.1485 / 0.173
ΔH = 12.42 KJ/mol
Answer:
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Explanation:
Step 1: Data given
Mass of sodium bicarbonate = 2.7 grams
Step 2: The balanced equation
HCl + NaHCO3 ⇔ NaCl + H2O + CO2
Step 3: Calculate moles NaHCO3
moles NaHCO3 =2.7 g / 84 g/mol= 0.032 moles
Step 4: Calculate moles HCl
For 1 mol NaHCO3 we need 1 mol HCl
For 0.032 moles NaHCO3 = 0.032 moles HCl
Step 5: Calculate mass HCl
Mass HCl = moles HCl * molar mass HCl
mass HCl = 0.032 * 36.46 g/mol= 1.17 grams
1.17 grams of HCl can neutralize 2.7 grams sodium bicarbonate
Answer:
Exosphere
Explanation:
it is found at the end reaching outer space
Answer:
Molarity of NaOH = 0.025 M
Explanation:
Given data:
Molarity of HCl = C₁ = 0.05 M
Volume of HCl = V₁= 50 mL
Molarity of NaOH = C₂=?
Volume of NaOH =V₂= 100 mL
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Molarity of HCl
V₁ = Volume of HCl
C₂ = Molarity of NaOH
V₂ = Volume of NaOH
Now we will put the values:
C₁V₁ = C₂V₂
0.05 M × 50 mL = C₂ × 100 mL
2.5 M.mL =C₂ × 100 mL
C₂ = 2.5 M.mL /100 mL
C₂ = 0.025 M
