Answer:
-154KJ/mol
Explanation:
mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter
= 0.2 × 100 / 1000 ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.
Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K
where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK
heat gain by calorimeter = (4.184 × 200 + 480) × 2.34 = 3081.3 J
ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol
Total mass of CaCO3 = 40 amu of Ca + 12amu of C + 16×3 amu of oxygen = 100amu of CaCO3
i.e 100 tonnes of CaCO3 .
mass of CO2 = 12amu of C + 2× 16amu of O = 44 amu of CO2
mass % of CO2 in CaCO3 = (44/100)×100 =44%
i.e
44% of 100 tonnes is CO2.
=44 tonnes of CO2.
therefore, 44% of CO2 is present in CaCO3.
Answer:
11482 ppt of Li
Explanation:
The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:
<em>Moles Mg²⁺:</em>
0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA
<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>
0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium
That means mass of lithium is (Molar mass Li=6.941g/mol):
4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:
0.00287g * (1000000μg / g) = 2870μg of Li
As ppt is μg of solute / Liter of solution, ppt of the solution is:
2870μg of Li / 0.250L =
<h3>11482 ppt of Li</h3>
Answer : The grams of
consumed is, 89.6 grams.
Solution : Given,
Mass of
= 265 g
Molar mass of
= 80 g/mole
Molar mass of
= 28 g/mole
First we have to calculate the moles of
.

The given balanced reaction is,

from the reaction, we conclude that
As, 1 mole of
produces from 1 mole of 
So, 3.2 moles of
produces from 3.2 moles of 
Now we have to calculate the mass of 


Therefore, the grams of
consumed is, 89.6 grams.