Answer:
<h3>The man's average body temp. will fall by 0.6°C to 38.4°C</h3>
Explanation:
The enthalpy (heat content) of the water, using a datum of 0°C, is
Hw = Mw kg x Cp,w (specific heat capacity kJ/kg °C) x Tw °C
= 1 kg x 4.18 kJ/kg°C x 3°C = 12.5 kJ
Hman (pre drink) = 68 kg x 3.6 kJ/kg/°C x 39°C = 9547 kJ
So Hman (post drink) = H (pre drink) + Hw = 9547 + 12.5 = 9559.5 kJ because no heat is lost immediately.
But Hman (after drink) has mass 68 + 1 = 69 kg
Also his new Cp will be approx (3.6 x 68/69) + (4.18 x 1/69) = 3.608 kJ/kg°C
So Hman = 9559.5 kJ (from above) = 69 kg x 3.608 kJ/kg°C x Tnew
Therefore the man's new overall temp. = 38.4°C which is a drop of 0.6°C
Volume of O2 : 168 L
<h3>Further explanation</h3>
Given
Reaction
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
5 moles of KCIO3
Required
volume of O2
Solution
From the equation, mol O2 :
= 3/2 x moles KClO3
= 3/2 x 5 moles
= 7.5 moles
Assumed at STP( 1 mol = 22.4 L) :
= 7.5 x 22.4 L
= 168 L
<span>Answer is: 2940 mL of
the HCL solution.</span>
c₁(HCl) = 10.0 M.
V₂(AgNO₃<span>) = ?.
c</span>₂(AgNO₃<span>) = 0.85 M.
V</span>₁(AgNO₃<span>) = 250 mL </span>÷ 1000 mL/L = 0.25 L.
<span>
c</span>₁<span> - original concentration of the solution, before it
gets diluted.
c</span>₂<span> - final concentration of the solution, after dilution.
V</span>₁<span> - volume to be diluted.
V</span>₂<span> - final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
V</span>₂(HCl) = c₁ · V₁ ÷ c₂.
<span>
V</span>₂(HCl) = 10 M · 0.25 L ÷ 0.85 M.
<span>
V</span>₂(HCl) = 2.94 L ·
1000 mL = 2940 mL.
Answer: 0.22 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
moles of 
Thus moles of 
= 0.011
According to stoichiometry:
2 moles of 
require 1 mole of 
Thus 0.011 moles of require=
moles of 
Thus Molarity of 
Therefore, the molarity of in the solution is 0.22 M
Number of moles ( n ) = ?
Volume ( V ) = 0.25 L
Molarity of solution ( M ) = 2.00 M
n = M x V
n = 2.00 x 0.25
n = 0.5 moles of Ba(NO₃)₂
hope this helps!
