Answer:
Explanation:
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Explanation:
Normal moles of
= volume × normal concentration
= 4.7 × 0.139 = 0.6533 mol
Moles of
in hyponatremia blood = volume × hyponatremia concentration
= 4.7 × 0.116 = 0.5452 mol
Moles of NaCl to be added = moles of extra
needed
= 0.6533 mol - 0.5452 mol = 0.1081 mol
Mass of NaCl = moles × molar mass of NaCl
= 0.1081 mol × 58.443
= 6.317g
= 6.32 g (approx)
Thus, we can conclude that mass of sodium chloride would need to be added to the blood is 6.32 g.
½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS