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
Ca(OH)2 (aq) because of elements oxygen and hydrogen
Ca(OH)2(aq)+H2SO4(aq)—→ CaSO4(s)+ 2H2O(l)
Answer:
3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.
Explanation:
The molecular mass of P2O5 is
P2 = 2* 31 = 62
O5 = 5 *<u> 16 = 80</u>
Molecular Mass = 142
Set up a Proportion
142 grams P2O5 supplies 62 grams of phosphorus
x kg P2O5 supplies 1.69 kg of phosphorus
Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.
142/x = 62/1.69 Cross multiply
142 * 1.69 = 62x combine the left
239.98 = 62x Divide by 62
239.98/62 = x
3.89 kg of P2O5 must be used.
Water. Arrhenius acids provide a H+ ion and a base provides a OH- ion, which react to form water.
