What mass of lp gas is necessary to heat 1.7 l of water from room temperature (25.0 ∘c) to boiling (100.0 ∘c)? assume that during heating, 16% of the heat emitted by the lp gas combustion goes to heat the water. the rest is lost as heat to the surroundings.

72.25 g

Explanation:

mass of 1.7 L water = 1.7 x 10⁻³ x 10³ kg

= 1.7 kg

heat required to raise its temperature from 25 degree to 100 degree

= mass x specific heat x rise in temperature

= 1.7 x 4.18 x 10³ x 75 J

= 532.95 kJ

Now calorific value of LP gas = 46.1 x 10⁶J / kg

Let required mass of LP gas be m kg

heat evolved from this amount of gas

=  46.1 x 10⁶ m

Heat utilized in warming water  

= 46.1 x 10⁶ m  x .16 J

So

46.1 x 10⁶ m  x .16  =  532.95 x 10³

7376 x 10³ m = 532.95 x 10³

m = 532.95 / 7376 kg

= 72.25 g

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz