<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
Answer:
This is the balanced equation:
Pb(NO₃)₂ (aq) + 2NaI (aq) → 2NaNO₃ (aq) + PbI₂ (s) ↓
Explanation:
This are the reactants:
PbNO₃
NaI
Iodide can react to Pb²⁺ to make a solid compound.
(B. 3) 172 All nonzero digits are significant.
(A. 4) 450.0 x 10^3 Trailing zeroes after the decimal point are significant.
(A. 4) 3427 All nonzero digits are significant.
(B. 3) 0.0000455 Leading zeroes are not significant.
(B. 3) 0.00456 Leading zeroes are not significant.
(C. 5) 2205.2 Zeroes between nonzero digits are significant.
(C. 5) 107.20 Trailing zeroes after the decimal point are significant.
(B. 3) 0.0473 Leading zeroes are not significant.
Answer:
-2.86x10³ kJ
Explanation:
The enthalpy of a reaction (ΔH) is defined as the heat produced or consumed by a reaction. In the reaction:
2 C₂H₆(g) + 7 O₂(g) → 4 CO₂(g) + 6 H₂O(g)
The ΔH is the heat envolved in the reaction per 2 moles of C₂H₆. 1.43x10³ kJ are involved when 1 mole reacts. Thus, when 2 moles react, involved heat is:
1.43x10³ kJ ₓ 2 = <em>2.86x10³ kJ</em>. As the reaction is a combustion reaction (Produce CO₂ and H₂O), the heat involved in the reaction is <em>PRODUCED, </em>that means ΔH is negative, <em>-2.86x10³ kJ</em>
