Answer:
38 canisters
Explanation:
The combustion of propane can be represented by the equation below:
C₃H₈ (g) + O₂ (g) → CO₂ (g) + H₂O(g) ΔHc = -103.85 kJ/mol
The boiling of water is represented below:
H₂O(l) → H₂O(g) ΔH = Q = mcΔT
The heat necessary to make water reach the boiling point is Q, which is calculated with the mass of water (m = 4.0 kg), its specif heat capacity ( c = 4.200 J/kg°C) and temperature variation (ΔT = 100 - 25)
Therefore, ΔH = Q = 4.0 kg x 4.200 kJ/kg°C x 75 °C
ΔH = Q = 1260.0 kJ
This is the amount of heat necessary to heat the water every day. Since the expedition will last 7 days, the total heat will be 8820 kJ
Now we need to calculate the amount of propane needed to generate this heat:
1 mol C₃H₈ _______ 103.85 kJ
X _______ 8820 kJ
x = 84.9 mol C₃H₈
1 mol C₃H₈ _____ 44 g
84.9 mol C₃H₈ ___ x
x = 3735.6 g
1 canister _____ 100 g C₃H₈
x _____ 3735.6 g
x = 37.3 canisters
The minimum number of fuel canisters I must bring is 38 canisters.
Options:
- NaOH(aq)
- NaOH(l)
- N2(g)
- NO(g)
Answer: <em>NaOH (aq)</em>
Explanation:
Defining state terminologies, '<em>aq' </em>means aqueous, meaning the analyte is in solution, as opposed to gaseous (g), or liquid (l).
Answer:
Explanation:
We are given the specific heat and change in temperature, so we should use this heat formula:
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
We know the mass is 150 grams. The specific heat of water is 4.184 J/g °C.
Let's find the change in temperature.
Subtract the initial temperature from the final temperature.
- ΔT= final temp - initial temp
- final= 95.0 °C and initial= 10.0 °C
- ΔT= 95.0 °C - 10.0 °C= 85.0 °C
Now we know all the values:
Substitute them into the formula.
Multiply all three numbers together. Note that the grams (g) and degrees Celsius (°C) will cancel out. Joules (J) will be the only remaining unit.
<u>53,346 Joules</u> of heat are required.
Answer:
It's probably to make the chemical symbols look smarter to other people.
Explanation:
Probably this is true.