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Answer:
Less than 7 so it is going to be acidic.
Explanation:
A buffer solution is one that resists changes in pH when small quantities of an acid or an alkali are added to it. An acidic buffer solution is simply one which has a pH of less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt.
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.
<span>83.9% is the weight percent of Al(C6H5)3 in the original 1.25 g sample.
First, look up the atomic weights of all elements involved.
Atomic weight of Aluminum = 26.981539
Atomic weight of Carbon = 12.0107
Atomic weight of Chlorine = 35.453
Atomic weight of Hydrogen = 1.00794
Now calculate the molar mass of Al(C6H5)3, and C6H6
Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 15.999
= 258.293239 g/mol
Molar mass of C6H6 = 6 * 12.0107 + 6 * 1.00794
= 78.11184 g/mol
Determine how many moles of C6H6 was produced
0.951 g / 78.11184 g/mol = 0.012174851 mol
Since the balanced formula indicates that 3 moles of C6H6 is produced for each mole of Al(C6H5)3 used, divide by 3 to get the number of moles of Al(C6H5)3 that was present.
0.012174851 mol / 3 = 0.004058284 mol
Now multiply by the molar mass of Al(C6H5)3 to get the mass of Al(C6H5)3 originally present.
0.004058284 mol * 258.293239 g/mol = 1.048227218 g
Finally, divide the mass of Al(C6H5)3 by the total mass of the original sample to get the weight percentage.
1.048227218 g / 1.25 g = 0.838582
Since all our measurements had 3 significant figures, round the result to 3 significant figures, giving 0.839 = 83.9%</span>
