Answer:
One gallon of octane produces approximately 7000 L of carbon dioxide.
Note:
I believe that the mass of octane should have been given as 2661 g. However, I understand that your instructor probably gave you this problem, so I will use 4000 g for the approximate mass of one gallon of octane. You can rework the problem on your own, substituting the correct masses of octane if you wish.
Step1. You must first determine the number of moles that are in 4000 g of octane, using the molar mass of octane. Step 2. Then you must determine the number of moles of carbon dioxide that can be produced by that number of moles of octane, based on the mole ratio between octane and carbon dioxide in the balanced equation. Step 3. Then use the ideal gas law to determine the volume in liters of carbon dioxide that can be formed.
pH of the buffer solution is 1.76.
Chemical dissociation of formic acid in the water:
HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq)
The solution of formic acid and formate ions is a buffer.
[HCOO⁻] = 0.015 M; equilibrium concentration of formate ions
[HCOOH] + [HCOO⁻] = 1.45 M; sum of concentration of formic acid and formate
[HCOOH] = 1.45 M - 0.015 M
[HCOOH] = 1.435 M; equilibrium concentration of formic acid
pKa = -logKa
pKa = -log 1.8×10⁻⁴ M
pKa = 3.74
Henderson–Hasselbalch equation: pH = pKa + log(cs/ck)
pH = 3.74 + log (0.015 M/1.435 M)
pH = 3.74 - 1.98
pH = 1.76
More about buffer: brainly.com/question/4177791
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Tap water and rain water are both homogeneous, even though they may have different levels of dissolved minerals and gases. A bottle of alcohol is a man-made homogeneous mixture, from a fine Italian wine to a glass of Scotch whisky. In the human body, blood plasma is an example of a homogeneous mixture.
First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
Explanation:
As the given data is as follows.
Height, H = 150 feet
Heat gain = 30,000 BTU/hr, and Heat loss = 25000 BTU/hr
m = mass of water heated = 700 gallons = 5810 lbs
is the heat capacity of water = 1 BTU/lb
(given)
= temperature difference = ![120^{o}F - 35^{o}F](https://tex.z-dn.net/?f=120%5E%7Bo%7DF%20-%2035%5E%7Bo%7DF)
Heat energy required to heat 700 gal can be calculated as follows:
Heat Required = ![5810 lbs \times 1 BTU/lb^{o}F \times (120^{o}F - 35^{o}F)](https://tex.z-dn.net/?f=5810%20lbs%20%5Ctimes%201%20BTU%2Flb%5E%7Bo%7DF%20%5Ctimes%20%28120%5E%7Bo%7DF%20-%2035%5E%7Bo%7DF%29)
Thus, water rises till
.