Octane on combustion yields CO₂ and H₂O;
2C₈H₁₈<span> + 2 5O</span>₂ → 16 CO₂<span> + 18 H</span>₂<span>O
</span>
According to equation,
2 moles Octane produces = 16 moles of CO₂
So,
498 moles Octane will produce = X moles of CO₂
Solving for X,
X = (498 mol × 16 mol) ÷ 2 mol
X = 3984 moles of CO₂
Now, Calculating for Volume,
As,
P V = n R T
Solving for V,
V = n R T / P ------- (1)
Pressure = P = 0.995 atm
<span>Volume = V = ? </span>
<span>Moles = n = 3984 </span>
<span>Gas Constant = R = 0.0821 L.atm.K</span>⁻¹.<span>mol</span>⁻¹<span> </span>
<span>Temperature = T = 23 + 273 = 296 K
</span>
Putting values in eq. 1
V = (3984 mol × 0.0821 L.atm.mol⁻¹.K⁻¹ × 296 K) ÷ 0.995 atm
Volume = 973040.95 L
A buffer system usually will need a weak acid, like H2CO3, which is carbonic acid, and a strong base, and the only compound that could figure a fairly strong base among the options given is Na2CO3. NH3 and NaHCO3 are weak bases and HF is a strong acid. Therefore the best answer will be letter A
Answer:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
Explanation:
The word equation for the combustion of propane can be obtained from the chemical equation;
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The word equation is therefore:
1 mole of propane combines with 5 moles of oxygen gas to produce 3 mole of carbon dioxide and 4 moles of water.
For such a combustion reaction, carbon dioxide and water are produced in the process.
Answer:
6.1×10^8
Explanation:
The reaction is;
Sn^2+(aq) + Cd(s) -----> Sn(s) + Cd^2+(aq)
E°cell = E°cathode - E°anode
E°cathode= -0.14 V
E°anode= -0.40 V
E°cell = -0.14-(-0.40)
E°cell= -0.14+0.40
E°cell= 0.26 V
But
E°cell= 0.0592/n log K
E°cell= 0.0592/2 log K
0.26= 0.0296log K
log K = 0.26/0.0296
log K= 8.7838
K= Antilog (8.7838)
K= 6.1×10^8
Answer:
Explanation:
Hello,
In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:
That in terms of the heat capacities, masses and temperature changes turns out:
Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:
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