Answer:
• 1.62432 moles of nitrogen
• Tire Pressure: 2.74 * 10⁵ Pa
• The tires will burst
• Pressure: 244 kPa
Explanation:
• We can determine the number of moles of nitrogen using the formula pV = nRT, where p = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature.
Now remember we have our initial pressure in kilopascals so let's convert to pascals (249 pascals). The volume is given in liters, so let's convert into m². And the initial temperature is given in Celsius ⇒ our absolute temperature in Kelvins.
Respectively the moles of nitrogen in each tire should be:
• We can solve this part similarly. All our values will be the same, besides the temperature, as we have to consider both the initial and final temperature here.
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• The text mentions that the tires will burst when the internal pressure reaches 269kP. From part #2 we know that the final pressure will be, in kilopascals, 274kP. As 274 > 269, the tires will burst in Death Valley.
• We would want the final temperature = breaking pressure. Therefore,
Please add a picture of what numbers ; )
The Balanced Chemical Equation is as follow;
4 KO₂ + 2 CO₂ → 2 K₂CO₃ + 3 O₂
First find out the Limiting Reagent,
According to equation,
284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
So,
27.9 g of KO₂ will react with = X L of CO₂
Solving for X,
X = (44.8 L × 27.9 g) ÷ 284 g
X = 4.40 L of CO₂
Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,
According to eq.
284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
27.9 g of KO₂ will form = X g of K₂CO₃
Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g
X = 13.57 g of K₂CO₃
So, 13.57 g of K₂CO₃ formed is the theoretical yield.
%age Yield = 13.57 / 21.8 × 100
%age Yield = 62.24 %
Answer:
3.4 × 10^23 molecules
Explanation:
To find the number of molecules present in C6H14, we multiply the number of moles in the compound by Avagadro's number (6.02 × 10^23 atoms).
number of molecules = number of moles (mol) × 6.02 × 10^23?
Number of molecules = 0.565 × 6.02 × 10^23
3.4 × 10^23 molecules
Answer:
(A) -117.48°C (B) 80.5368°C
Explanation:
It is given that density of ethanol is 0.789
volume of ethanol=89.8 mL
mass of ethanol = density of ethanol×volume = 0.789×89.8 = 70.8522 g
molar mass of ethylene glycol = 62 g/mole
therefore moles of ethylene glycol in 7.6 g of it = 7.7/62 = 0.1241 mole
molality = moles of ethylene glycol/mass of ethanol in kg
also Kf for ethanol = 1.99 C/m
And Kb for ethanol = 1.22 C/m
boiling point of ethanol = 78.4 °C
freezing point of ethanol = -114° C
(A) depression in freezing point = molality×Kf = 1.75153×1.99 = 3.4855
thus freezing point of solution = -114-3.4855= -117.48°C
(B) elevation in boiling point = molality×Kb = 1.75153×1.22 = 2.13686
thus boiling point of solution = 78.4+2.13686=80.5368°C