The correct answer would be 32/16s
Answer:
0.56 liters
Explanation:
First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:
- 0.80 g ÷ 32 g/mol = 0.025 mol
At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:
- 0.025 mol * 22.4 L/mol = 0.56 L
Thus the answer is 0.56 liters.
Answer:
2-4 minutes
Explanation:
Fastest changing temperature means larger change in temperature when subtracting final temperature from initial temperature in a given time period (given time period is 2 minutes for all the options)
For 0-2 minutes, our final temperature was 40 (at 2 min) and initial temperature was 20 (at 0 min), thus temperature change was only 20 C.
For 2-4, our final temperature was 80 (at 4 min) and initial temperature was 40 (at 2 min) thus temperature change was 40 C.
For 4-6, our final temperature was 100 (at 6 min) and initial temperature was 80 (at 4 min) thus temperature change was 20 C.
We are not given temperature at 8 min so option D is invalid.
As we can clearly see that in a given 2 minute period, option B has the fastest change because it changed 40C when compared to other options that changed only 20C from starting temperature.
Hope that makes sense.
Answer:
The answer to your question is 8.28 g of glucose
Explanation:
Data
Glucose (C₆H₁₂O₆) = ?
Ethanol (CH₃CH₂OH)
Carbon dioxide (CO₂) = 2.25 l
Pressure = 1 atm
T = 295°K
Reaction
C₆H₁₂O₆ ⇒ 2C₂H₅OH(l) +2CO₂(g)
- Calculate the number of moles
PV = nRT
Solve for n

Substitution

Simplification
n = 0.092
- Calculate the mass of glucose
1 mol of glucose --------------- 2 moles of carbon dioxide
x --------------- 0.092 moles
x = (0.092 x 1) / 2
x = 0.046 moles of glucose
Molecular weight of glucose = 180 g
180 g of glucose --------------- 1 mol
x g ---------------0.046 moles
x = (0.046 x 180) / 1
x = 8.28 g of glucose
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