439.3 g CO2
Explanation:
First find the # of moles of CO2 that results from the combustion of 3.327 mol C3H6:
3.227 mol C3H6 × (6 mol CO2/2 mol C3H6)
= 9.981 mol CO2
Use the molar mass of CO2 to determine the # of grams of CO2:
9.981 mol CO2 x (44.01 g CO2/1 mol CO2)
= 439.3 g CO2
Answer:
It's B !
Explanation:
Formulas. The molecular formula for glucose is C6H12O6. This means that there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms bonded together to make one molecule of glucose.
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Answer:
D. Supported by observations
Explanation:
Because theories change over time, they both need to be proven, they do not go off of consensus, but they do need to be supported by observations.
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Answer:
The volume of the gas will be 78.31 L at 1.7 °C.
Explanation:
We can find the temperature of the gas by the ideal gas law equation:
Where:
n: is the number of moles
V: is the volume
T: is the temperature
R: is the gas constant = 0.082 L*atm/(K*mol)
From the initial we can find the number of moles:
Now, we can find the temperature with the final conditions:
The temperature in Celsius is:
Therefore, the volume of the gas will be 78.31 L at 1.7 °C.
I hope it helps you!
Answer:
1 gram
Explanation:
Half life = 25 years
Starting mass = 16 grams
Time = 100 years
Number of half lives = Time / Duration of Half life = 100 / 25 = 4
After first Half life;
Remaining mass = 16 / 2 = 8 g
After Second Half life;
Remaining mass = 8 / 2 = 4 g
After Third Half life;
Remaining mass = 4 / 2 = 2 g
After Fourth Half life;
Remaining mass = 2 / 2 = 1 g
