Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm
I think it’s C but not certainly positive
Answer:
P2 = 352 mm Hg (rounded to three significant figures)
Explanation:
PV = nRT
where P is the pressure,
V is the volume,
n is the moles of gas,
R is the gas constant,
and T is the temperature.
We must relate this equation to a sample of gas at two different volumes however. Looking at the equation, we can relate the change in volume by:
P1V1 = P2V2
where P1 is the initial pressure,
V1 is the initial volume,
P2 is the final pressure,
and V2 is the final volume.
Looking at this relationship, pressure and volume have an indirect relationship; when one goes up, the other goes down. In that case, we can use this equation to solve for the new pressure.
P1V1 = P2V2
(759 mm Hg)(1.04 L) = P2(2.24 L)
P2 = 352 mm Hg (rounded to three significant figures)
Answer:
<em>b</em>
<em>an increase in non-biodegradable wastes</em>
Explanation:
The error caused will lower the number of moles when he did not do the second heating.This question is related to the empirical formula.
<h3>What is Empirical formula ?</h3>
A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.
The empirical formula is the simplest ratio of atoms in a given compound.
we know that heating is used to eliminate most of the water molecules which are attached to the compound in the question, it's given that the second heating was not performed by the students.
This means that there are still some water molecules attached to the compound. This will imply that higher mass will be calculated for the compound because the water molecules are also considered.
So now, if we look at the calculation for the number of moles of water in this, we can say that this will be lower than the actual amount.
This is because there are some unaccounted moles of water which are still attached to the compound and these will be accounted in the compound mass rather than the waters. Therefore, the number of moles will be lower.
Hence, The error caused will lower the number of moles when he did not do the second heating.
Learn more about Empirical formula here ;
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