<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass.
For this, we look at the atomic masses of the elements present in the compound.
Cu has an atomic mass of 63.546 amu
Fe has 55.845 amu
and S has 36.065 amu
Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu
So we have not established the mass of the compound in amus
63.546 + 55.845 + 72.13 = 191.521
That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845
So to get the percentage, or fraction of iron, we take 55.845 / 191.521
Which comes out to 29.15% by mass
Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
Answer:
Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.
Explanation:
I just took the test and i got it right
Explanation:
The given data is as follows.
Thickness (dx) = 0.87 m, thermal conductivity (k) = 13 W/m-K
Surface area (A) = 5
, 
According to Fourier's law,
Q = 
Hence, putting the given values into the above formula as follows.
Q = 
= 
= 5902.298 W
Therefore, we can conclude that the rate of heat transfer is 5902.298 W.
The energy of the carbide released is 7262.5MJ.
<h3>What is the energy?</h3>
We know that the reaction between calcium oxide and carbon occurs in accordance with the reaction;
. The reaction is seen to produce 464.8kJ of energy per mole of carbide produced.
Number of moles of
produced = 1000 * 10^3 g/64 g/mol
= 15625 moles of calcium carbide
If 1 mole of
transfers 464.8 * 10^3 J
15625 moles of calcium carbide transfers 15625 moles * 464.8 * 10^3 J/ 1 mol
= 7262.5MJ
Learn more about reaction enthalpy:brainly.com/question/1657608
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Answer:
b
Explanation:
it's b because I just went over that frome my class and got it correct