Answer:
The empirical formula is CH2O, and the molecular formula is some multiple of this
Explanation:
In 100 g of the unknown, there are 40.0⋅g12.011⋅g⋅mol−1 C; 6.7⋅g1.00794⋅g⋅mol−1 H; and 53.5⋅g16.00⋅g⋅mol−1 O.
We divide thru to get, C:H:O = 3.33:6.65:3.34. When we divide each elemental ratio by the LOWEST number, we get an empirical formula of CH2O, i.e. near enough to WHOLE numbers. Now the molecular formula is always a multiple of the empirical formula; i.e. (EF)n=MF.So 60.0⋅g⋅mol−1=n×(12.011+2×1.00794+16.00)g⋅mol−1.Clearly n=2, and the molecular formula is 2×(CH2O) = CxHyOz.
It’s charge was neutral due to the equal number of protons and electrons. when it becomes an ion it loses 3 electrons leaving behind only 10. the answer is 10. the equation is +13 +(-10)=+3
The copper wire was sanded before burning in order to make sure that copper metal was exposed on the surface of the wire.
Answer: B
Explanation
The copper wire when placed in atmosphere without coating leads to oxidation of copper metal with respect to the impurities present in the atmosphere.
As copper is electropositive in nature, so electronegative ions present in the universe will try to react with copper and the copper will react easily with other elements.
So generally copper wire is coated with color or polymer coating.
In this case, the copper wire without any coating is sanded, so that the eddy sheets or polishing materials on friction with copper wire will remove the impurities by the electrostatic law of conservation of charges and charge transfer.
As the impurities are removed when copper wire is sanded, the copper atoms will be exposed on the surface of the wire leading to burning of copper in the copper wire.
Answer:
Explanation:
Hello,
In this case, for a first-order reaction, we can firstly compute the rate constant from the given half-life:
In such a way, the integrated first-order law, allows us to compute the final mass of the substance once 10.0 minutes (600 seconds) have passed:
Best regards.
Answer:
A = True
B = False
C = True.
I think this is the answer
