We know that the oxygen (O2) causes a glowing
splint to reignite, however, let us check what is missing on the products side
of the chemical equation: <span>
KBrO3 → KBr + ?
As we see, oxygen (O2) is the element missing from the other
side. Therefore the balanced chemical
equation for this decomposition is as follows: </span>
<span>2KBrO3 → 2KBr + 3O2 </span>
Answer:
Almost all corn oil is expeller-pressed
Explanation:
then solvent-extracted using hexane or 2-methylpentane (isohexane). The solvent is evaporated from the corn oil, recovered, and re-used. After extraction, the corn oil is then refined by degumming and/or alkali treatment, both of which remove phosphatides.Oct 16, 2020
When an element losses its electron its called a cation. When an element accepted that electron it called anion. This is called an ionic bond.
Answer:
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
(Ans)
ΔHf° of CaC2 = -59.0 kJ/mol
Explanation:
CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ
ΔHrxn = −127.2kJ
ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);
ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn
Where
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol
ΔHf°(CaC2) = -59.0 kJ/mol
Explanation:
It is known that in one day there are 24 hours. Hence, number of seconds in 24 hours are as follows.
Hence, total charge passed daily is calculated as follows.
And, number of Faraday of charge is as follows.
= 134300.52 F
The oxidation state of aluminium in 
is +3.
So, if we have to produce 1 mole of Al(s) we need 3 Faraday of charge.
Therefore, from 134300.52 F the moles of Al obtained with 89% efficiency is calculated as follows.
= 39842.487 mol
or, = 
Molar mass of Al = 27 g/mol
Therefore, mass in gram will be calculated as follows.
Mass in grams = 
= 
= 1075.7 kg/day
Thus, we can conclude that the daily aluminum production of given aluminium is 1075.7 kg/day.
