Instruments
The specific type of instruments depends on the type of laboratory that you're working in: some labs for example use electron microscopes, others use mass spectrophotometers, others use multiplex biochemical analyzers, etc. But very broadly, the specialized tools we use in the laboratory are usually referred to as "instruments"
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
I'm not exactly sure which one but I do know that an acid and a base react in a aqueous solution to form water, so i would probably eliminate the ones that aren't aqueous solutions.
Answer:
chemical symbol= Cl ( chlorine)
electrons= 17
2p electrons = 6
Answer:
Explanation:
1 = The given chemical reaction does not follow the law of conservation of mass because,
2 = Four hydrogen atoms are present in reactant side and two hydrogen atoms are present in product side.
3 = 1 ) The given chemical reaction does not follow the law of conservation of mass because,
CH₄ + O₂ → CO₂ + H₂O
16 g + 32 g 44 g + 18 g
48 g 62 g
Law of conservation of mass:
This law stated that mass can not be created or destroyed in chemical reaction. It just changed from one to another form.
For example:
C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
28 g + 96 g = 88 g + 36 g
124 g = 124 g
