Answer:
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Explanation:
Answer:
- 2Al (s) + 3Br₂ (l) → 2AlBr₃(s)
Explanation:
<u>1) Word equation:</u>
- aluminum metal + liquid bromine → aluminum bromide
<u>2) Chemical equation:</u>
- Al (s) + Br₂ (l) → AlBr₃(s)
<u>3) Balanced chemical equation:</u>
- 2Al (s) + 3Br₂ (l) → 2AlBr₃(s)
<u>4) Notes:</u>
- The most common oxidation state of aluminum is +3.
- Bromine has 5 common oxidation states: -1, +1, +3, +5, and +7.
- Since aluminum has a positive oxidation state, bromine must have the negative oxidation state, - 1.
- In the molecular formula the valences (oxidation numbers) are exchanged as subscripts, that is the reason Al has subscript 1 (understood) while Br has subscript 3.
- This is combination or synthesis reaction: two elements combine to form a new compound.
- This is a very exothermic reaction which yields light (sparks) and heat.
- The product, AlBr₃ is a molecular compound (not ionic).
I am assuming that this compound contains carbon, hydrogen and oxygen. The molar mass of carbon, hydrogen and oxygen is 12, 1 and 16 grams/mol. Given these molar mass we now need to know the number of C H and O in the compound. To have 132 the compound must be C6H12O3
<u>Answer:</u> The average atomic mass of copper is 63.546 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
- <u>For isotope 1 (Cu-63):</u>
Mass of isotope 1 = 62.93 amu
Percentage abundance of isotope 1 = 69.2 %
Fractional abundance of isotope 1 = 0.692
- <u>For isotope 2 (Cu-65):</u>
Mass of isotope 2 = 64.93 amu
Percentage abundance of isotope 2 = 30.8 %
Fractional abundance of isotope 2 = 0.308
Putting values in equation 1, we get:
Hence, the average atomic mass of copper is 63.546 amu
Answer:
Claim: The chemical equation presented in the data table does NOT follow the law of conservation of matter
Evidence: Since we initially have 14.25 grams of Na and 9.5 grams of HCl
we will find the number of moles in each, to further apply stoichiometry
So,
Moles of Na: Given mass/Molar Mass = 14.25/23 = 0.62 Moles of Na
Moles of HCl: Given mass/Molar Mass = 9.5/36.5 = 0.26 Moles of HCl
Since HCl is the limiting reagent in this reaction,
0.26 Moles of Na will be consumed, which is equal to
5.98 grams of Na
Reasoning:
We are given that the total mass of the product is 22.98 grams
but through stoichiometry, we have found that only (9.5 + 5.98) = 15.48 grams of product can possibly be formed
Hence, the reaction presented in the table does NOT follow the law of conservation of matter