Answer:
<u>It</u><u> </u><u>is</u><u> </u><u>a</u><u> </u><u>combustion</u><u> </u><u>reaction</u><u>.</u>
Answer:
Explanation:
By simple distillation, we can separate tow miscible liquids having sufficient difference in their boiling points.
By fractional distillation, we can separate the mixture of two or more miscible liquids for which the difference in their boiling point is less than 25k
Simple distillation is done using an air condenser or water condenser.
Fractionating column is fitted in between the distillation flask and the condenser for the process.
Simple distillation eg: mixture of water and acetone
fractional distillation eg: crude oil
hope this helps
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Answer:
5.81L
Explanation:
N1 = 1.70 moles
V1 = 3.80L
V2 = ?
N2 = 2.60 moles
Mole - volume relationship,
N1 / V1 = N2 / V2
V2 = (N2 × V1) / N1
V2 = (2.60 × 3.80) / 1.70
V2 = 9.88 / 1.70
V2 = 5.81 L
The volume of the gas is 5.81L
The composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu is 29.4 at % and 70.6 at % for Zn and Cu respectively.
<h3>How to convert weight % to atom %?</h3>
The weight percentage of an element can be converted to atom percentage as follows:
- C(Zn) = C1A2 / (C2A1) + (C1A2) × 100
- C(Cu) = C2A1 / (C1A2) + (C2A1) × 100
Where;
- C1 = 30
- C2 = 70
- A1 = 65.4g/mol
- A2 = 63.55g/mol
- C(Zn) = {30*63.55} / (70*65.4) + (30*63.55) × 100
- C(Cu) = {70*65.4} / (30*63.55) + (70*65.4) × 100
- C(Zn) = 29.4 at %
- C(Cu) = 70.6 at %
Therefore, the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu is 29.4 at % and 70.6 at % for Zn and Cu respectively.
Learn more about atomic percent at: brainly.com/question/10018123
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