Answer:
Particle Symbol Mass
electron e- 0.0005486 amu
proton p+ 1.007276 amu
neutron no 1.008665
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Answer:
<u>2.26 </u>
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Explanation:
First you should know what are significant figures,
1 . All non - zero digits are significant.
2 . The zero between two non- zero are significant.Ex 302 has 3 significant figures.
3. The zero before the decimal and before any non-zero digit is non significant. Example : 0.003 has only 1 significant figure.
4. The zero after non zero are non - significant . But the zeros after the decimal point are significant.
300 has only 1 significant figure
300.0 has 4 significant figure
The least number of significant figures present in the number should be there in the final answer of the calculation.
Look at the number having minimum significant digits :
It is 3 (every number has 3- significant figures ) So the answer should also contain 3 - significant figures.

First solve the numerator part


Round off this number to 3 significant figures.
Answer is =<u> 2.26</u>
It has 3 - significant digit since all the digits are non-zero.
Answer:
Chromosphere
Explanation:
You see the middle layer of the sun’s atmosphere, the Chromosphere, at the start and end of a total eclipse.