Answer:
0.65moles of HCl was produced
Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
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Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Answer:
Chromosomes and I think its too many
Explanation:
The Patch's area of the space shuttle in km² is 2.07 × 10⁻⁹ km²
Given, that a space shuttle requires a 20.7 cm² patch
We have to convert the patch's area from cm² into km².
Unit conversion is a method in which we multiply or divide with a particular numerical factor and then finally round off to the nearest significant digits.
Patch area of the space shuttle is 20.7 cm²
1 cm = 0.00001 km
or, 1 cm² = (0.00001 km)²
or, 1 cm² = 10⁻¹⁰km²
20.7 cm² = 20.7 × 10⁻¹⁰km²
20.7 cm² = 2.07 × 10⁻⁹ km²
The patch area in square kilometers is 2.07 × 10⁻⁹ km²
To learn more about unit conversion, visit: brainly.com/question/11543684
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