Answer:
The first one is correct.
Explanation: Eh Are you cheating on quiz?
Answer:
Explanation:
The Unit Of Current is defined as the flow of 1 coulomb of charge in one second . Its unit is Ampere.
An ammeter is a measuring device used to measure the electric current in a circuit. A voltmeter is connected in parallel with a device to measure its voltage, while an ammeter is connected in series with a device to measure its current.
Answer:
<em>The N given in the question is normality while the M given in the question designates morality.
</em>
Explanation:
There are minor differences between both the terms normality and morality given in the question and both are inter convertible.
When one litre of solution is considered when the particles of solute in that litre would show the morality of the solution while when one litre of Solution is considered and equivalent are given then it is known as normality.
Answer:
The isotope with mass 6.015amu =7.5% abundance
The isotope with mass 7.016amu = 92.5%
Explanation:
<u>Step 1:</u> Given data
Lithium has an average atomic mass of 6.941 amu
Lithium has 2 isotopos with a mass, respectively 6.015amu and 7.016amu
<u>Step 2: </u>Calculate the abundances
Isotope with mass 6.015 amu has an abundance of X
Isotope with mass 7.016 amu has an abundance of Y
The average atomic mass is the average mass of the isotopes with its respectively abundance.
X+Y = 1 or X = 1 - Y
6.941 = 6.015X +7.016Y
This gives us:
6.941 = 6.015(1-Y) +7.016Y
6.941 = 6.015 - 6.015Y + 7.016Y
6.941 - 6.015 = -6.015Y +7.016Y
0.926 = 1.001Y
Y = 0.925
Y = 92.5%
X = 1- 92.5 = 7.5%
To control we can do the following equation:
6.015*0.075 + 7.016*0.925 = 6.941
This means the abundances of the 2 isotopes are:
The isotope with mass 6.015amu =7.5% abundance
The isotope with mass 7.016amu = 92.5%
Answer:
Diversification and adaptive radiation.
Explanation: