To solve for the weighted average atomic weight of X, all
we have to do is to calculate for the contribution of each isotope. The
contribution of each isotope on the atomic weight of x is calculated by
multiplying their respective atomic weights with the abundance, therefore:
X – 28 = 27.977 amu * (92.23 / 100)
X – 28 = 25.803 amu
X – 29 = 28.976 amu * (4.67 / 100)
X – 29 = 1.353 amu
X – 30 = 29.974 amu * (3.10 / 100)
X – 30 = 0.929 amu
Hence:
X = X – 28 + X – 29 + X – 30
X = 25.803 amu + 1.353 amu + 0.929 amu
<span>X = 28.085 amu</span>
Explanation:
<h2><u>Steps </u><u>:</u></h2>
- <u>Move </u><u>decimal</u><u> </u><u>from</u><u> </u><u>left </u><u>to </u><u>right</u><u> </u><u>=</u><u>0</u><u> </u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>0</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u>
- <u>Then </u><u>count </u><u>the</u><u> </u><u>numbers</u><u> </u><u>before</u><u> </u><u>decimal </u><u>and </u><u>w</u><u>rite </u><u>it </u><u>like</u><u> </u><u>this </u><u>=</u><u>2</u><u>4</u><u>0</u><u>.</u><u>0</u><u>x</u><u>1</u><u>0</u><u> </u><u>power-</u><u>9</u><u> </u>
- <u>That's</u><u> </u><u>all </u>
<u>hope</u><u> it</u><u> </u><u>help</u>
<h2><u>#</u><u>H</u><u>o</u><u>p</u><u>e</u></h2>
Answer:
a) I₁ = 11.2 Lux
, vertical direction
, b) I₂ = 1.44 Lux
Explanation:
a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed
consequently the above the processed light has half of the incident intensity and the directional of the polarized
I₁ = I₀ / 2
I₁ = 22.4 / 2
I₁ = 11.2 Lux
is polarized in the vertical direction
b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus
I₂ = I₁ cos² θ
I₂ = 11.2 cos² 69
I₂ = 1.44 Lux
Answer:
Explanation:
We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.
By using energy conservation
KE₁ + U₁ = KE₂ + U₂
KE₁=Kinetic energy at location 1
U₁ =Potential energy at location 1
KE₂=Kinetic energy at location 2
U₂=Potential energy at location 2
Therefore, Raymond is thinking in a right way.