Answer:
Wavelength of this beam of light: 
.
Explanation:
The speed of light in vacuum is approximately 
.
Light behaves like a wave. The wavelength of a wave is equal to the distance that it travels (in the given medium) in each period of oscillation.
On the other hand, the frequency of a wave is the number of periods in unit time. 
means one oscillation per second. The frequency of this particular wave is 
. In other words, there are 
oscillations in each second.
The period of oscillation will be equal to
.
In that period of time, a beam of light in vacuum would have traveled
.
In other words, if this beam of light of frequency is in vacuum, its wavelength will be equal to .
Answer:
Al
Explanation:
4 Al + 3 O₂ → 2 Al₂O₃
You need to figure out which one has the smaller mole ratio. Convert both substances from grams to moles.
(10.0 g Al)/(26.98 g/mol) = 0.3706 mol Al
(19.0 g O₂)/(32.00 g/mol) = 0.5938 mol O₂
Now, use the mole ratios of reactant to product to see which substance produces the least amount of product.
(0.3706 mol Al) × (2 mol Al₂O₃/4 mol Al) = 0.1853 mol Al₂O₃
(0.5938 mol O₂) × (2 mol Al₂O₃/3 mol O₂) = 0.3958 mol Al₂O₃
Since aluminum produces the least amount of product, this is the limiting reagent.
Answer:
4.86×10^23 molecule of Pb
Explanation:
Based on that equation, for every 2 moles of ammonia, you get 3 moles of lead.
So:
2 mol NH3/ 3 mol Pb
Using this ratio we can find the amounts of either molecule. Given 5.38 mol NH3:
(5.38 NH3)(3 Pb/ 2 NH3) = (5.38)(3/2) mol Pb = 8.07 mol Pb
Then, we just need to use Avagadro's number to get the number of molecules.
(8.07)(6.02×10^23) = 4.86×10^23 molecule of Pb
let me know when u find out plz because i would like to know as well its one of my chemistry qustions in an assiment. :)
Element with an atomic number of 58 is actually Cerium, so the symbol should be Ce, not Co because that is Cobalt which has an atomic number of 27. With that being said, the notation for isotopes is the symbol of the element with a superscript and a subscript that are aligned. The superscript represents the mass number.
Mass number = protons + neutrons = 58 + 33 = 91
The subscript is the atomic number which is 58. The notation is written in the picture attached.
