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 .
The balanced equation given is:
4NH3 + 3O2 .....> 2N2 + 6H2O
From this equation, we can note that 4 moles of NH3 are required to produce 2 moles of N2.
Therefore, the mole ratio of NH3 to N2 is 4:2 which can be simplified into 2:1
<h2>Answer:</h2>
The correct answer is option C which is, "Electrons in the orbit closest to the nucleus have the least amount of energy".
<h3>
Explanation:</h3>
- There are different orbitals around the nucleus on which the electrons moves around the nucleus.
- These orbitals have a specific energy, due to which they are known as energy levels.
- The energy level near to the nucleus has least amount of the energy and the energy of the orbitals increase as the distance of the orbitals increase to the nucleus.
Answer:
The mass percentage of calcium carbonated reacted is 2.5%.
Explanation:
The reaction is:
Thus the Kp of the equilibrium will be:
Kp = partial pressure of carbon dioxide [as the other are solid]
Moles of calcium carbonate initially present = 
Let us apply ICE table to the equilibrium given:
Initial 0.2 0 0
Change -x +x +x
Equilibrium 0.2-x x x
Kp = partial pressure of carbon dioxide
Kp = Kc(RT)ⁿ
where n = difference in the number of moles of gaseous products and reactants
for given reaction n = 1
R = gas constant = 8.314 J /mol K
T = temperature = 800 ⁰C = 1073 K
Putting values
Kc =
Kc = ![\frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B2%7D%5D%5BCaO%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D%3D%20%5Cfrac%7Bx%5E%7B2%7D%20%7D%7B%280.2-x%29%7D%3D1.3X10%5E%7B-4%7D)
On calculating
x = 0.005
where x = the moles of calcium carbonate dissociated or reacted.
Percentage of the moles or mass reacted = 
%
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
