Answer:
The hydrogen atom has just one electron, but many spectral lines. However it contains many shells and the movement of that electron from one shell to another causes the release of energy and also an emission of photons.
A spectral line are dark or bright lines formed within a specific frequency range which differ from other frequencies.Because of the difference of energy for the various shells, it produces different wavelengths and this is the reason for the many spectral line for hydrogen.
Answer: must have THE SAME number of atoms for each element
Explanation: Chemical equations must be balanced -- they must have the same number of atoms of each element on both sides of the equation. As a result, the mass of the reactants must be equal to the mass of the products of the reaction.
Answer:
800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
Explanation:
3000 lb of 13% solution is required .
Total adhesive in weight = 3000 x .13 = 390 lb of adhesive
Available = 500 lb of 10% solution = 50 lb of adhesive
Rest = 390 - 50 = 340 lb required .
rest mass of solution = 3000 - 500 = 2500 lb
mass of adhesive required = 340 lb
Let the mass of 20% required be V
mass of adhesive = .20 V
.20 V = 340
V = 1700
rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent
So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .
Answer:
=1.666 liters
Explanation:
1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.
0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1
=11.2 liters.
Standard pressure= 1 atmosphere (Atm)
Standard temperature = 273.15 Kelvin
According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂
Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.
Therefore P₂ =P₁V₁T₂/T₁V₂
Substituting for the values we get:
P₂= (1 atm× 11.2L ×203K)/ (273K×5L)
=1.666 atm
Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is 
= 283.725 kJ ⋅ mol − 1
