Answer:
439.7nm
Explanation:
Energy of a quantum can be calculated using below formula
E=hv...........eqn(1)
But v=λ/ c .........eqn(2)
If we substitute eqn(2) into eqn(1) we have
E= hc/(λ)
Where E= energy
h= Plank's constant= 6.62607004 × 10-34 m2 kg / s
c= speed of light
c= 2.998 × 10^8 m/s
λ= wavelength= ?
But the energy was given in Kj , it must be converted to Kj/ photon for unit consistency.
Energy E= 272 kJ/mol × 1mol/6.02× 10^23
Energy= 451.83× 10^-24 Kj/ photon
E= hc/(λ)...........eqn(1)
If we make λ subject of the formula
λ= hc/E
Then substitute the values we have
λ= [(6.626 × 10^-34) × (2.998 × 10^8)]/451.83× 10^-24
λ=(0.00043965) × (1Kj/1000J) × (10^9nm/1m)
λ=439.7nm
Hence, the longest wavelength of radiation with enough energy to break carbon-sulfur bonds is 439.7nm
Answer:
21.02moles of KBr
Explanation:
Parameters given:
Number of moles BaBr₂ = 10.51moles
Complete reaction equation:
BaBr₂ + K₂SO₄ → KBr + BaSO₄
Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:
The balanced equation is:
BaBr₂ + K₂SO₄ → 2KBr + BaSO₄
From the equation:
1 mole of BaBr₂ produces 2 moles of KBr
∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr
A.) a rotation is the earth "rotating" on its axis while a revolution is the earth "revolving" around the sun so 1 rotation is from sun rise to sun rise while one revolution is from January 1st to the next January 1st
Answer:oxygen Explanation:The medical condition described here is anaemia. It is a blood cell disorder whereby the red blood cell doesn't function properly and hence doesn't carry enough oxygen to the tissues. This is usually caused when ones body is deficient of iron.The symptoms that may occur to such patients are weakness, fatigue, headache and pale skin.Based on the explanation, the answer is oxygen
Explanation:
In the physical properties, it is mentioned that the element has 4 valence electrons. The elements in the periodic table are arranged such that, the number of valence electrons present in the neutral atoms belonging to a particular group is equal to the group number.
Thus, the unidentified element can be best classified as a nonmetal in period 4
Ans C)
