C. alpha = nucleus of a helium atom
beta = electron
gamma = photon
Answer:
N- 1s2 2s2 2p3
Mg- 1s2 2s2 2p6 3s2
O- 1s2 2s2 2p4
F- 1s2 2s2 2p5
Al-1s2 2s2 2p6 3s2 3p1
Explanation:
Order of decreasing atomic radius
Mg,Al, N,O,F
Order of increasing ionization energy
Mg,Al, N,O,F
Reason:
Atomic radius decreases with increase in nonmetallic character. Looking at the electronic configurations, as effective nuclear charge increases, the atom becomes smaller and the attractive force between the nucleus and the outermost electrons increases. Hence, the radius of the atom decreases and ionization energy increases. Note that the addition of more orbital electrons implies addition of more nuclear charge since the both must exactly balance for the atom to remain electrically neutral. The more the electrons in the outermost shell, the higher the first ionization energy.
One way of expressing concentration is by percent. It may be on the basis of mass, mole or volume. Percent is expressed as the amount of solute per amount of the solution. For this case, we are given the percent by mass. In order to solve the amount of solute, we multiply the percent with the amount of the solution.
Mass of solute = percent by mass x mass solution
Mass of solute = 0.0350 x 2.50 x10^2 = 8.75 grams of solute
<h3>
Answer:</h3>
0.89 J/g°C
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Explanation:</h3>
Concept tested: Quantity of heat
We are given;
- Mass of the aluminium sample is 120 g
- Quantity of heat absorbed by aluminium sample is 9612 g
- Change in temperature, ΔT = 115°C - 25°C
= 90°C
We are required to calculate the specific heat capacity;
- We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.
That is;
Q = m × c × ΔT
- Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.
Specific heat capacity, c = Q ÷ mΔT
= 9612 J ÷ (120 g × 90°C)
= 0.89 J/g°C
Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C