Answer:
A) 54.04%
B) 13-karat
Explanation:
A) From the problem we have
<em>1)</em> Mg + Ms = 9.40 g
<em>2)</em> Vg + Vs = 0.675 cm³
Where M stands for mass, V stands for volume, and g and s stand for gold and silver respectively.
We can rewrite the first equation using the density values:
<em>3)</em> Vg * 19.3 g/cm³ + Vs * 10.5 g/cm³ = 9.40
So now we have<em> a system of two equations</em> (2 and 3) <em>with two unknowns</em>:
We <u>express Vg in terms of Vs</u>:
We <u>replace the value of Vg in equation 3</u>:
- Vg * 19.3 + Vs * 10.5 = 9.40
- (0.675-Vs) * 19.3 + Vs * 10.5 = 9.40
- 13.0275 - 19.3Vs + 10.5Vs = 9.40
Now we <u>calculate Vg</u>:
- Vg + 0.412 cm³ = 0.675 cm³
We <u>calculate Mg from Vg</u>:
- 0.263 cm³ * 19.3 g/cm³ = 5.08 g
We calculate the mass percentage of gold:
- 5.08 / 9.40 * 100% = 54.04%
B)
We multiply 24 by the percentage fraction:
- 24 * 54.04/100 = 12.97-karat ≅ 13-karat
Explanation:
According to Bohr's postulates, the electron in the present in the lower energy level can absorb energy and exits to higher energy level. Also, when this electron returns back to its orbit, it emits some energy.
Since the hydrogen consists of 1 electron and 1 proton. The lowest energy configuration of the hydrogen is when n =1 or, when the electron is present in the K-shell or the ground state.
The possible transition for the electron given in the question is :
n = 2, 3 and 4
The schematic diagram of the hydrogen atom consisting of these four quantum levels in which the electron can jump (Absorption) and comeback to from these energy levels (emission) .
The simple formula is C = n/V
n = mols
C = Concentration or Molarity
V = Volume in Liters.
n = 2
V = 4
C = 2 / 4
C = 0.5 mol/Litre
Families are another names for the columns
Answer:
Explanation:
Bromine >Tellurium > Phosphorus > Helium > Sodium
Electron affinity of Bromine , Tellurium , Phosphorus are positive , of helium is zero and of sodium is negative .