According to Bohr's model of the atom, the higher the orbital in which the electrons are found, the higher their energy or excitation state. Therefore, the electrons with the least amount of energy are those at the lowest orbitals, which are closer to the nucleus.
These orbitals are characterized by 4 quantum numbers, namely the principal quantum number (n), orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms). The principal quantum number reflects the distance of the electrons from the nucleus with n=1 as the orbital closest to the nucleus. Thus, according to Bohr's model, electrons in the orbital with n=1 have the lowest energy.
Answer:
0.0890 M
Explanation:
Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:
C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL
C2 = 0.149 x 150/250
= 0.089 M
To determine the concentration of S2O32- (aq), consider the equation:

The concentration of Na2S2O3 and S2O32- (aq) is 1:1
Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.
To 3 significant figures = 0.0890 M
H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O
v(NaOH)=46 ml=0.046 l
c(NaOH)=1.0 mol/l
v(H₂SO₄)=55 ml=0.055 l
n(NaOH)=v(NaOH)*c(NaOH)
n(H₂SO₄)=0.5n(NaOH)
c(H₂SO₄)=n(H₂SO₄)/v(H₂SO₄)=0.5*v(NaOH)*c(NaOH)/v(H₂SO₄)
c(H₂SO₄)=0.5*0.046*1.0/0.055=0.418 mol/l
The concentration of the H₂SO₄ is 0.418M.
Malleable, conducts electricity, shiny, hard
hope this helps!
Answer : The speed in meters per second is 17.89 m/s
Explanation :
The conversion used from miles to kilometer is:
1 mile = 1.61 km
The conversion used from kilometer to meter is:
1 km = 1000 m
So, 
The conversion used from hour to second is:
1 hr = 3600 s
So, 
As we are given the speed 40.00 miles per hour. Now we have to determine the speed in meter per second.
As, 
So, 
Therefore, the speed in meters per second is 17.89 m/s