Answer:
c) 9.03 x 10^23
Explanation:
find the molar mass of Al
Al is 27.0 grams
Then use that, to find the number of moles in Aluminum.
Then use Avogadro's number which is 6.02 * 10^23
After that, write all of that down with dimensional analysis.
40.5 g * 1 mol/ 27.0 g of Al * 6.02 x 10^23 / 1mol
As your final answer, you will get 9.03 * 10^23 atoms with sig figs.
Hope it helped!
Answer:
Large molecules tend to have greater boiling points because the London dispersion forces are stronger within.
Explanation:
Try to grow a plant with and without soil
Answer:
Sc (Scandium) has the given electronic configuration.
Explanation:
The given electronic configuration is [Ar]
.
The last electron enters the d-subshell and hence is a d-block element known as Scandium with chemical symbol Sc.
For 4s subshell
n=4,l=0 and m ranges from -l to +l so m=0.
For 3d subshell
n=3,l=2 and m ranges from -l to +l so m can take values -2,-1,0,+1,+2
Note:
l values for subshells:
s : 0
p : 1
d : 2
f : 3 and so on.
0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
