A) sodium fluoride
B) rubidium oxide
C) boron trichloride
D) dihydrogen selenide
E) tetraphosphate hexoxide
F) iodine trichloride
Ppm = mass of solute mg / mass of solvent kg
0.008 * 1000 = 8.0 mg ( solute )
1000 / 1000 = 1.0 kg (solvent )
ppm = 8 / 1
= 8.0 ppm
hope this helps!
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
Answer:
z≅3
Atomic number is 3, So ion is Lithium ion (
)
Explanation:
First of all
v=f*λ
In our case v=c
c=f*λ
λ=c/f
where:
c is the speed of light
f is the frequency
Using Rydberg's Formula:
Where:
R is Rydberg constant=
z is atomic Number
For highest Energy:
n_1=1
n_2=∞
z≅3
Atomic number is 3, So ion is Lithium ion ()
