Answer:
0.30 mole NaCl needed.
Explanation:
From the definition of solution molarity => Molarity = moles solute / Volume of Solution in Liters => Molarity = moles/Volume(L) => solve for moles =>
moles = Molarity X Volume(L) = (0.1M)(3.0L) = 0.30 mole NaCl needed.
Answer:
c. 3.00 M HCl
Explanation:
From dilution formula
C1V1 = C2V2
C1=?, V1= 10.0ml, C= 1.5, V2= 20.0ml
Substitute and Simplify
C1×10= 1.5×20
C1= 3.00M
Answer:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
Explanation:
