A-Aluminium sulphate.
B-Calcium Chloride.
C-Potassium sulphate.
D-Potassium Nitrate .
E-Calcium Carbonate.
Answer:
atomic mass 112.4 and Cadmium (Cd)
Explanation:
You have 73.35 g of MO.
After the reaction the O is removed and you only have M which the mass is 64.21 g.
With that you can calculate the mass of O removed:
Mass of O = 9.14 g
Mass = AM * moles ; (AM : Atomic Mass)
9.14 = 16 * moles
moles = 9.14 / 16
moles = 0.57125
The formula of the metal oxide is MO, meaning it has 1 mole of M per mole of O. 73.35 had 0.57125 moles of O, then it also had 0.57125 moles of M, and the remaining mass of 64.21 g represents those moles
Mass = AM * moles ; (AM : Atomic Mass)
64.21 = AM * 0.57125
AM = 64.21/0.57125 = 112.4 amu
The metal with an atomic mass of 112.4 is Cadmium (Cd), therefore the metal oxide is CdO
The arrangement of the periodic table is related to the electron configuration of the atoms of each element. Electron configurations represents how electrons are distributed. For instance, the neon atom has the electron configuration of 1s^2 2s^2 2p^6.
Answer:
6.0 L
Explanation:
Use the dilution equation M1V1 = M2V2
M1 = 0.075 M
V1 = 200 L
M2 = 2.5 M
V2 = ?
Solve for V2 --> V2 = M1V1/M2
V2 = (0.075 M)(200 L) / (2.5 M) = 6.0 L