Answer:
b)5l x 10kg c)10kg + 9l (Not sure for the last 1)
Explanation:
Atomic number of magnesium is 12 and its electronic distribution is 2, 8, 2. On the other hand, atomic number of iodine is 53 and its electronic configuration is ![[Kr]4d^{10}5s^{2}5p^{5}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B10%7D5s%5E%7B2%7D5p%5E%7B5%7D)
.
Hence, there are 7 valence electrons in an iodine atom and there are 2 valence electrons in a magnesium atom.
So, one atom of iodine requires one electron from a donor atom to complete its octet. But one magnesium atom contains two valence electrons.
Therefore, one magnesium atom will combine with two iodine atoms to result in the formation of magnesium iodide as follows.
Therefore, an ionic bond will be formed when magnesium reacts with iodine to make magnesium iodide.
Answer:
their all small and arent considered planets
hope this could help
Explanation:
Answer:
The minimum pressure should be 901.79 kPa
Explanation:
<u>Step 1: </u>Data given
Temperature = 25°C
Molarity of sodium chloride = 0.163 M
Molarity of magnesium sulfate = 0.019 M
<u>Step 2:</u> Calculate osmotic pressure
The formula for the osmotic pressure =
Π=MRT.
⇒ with M = the total molarity of all of the particles in the solution.
⇒ R = gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 25 °C = 298 K
NaCl→ Na+ + Cl-
MgSO4 → Mg^2+ + SO4^2-
M = 2(0.163) + 2(0.019 M)
M = 0.364 M
Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)
Π = 8.90 atm
(8.90 atm)(101.325 kPa/atm) = 901.79 kPa
The minimum pressure should be 901.79 kPa
Answer:
(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.
Explanation:
Without mincing words let's dive straight into the solution to the question.
(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;
The freezing point depression = [ 1 - (-3)]° C = 4°C.
(b). The molality can be Determine by using the formula below;
Molality = the number of moles found in the solute/ solvent's weight(kg).
Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.
(c). The mass of acetone that was in the decanted solution = 11.1 g.
(d). The mass of water that was in the decanted solution = 89.01 g.
(e). 2.4 = x/ 58 × (1000/1000).
x = 2.4 × 58 = 139.2 g.
(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.
