2Na + O2 -> NaO2
that name is :
sodium peroxide
Answer:
One mole of cadmium (6multiply1023 atoms) has a mass of 112 grams, as shown in the periodic table on the inside front cover of the textbook. The density of cadmium is 8.65 grams/cm3.
Explanation:
From the calculation, the molar mass of the solution is 141 g/mol.
<h3>What is the molar mass?</h3>
We know that;
ΔT = K m i
K = the freezing constant
m = molality of the solution
i = the Van't Hoft factor
The molality of the solution is obtained from;
m = ΔT/K i
m = 3.89/5.12 * 1
m = 0.76 m
Now;
0.76 = 26.7 /MM/0.250
0.76 = 26.7 /0.250MM
0.76 * 0.250MM = 26.7
MM= 26.7/0.76 * 0.250
MM = 141 g/mol
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Answer:
Fruit punch with a very dark blue color
Explanation:
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
