Koalas are not crusty, but their fur is very coarse, like wool.
Hope this helps.
Answer:
1. 266.22 g/mol
2. 168.81 g/mol
3. 223.35 g/mol
4. 199.88 g/mol
Explanation:
For you to calculate the molar mass of the salt you need to sum the molar masses of every element in the salt.
In the first salt, PdBr
, the subscript 2 means that there are 2 atoms of Br. So for you to calculate the molar mass of the salt you need to sum the molar mass of Pd and 2 times the molar mass of Br, as follows:
106 g/mol + 2(79.90 g/mol) = 266.22 g/mol
In the second salt BeBr there are 2 atoms of Br and 1 of Be, so the molar mass is:
9.012 g/mol +2(79.90 g/mol) = 186.22 g/mol
In the third salt CuBr there are 2 atoms of Br and 1 of Cu, so the molar mass is:
63.55 g/mol + 2(79.90 g/mol) = 223.35 g/mol
And in the fourth salt CaBr there are 2 atoms of Br and 1 of Ca, so the molar mass is:
40.08 g/mol + 2(79.90 g/mol) = 199.88 g/mol
The answer is B! Chromatography involves solvent separation on a solid medium. Distillation takes advantage of differences in boiling points. Evaporation removes a liquid from a solution to leave a solid material
The best definition of low electromagnetic energy would be A. Low electromagnetic energy has long wavelengths and low frequency.
Answer:
Option d.
1 mole AlCl3in 500 g water
Explanation:
ΔT = Kf . m . i
Freezing T° of solution = - (Kf . m . i)
In order to have the lowest freezing T° of solution, we need to know which solution has the highest value for the product (Kf . m . i)
Kf is a constant, so stays the same and m stays also the same because we have the same moles, in the same amount of solvent. In conclussion, same molality to all.
i defines everything. The i refers to the Van't Hoff factor which are the number of ions dissolved in solution. We assume 100 & of ionization so:
a. Glucose → i = 1
Glucose is non electrolytic, no ions formed
b. MgF₂ → Mg²⁺ + 2F⁻
i = 3. 1 mol of magnessium cation and 2 fluorides.
c. KBr → K⁺ + Br⁻
i = 2. 1 mol potassium cation and 1 mol of bromide anion
d. AlCl₃ → Al³⁺ + 3Cl⁻
i = 4. 1 mol of aluminum cation and 3 mol of chlorides.
Kf . m . 4 → option d will has the highest product, therefore will be the lowest freezing point.
