4 protons the number of proton has the same number of electron [which is the same as atomic number]
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
Answer: 2.48×10^-17 J
Explanation:
Given the following :
Wavelength = 8nm (8 x 10^-9 m)
Energy(e) of X-ray =?
Energy=[speed of light(c) × planck's constant (h)] ÷ wavelength
Speed of light = 3×10^8m/s
Planck's constant = 6.626×10^-34 Js
Wavelength = 8 x 10^-9 m
Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9
Energy = [19.878×10^(8-34)] / 8 x 10^-9
Energy = 2.48475 × 10^(-26+9)
Energy = 2.48×10^-17 J
Answer:
A. PO43-
Explanation:
in the given is buffer . so H2PO4- and HPO42- both are present with equal concentration . Na+ is spectator ion it is also present in the concentration higher than the given species above .
but PO4-3 is not present . so it is lowest concentration
Answer:
At anode - 
At cathode - 
Explanation:
Electrolysis of NaBr:
Water will exist as:

The salt, NaBr will dissociate as:

At the anode, oxidation takes place, as shown below.

At the cathode, reduction takes place, as shown below.
