Answer:
C3H6 + Br2 → C3H6Br2
Explanation:
The reaction in which C3H6Br2 (1,2-Dibromopropane) is created is:
We can see that the only difference between the product (C3H6Br2) and the known reactant (C3H6) of the reaction is two bromine atoms (Br2). Br2 is diatomic bromine - a molecule we get after combining two bromine atoms. This compound is a red-brown liquid at room temperature, which means that that is the liquid described in your question.
Moles of water atoms = mass/molecular weight = 105/18 = 5.83 mol. Number of moles of hydrogen in water = 2 x moles of water = 11.66. Number of H atoms in water = moles of hydrogen x 6.02 x 10^23 = 7.019 x 10^24 ~ 7.02 x 10^24 atoms. Hope this helps.
I'm pretty sure it's 9726 milligrams of iodine. Hope this helps.
When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
B. A gram would have a lot more molecules of propane than a mole
