Answer:
The answer to your question is 0.22
Explanation:
Data
Acetonitrile (CH₃CN) density = 0.786 g/ml
Methanol (CH₃OH) density = 0.791 g/ml
Volume of CH₃OH = 22 ml
Volume of CH₃CN = 98.4 ml
Process
1.- Calculate the mass of Acetonitrile and the mass of Methanol
density = mass/ volume
mass = density x volume
Acetonitrile
mass = 0.786 x 98.4
= 77.34 g
Methanol
mass = 0.791 x 22
= 17.40 g
2.- Calculate the moles of the reactants
Acetonitrile molar mass = (12 x 2) + (14 x 1) + (3 x 1)
= 24 + 14 + 3
= 41 g
Methanol molar mass = (12 x 1) + (4 x 1) + (16 x 1)
= 12 + 4 + 16
= 32 g
Moles of Acetonitrile
41 g ----------------- 1 mol
77.34g ------------ x
x = (77.34 x 1) / 41
x = 1.89 moles
Moles of Methanol
32 g -------------- 1 mol
17.40 g --------- x
x = (17.40 x 1)/32
x = 0.54 moles
3.- Calculate the mole fraction of Methanol
Total number of moles = 1.89 + 0.54
= 2.43
Mole fraction = moles of Methanol / total number of moles
Mole fraction = 0.54/ 2.43
Mole fraction = 0.22
Answer: The ion formed after the reduction of bromine is 
Explanation:
The electronic configuration of Sodium (Na) = ![[Ne]3s^1](https://tex.z-dn.net/?f=%5BNe%5D3s%5E1)
The electronic configuration of Bromine (Br) = ![[Ar]3d^{10}4s^24p^5](https://tex.z-dn.net/?f=%5BAr%5D3d%5E%7B10%7D4s%5E24p%5E5)
From the above configurations, Sodium ion will loose 1 electron in order to gain stable electronic configuration and that electron is accepted by the Bromine atom because it is 1 electron short of the stable electronic configuration.
(oxidation reaction)
(Reduction reaction)
Bromine atom is reduced to form
Reduction reactions are the reactions in which the element gain electrons.
Oxidation reactions are the reactions in which the element looses its electrons.
The answer is C) catalyst speed up reactions and activation energy
Answer:
Yes chemistry. Try to add then multiply the top. Get the moles and you will find it.
Explanation:
Try to add then multiply the moles in the equation
