Answer:
6.02 × 10²³ atoms
Explanation:
The number 6.02 × 10²³ is called Avogadro number. It is the number of atoms, ions and molecules in one gram atoms of an element, one gram ions of substance and one gram molecule of a compound.
For example:
32 g of oxygen = one mole = 6.02 × 10²³ atoms O.
1.008 g of hydrogen = one mole = 6.02 × 10²³ atoms of H.
or
18 g of H₂O =one mole = 6.02 × 10²³ molecules of H₂O
44 g of CO₂ = one mole = 6.02 × 10²³ molecules of CO₂
or
62 g of NO₃⁻ = one mole of NO₃⁻ = 6.02 × 10²³ ions of NO₃⁻
Answer:
462g
Explanation:
First, let us calculate the molar mass of Cu(CN)2. This is illustrated below:
Molar Mass of Cu(CN)2 = 63.5 + 2(12+14) = 63.5 + 2(26) = 63.5 + 52 = 115.5g/mol
Number of mole of Cu(CN)2 given from the question = 4moles
Mass = number of mole x molar Mass
Mass of Cu(CN)2 = 4 x 115.5
Mass of Cu(CN)2 = 462g
Answer:
Oxygen
Explanation:
First, list out all the ions in the aqueous solution:
Ag+, NO3-
H+, OH-
In the anode, the substance lose electrons to undergo oxidation.
From the 4 ions, only OH- can lose electrons to form water and oxygen,
4OH- ---> O2 + 2H2O + 4e-
While others tend to gain electrons to form new substances instead (they undergo reduction).
Oxygen is the gas produced.
Answer : The equilibrium concentration of
at
is,
.
Solution : Given,
Equilibrium constant, 
Initial concentration of
= 0.260 m
Let, the 'x' mol/L of
are formed and at same time 'x' mol/L of
are also formed.
The equilibrium reaction is,

Initially 0.260 m 0 0
At equilibrium (0.260 - x) x x
The expression for equilibrium constant for a given reaction is,
![K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_2H_3O_2%5E-%5D%7D%7B%5BHC_2H_3O_2%5D%7D)
Now put all the given values in this expression, we get

By rearranging the terms, we get the value of 'x'.

Therefore, the equilibrium concentration of
at
is,
.