The Equation of reaction
C + O2 === CO2
First Blank spot would be
---- 3.5moles of O2
2nd spot
Up.... 1mole of CO2
Down ---- 2moles of Oxygen atoms
3rd spot
Up... --- 22.4L of CO2(1mole of any gas at STP contains 22.4L)
Down---- 1mole of CO2
Final answer --- 3.5 x 1/2 x 22.4
=39.2L of CO2
1 mole of nacl has 1 mole of na ions
therefore 2 moles of nacl has 2 moles of na ions
Mole x L
= 2x6.02x10^23
=12.04 x 10^23
Let's investigate the substances involved in the reaction first. The compound <span>CH3NH3+Cl- is a salt from the weak base CH3NH2 and the strong acid HCl. When this salt is hydrated with water, it will dissociate into CH3NH2Cl and H3O+:
CH3NH3+Cl- + H2O </span>⇒ CH3NH2Cl + H3O+
Nest, let's apply the ICE(Initial-Change-Equilibrium) table where x is denoted as the number of moles used up in the reaction:
CH3NH3+Cl- + H2O ⇒ CH3NH2Cl + H3O+
Initial 0.51 0 0
Change -x +x +x
-------------------------------------------------------------------------------
Equilibrium 0.51 - x x x
Then, let's find the equilibrium constant of the reaction. Since the reaction is hydrolysis we use KH, which is the ratio of Kw to Ka or Kb. Kw is the equilibrium constant for water hydrolysis which is equal to 1×10⁻¹⁴. Since the salt comes from the weak base, we use Kb. Since pKb = 3.44, then. 3.44 = -log(Kb). Thus, Kb = 3.6307×10⁻⁴
KH = Kw/Kb = (x)(x)/(0.51 - x)
1×10⁻¹⁴/ 3.6307×10⁻⁴ = x²/(0.51-x)
x = 3.748×10⁻⁶
Since x from the ICE table is equal to the equilibrium concentration of H+, we can find the pH of the aqueous solution:
pH = -log(H+) = -log(x)
pH = -log ( 3.748×10⁻⁶)
pH = 5.43
The answer is c
i hope I'm right sorry if I'm wrong
Answer:
The more fraction NaoH is 39.998g/mol
