2 NaCl --------> 2 Na + Cl₂
2 mol * 23 g Na = 46 g
2*35,5 g Cl = 71
46 g Na --------- 71 g Cl₂
460 g Na -------- ?
mass ( Cl₂) = 460 . 71 / 46
Mass (Cl₂) = 32660 / 46
m= 710 g of Cl₂
<span>hope thips helps</span>
<h3>
Answer:</h3>
812 kPa
<h3>
Explanation:</h3>
- According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
- Mathematically,
At varying pressure and volume;
P1V1=P2V2
In this case;
Initial volume, V1 = 2.0 L
Initial pressure, P1 = 101.5 kPa
Final volume, V1 = 0.25 L
We are required to determine the new pressure;
Replacing the known variables with the values;
= 812 kPa
Thus, the pressure of air inside the balloon after squeezing is 812 kPa
Answer:
Explanation:
Hello,
In this case, given the reaction, we can directly compute the theoretically yielded grams of carbon dioxide, considering the 2:2 molar ratio between carbon monoxide (molar mass = 28 g/mol) and carbon dioxide (molar mass = 44 g/mol) and the initial reacting grams of carbon monoxide in excess oxygen:
Thus, as only 8 g were actually yielded, we compute the percent yield:
Best regards.
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
