The mass of zinc required to react with HNO3 is 3.45 g
<h3>What is moles of HNO3 reacting?</h3>
The moles of HNO3 reacting is calculated using the formula below:
- moles = concentration × volume
volume = 0.850
volume = 125.0 mL = 0.125 L
Moles of HNO3 = 0.850 × 0.125 = 0.10625 moles
Equation of reaction:
Zn + 2 HNO3 --> Zn(NO3)2 + H2
2 moles of HNO3 reacts with 1 mole of Zn.
moles of zinc reacting = 0.10625/2 = 0.053125
Mass of zinc = moles × molar mass
molar mass of Zn = 65
Mass of zinc = 0.053125 × 65
mass of Zn = 3.45 g
Therefore, the mass of zinc will completely react with 125.0 mL of a solution of nitric acid (HNO3) which has concentration of 0.850M is 3.45 g
Learn more about moles at: brainly.com/question/15356425
Hi !
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Hood it helps and have a wonderful day!
This I believe would be double covalent bonds
That both of them are warm....at a certain temputurature causing one to fell,cozy
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
