The number of mole of Ca reacted is:
4.86 g Ca/ (40.08 g/mol Ca)= 0.121 mol Ca
Because Ca reacted completely with oxygen and there is 2 mol Ca, there is 1 mol O2 reacted.
Total mass of oxygen that reacted is:
0.121 mol Ca* (1mol O2/ 2 mol Ca)* (32 g O2/ 1 mol O2)= 1.94 g O2 reacted.
Hope this would help~
<span>A substance is composed entirely of one type of atom, it is known as a Chemical element
because mixture contain more than one type of atom and it cannot be neutron or isotope
so correct option is B
hope it helps</span>
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
Answer:
Ammonia gas(an alkaline gas with characteristics of choking or irritating smell) is not liberated when 6mole of HCl is added to the solution instead of 6mole of NaOH, to test for the presence of ammonium ion in the solution
Explanation:
As expected, when testing for ammonium ion in a solution (precisely ammonium salt solution), Sodium Hydroxide (NaOH) is required as the test reagent.
When NaOH is added to the solution, A gas with characteristics of choking or irritating smell is liberated.
This gas turn red litmus paper blue.
This liberated gas is an alkaline gas, which is confirmed as an ammonia gas(NH3).
If HCl is added instead of NaOH, the ammonia gas will not be liberated, which indicates that the test reagent used is wrong.
<span>Answer:
</span><span>
</span><span>
</span><span>Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)</span><span />
<span>Explanation:
</span>
<span>1) Combine the cation Li⁺ (aq) with the anion Cl- (aq) to form LiCl(s).
</span>
<span>LiCl is a solid soluble substance, a typical ionic compound. So, it will reamain as separate ions in the product side: Li⁺ + CL⁻</span>
<span>2) Combine the anion OH⁻ with the cation H⁺ to form H₂O(l).
</span>
<span>Since, the ionization of H₂O is low, it will remain as liquid in the product side: H₂O(l)</span>
<span>3) Finally, you can wirte the total ionic equation:
</span>
Li⁺ (aq) + OH⁻ (aq) + H⁺ (aq) + Cl⁻(aq) → Li⁺ (aq) + Cl⁻ (aq) + H₂O(l)
