Answer:
6.23 KOH 90% son necesarios
Explanation:
Una solución 1N de KOH requiere 1equivalente (En KOH, 1eq = 1mol) por cada litro de solución.
Para responder esta pregunta se requiere hallar los equivalentes = Moles de KOH para preparar 100mL = 0.100L de una solución 1N. Haciendo uso de la masa molar de KOH y del porcentaje de pureza del KOH se pueden calcular los gramos requeridos para preparar la solución así:
<em>Equivalentes KOH:</em>
0.100L * (1eq / L) = 0.100eq = 0.100moles
<em>Gramos KOH -Masa molar: 56.1056g/mol-:</em>
0.100moles * (56.1056g/mol) = 5.61 KOH se requieren
<em>KOH 90%:</em>
5.61g KOH * (100g KOH 90% / 90g KOH) =
<h3>6.23 KOH 90% son necesarios</h3>
I don’t know if this is right but I thing It’s B
As long as there is some oxygen, the reaction will occur. If there is an insufficient amount of oxygen, then the oxygen will be the limiting reacting and there will be excess of the other reactants.
Answer:
6.78 × 10⁻³ L
Explanation:
Step 1: Write the balanced equation
Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)
Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)
At STP, 1 mole of H₂O(g) has a volume of 22.4 L.
0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol
Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)
The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.
Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃
At STP, 1 mole of NH₃(g) has a volume of 22.4 L.
3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L
Answer:
Ag+(aq) + Cl-(aq) —> AgCl(s)
Explanation:
2AgNO3(aq) + CaCl2(aq) —>2AgCl(s) + Ca(NO3)2(aq)
The balanced net ionic equation for the reaction above can be obtained as follow:
AgNO3(aq) and CaCl2(aq) will dissociate in solution as follow:
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
CaCl2(aq) —> Ca2+(aq) + 2Cl-(aq)
AgNO3(aq) + CaCl2(aq) –>
2Ag+(aq) + 2NO3-(aq) + Ca2+(aq) + 2Cl-(aq) —> 2AgCl(s) + Ca2+(aq) + 2NO3-(aq)
Cancel out the spectator ions i.e Ca2+(aq) and 2NO3- to obtain the net ionic equation.
2Ag+(aq) + 2Cl-(aq) —> 2AgCl(s)
Divide through by 2
Ag+(aq) + Cl-(aq) —> AgCl(s)
The, the net ionic equation is
Ag+(aq) + Cl-(aq) —> AgCl(s)
