Answer:
(A) NaOH and Ca(OH)2: Ca(OH)2
(B) MgCl2 and MgF2: MgF2
(C) Agl and KI: AgI
(D) NH4Cl and PbCl2: PbCl2
Explanation:
We need to see the solubility in water, at similar temperatures, for each compound and see which one is less soluble than the other:
NaOH: 1000 g/L (25 °C)
Ca(OH)2: 1.73 g/L (20 °C)
MgCl2: 54.3 g/100 mL (20 °C)
MgF2: 0.013 g/100 mL (20 °C)
Agl: 3×10−7g/100mL (20 °C)
KI: 140 g/100mL (20 °C)
NH4Cl: 383.0 g/L (25 °C)
PbCl2: 10.8 g/L (20 °C)
After the comparison made we can conclude that the less soluble, after saturation of water, will precipitate first.
The net ionic equation is
Ag⁺(aq) +Cl⁻(aq) → AgCl(s)
Explanation
AgNO₃ (aq) + KCl (aq)→ AgCl(s) +KNO₃(aq)
from above molecular equation break all soluble electrolyte into ions
Ag⁺(aq) +NO₃⁻ (aq) + K⁺(aq) +Cl⁻(aq) → AgCl (s) + K⁺(aq) + No₃⁻(aq)
cancel the spectator ions in both side of equation =K⁺ and NO₃⁻ ions
The net ionic equation is therefore
= Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Primary consumer just remember the “primary” (first) consumer so the one that eats a heterotroph
Answer:
The final volume will be 3.39 liters
Explanation:
P1V1/T1 = P2V2/T2
P1 = 3 atm
T1 = 19( degrees celsius) + 273 = 292 K
V2 = TO BE DETERMINED
P2 =1 atm
T2 = 22( degrees celsius) + 273 = 295 k
P1V1/T1 = P2V2/T2
V2 = P1V1T2/P2T1 = (3)(1.12)(295)/(1)292) = 3.39 Liters
Answer: metal
Explanation: it says element so conductor and semiconductor are out and a halogen is a group of gases on the periodic table so this brings up to a conclusion of metal.
NOTE: I am not a professional***
