Answer:
B
Explanation:
If you are comparing an amount you are trying to find the quantity or the total of something or the size therefore answer B is the right answer
Answer:
1.) Tin oxide + Carbon → Tin + Carbon monoxide.
2.) Lead oxide + Carbon → Lead + Carbon monoxide.
These equations show that the metals are being extracted by their oxides by heating with carbon. This is because the metals (Tin and Lead) are less reactive than carbon. Remember, extraction methods of metals are determined by their reactivity. Since the elements tin and lead are less reactive than carbon, the extraction process of heating with carbon is needed. Remember that when a metal heats with carbon, the carbon displaces the metal from the compound and removes the oxygen from the oxide. Leaving you with the metal and carbon monoxide.
Answer:
The correct answer is 0.165 g NaCl.
Explanation:
The following is the precipitation reaction taking place between sodium chloride and silver nitrate:
NaCl (aq) + AgNO₂ (aq) ⇒ NaNO₃ (aq) + AgCl (s)
The complete ionic reaction of the reaction will be,
Na⁺ + Cl⁻ + Ag⁺ + NO₃⁻ ⇒ AgCl (s) + Na⁺ + NO₃⁻
Hence, the net ionic equation for the mentioned reaction is:
Ag⁺ (aq) + Cl⁻ (aq) ⇒ AgCl (s)
Thus, it can be witnessed that one mole of chloride ion is needed so that one mole of Ag⁺ ion get neutralized. There is a need to find the moles of silver ions present in the solution of AgNO₃ and then transform these moles to the moles of chloride ion. Ultimately, these moles can be converted to the concentration of sodium chloride needed.
The no. of moles of silver ions found in silver nitrate solution is,
(2.50 × 10² mL) × (0.0113 mol Ag⁺/1000 ml solution) = 2.83 × 10⁻³mol Ag⁺
Now the moles of chloride ions needed to precipitate the silver ions is,
(2.83 × 10⁻³ mol Ag⁺ ) × (1 mol Cl⁻/1 mol Ag⁺) = 2.825 × 10⁻³mol Cl⁻
The mass of sodium chloride needed for precipitating the silver ions will be,
mass of NaCl = (2.83 × 10⁻³ mol Cl⁻) × (1 mol NaCl / 1 mol Cl⁻) × (58.44 grams / 1 mol NaCl)
= 0.165 gram NaCl.
