Answer:
Indeed, most ionic compounds contain polyatomic ions. Well-known examples are sodium hydroxide (NaOH) with OH- as the polyatomic anion, calcium carbonate (CaCO3), and ammonium nitrate (NH4NO3), which contains two polyatomic ions: NH+ and NO3-.
Explanation:
hope it helps
Answer: Oil: covalent
Cornstarch: Covalent
Sodium chloride: Ionic
Sodium bicarbonate: Ionic
Explanation: Covalent compounds are formed by sharing of electrons between non metals whereas ionic compounds are formed by transfer of electrons from metals to non metals.
1. Oil, which is built from the nonmetals hydrogen, carbon, and oxygen: forms a covalent compound by sharing of electrons between non metals hydrogen, carbon, and oxygen. Covalent compounds are insoluble in water.
2. Cornstarch, a carbohydrate consisting of hydrogen, carbon, and oxygen: forms a covalent compound by sharing of electrons between non metals hydrogen, carbon, and oxygen. Covalent compounds are insoluble in water.
3. Sodium chloride (table salt), whose formula is NaCl is formed by transfer of electrons from sodium to chlorine.Ionic compounds are soluble in water.
4. Sodium bicarbonate, whose formula is is formed by transfer of electrons from sodium to .Ionic compounds are soluble in water.
Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂ * * 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂ * = 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be:
As i am reading the problem, i notice they give two concentrations (M), a volume and asking for the other volume; this should be a hint that we need to use the dilution formula---> M1V1 = M2V2
M1= 2.55 M
V1= ?
M2= 0.75 M
V2= 125 mL
now, we plug in the values into the formula
(2.55 x V1) = (0.75 x 125)
<span>V1= 36.8 mL</span>