1 Answers
Call
Talk Doctor Online in Bissoy App
Reaction between lead(ii) nitrate and sodium iodide to produce lead(ii) iodide and sodium nitrate
Pb(NO3)2 (aq) + 2 NaI (aq) → PbI2 (s) + 2 NaNO3 (aq)2Na(+) +2I(-) + Pb(2+) + 2NO3(-) = 2Na(+) +PbI2(s) + 2NO3(-)2I(-) + Pb(2+) = PbI2 (s)Lead iodide appears as a yellow crystalline solid. Insoluble in water.
1 Answers 1 views
Which one is more reactive? A- Methyl iodide B- Ethyl iodide C- Propyl iodide D- Isopropyl iodide Please address the query as soon as possible. Thank You!
B- Ethyl iodide
1 Answers 1 views
write balanced chemical equations for each, remember to make your molecules neutral first. Sodium iodide and potassium hydroxide form when aqueous iodide is mixed with sodium hydroxide solution.
KI+NaOH=NaI+KOH
1 Answers 1 views
If 86.33 grams of sodium reacts with an excess of chlorine, what amount of Salt (sodium chloride) will be produced? Sodium + chlorine ® sodium chloride Na + Cl2 ® NaCl
Solution:Sodium + chlorine → sodium chloride The balanced chemical equation:2Na + Cl2 → 2NaClAccording to the equation above: n(Na) = n(NaCl)Moles of Na = Mass of Na / Molar mass of NaThe molar...
1 Answers 1 views
If you start with 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?
Pb(NO3)2 + 2NaI = 2NaNO3 + PbI2n = m / MM (Pb(NO3)2) = 331.2 g/molM (NaI) = 150 g/molM (NaNO3) = 85 g/moln (Pb(NO3)2) = 25 / 331.2 = 0.08...
1 Answers 1 views
A 0.4230 g sample of impure sodium nitrate was heated, converting all the sodium nitrate to .1235 g of sodium nitrate and oxygen gas. Determine the percent of sodium nitrate in the original sample.
Answer:2HNO3 = 2HNO2 + O2m= 85 x 0.1235 / 69= 0.1521W%= 0.1521 / 0.4230 = 36%
1 Answers 1 views
lead (II) sulfide has a composition of 86.6% lead and 13.4% sulfur . if 10.5 g of lead reacts to form lead (II) sulfide . What mass of sulfur reacts?
86.6/201:13.4/32=1:1.PbS.Pb+S=PbS.(10.5/201) x 32= 1.67164179 g of S
1 Answers 1 views
Which of the above equilibria is forced to shift furthest to the left? Explain.
- Lead nitrate and Sodium chloride
- Lead chloride and Sodium chromate
-Lead chromate and Sodium sulphide
Answer: Combination which makes less soluble product shift towards left side of the equation.Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3PbCl2 + Na2CrO4 = PbCrO4 + 2NaClPbCrO4 + Na2S = PbS...
1 Answers 1 views
28.7 mL of 2.3 M lead (II) nitrate solution reacted with excess potassium iodide. What mass of yellow precipitate formed?
15.2153
1 Answers 1 views
If 58.26 g of iodine reacts with excess aluminum. If 56.11 g of aluminum iodide is actually formed in the reaction percent yield of aluminum iodide
The reaction is as following:2Al + 3I2 = 2AlI3The theoretical yield of the reaction is:m(AlI3) = m(I2) × [2 × Mr(AlI3)] / [3 × Mr(I2)] = 58.26 g × 2...
1 Answers 1 views