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Why was it necessary to dispose the contents of the potassium iodide solution lead(II) nitrate solution vial into a separate waste container?
It is because of it's side effects like stomach or gastro-intestinal upset, allergic reactions, rashes, and inflammation of the salivary glands. When taken as recommended, KI (potassium iodide) can cause...
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lead-204 203.973amu, 1.40% abundance. lead-206 205.974 amu, 24.10% abundance. lead-207???, abundance 22.10%. lead-208 207.977, abundance 52.40%. Calculate the isotope mass of lead-207
=207.2 amu (Pb)
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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.
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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
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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...
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5.2 moles of calcium nitrate solution are mixed with excess solution of sodium nitrate are formed
10.4
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If 50.0 mL of 0.125 mol/L sodium phosphate solution is reacted with 50.0 mL of 0.125 mol/L magnesium nitrate solution, what is the predicted mass of precipitate in g?
Solution. 0.00625 0,00625 y2Na3PO4 + 3Mg(NO3)2 → Mg3(PO4)2↓ + 6NaNO3 2 3 1V(Na3PO4) = 50,0 mL = 0,05 L; V(Mg(NO3)2) = 50,0 mL = 0,05 L;n(Na3PO4) = c*V = 0,125*0,05 = 0,00625 mol;n(Mg(NO3)2) = c*V = 0,125*0,05 = 0,00625 mol;Let n(Mg(NO3)2)...
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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...
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5kg of potassium is reacted with excess sulfur to make potassium sulfide. Work out the mass of potassium sulfide produced.
128.2 mole x mole2K + S -> K2S2 mole 1 molen(K) = m(K)/M(K) = 5 000 g/39 g/mole = 128.2 mole128.2 mole/2 mole = x mole/1 molex = (128.2 *...
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When lead nitrate reacts with sodium iodide, sodium nitrate and lead iodide are formed.
This is the balanced equation of the reaction Pb(NO3)2 +2NaI → 2NaNO3 + PbI2
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