Answer:
The swirling yellow solid formed is lead iodide (PbI₂).
Explanation:
- The reaction of potassium iodide (KI) with lead nitrate (Pb(NO₃)₂) will produce lead iodide (PbI₂) and potassium nitrate (KNO₃) according to the equation:
2KI + Pb(NO₃)₂ → PbI₂↓ + 2KNO₃
- Lead iodide (PbI₂) is a yellow swirling precipitate that is formed from the reaction.
Answer:
<h3>I think, answer is threshold energy.</h3>
OR
<h3>activation energy. </h3>
Explanation:
<h3>Hope it helps you....</h3>
<h2>Thank you..</h2>
<u>We are given:</u>
Mass of ice = 21 grams
The ice is already at 0°c, the temperature at which it melts to form water
Molar heat of fusion of Ice = 6.02 kJ/mol
<u>Finding the energy required:</u>
<u>Number of moles of Ice: </u>
Molar mass of water = 18 g/mol
Number of moles = given mass/ molar mass
Number of moles = 21 / 18 = 7/6 moles
<u>Energy required to melt the given amount of ice:</u>
Energy = number of moles * molar heat of fusion
Energy = (7/6) * (6.02)
Energy = 7.02 kJ OR 7020 joules
Reaction of option c produces precipitate.
Rhodium on reacting with potassium phosphate produces rhodium phosphate which remain in solution due to low lattice energy for rhodium phosphate.
Niobium on reacting with lithium carbonate produces niobium carbonate and it will remain in aqueous form.
Cobalt on reacting with zinc nitrate produces cobalt nitrate. This, Co(NO3 )2 is insoluble precipitate and settles at bottom whereas zinc ion will remain in solution as follows:
Potassium ion on reacting with sodium sulfide produces potassium sulfide which remain in solution
