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:
54.30 grams acetone
Explanation:
(0. 935 mol)*(58.08 g/mole) = 54.30 grams acetone
Answer:
238,485 Joules
Explanation:
The amount of energy required is a summation of heat of fusion, capacity and vaporization.
Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)
m (mass of water) = 75 g
Lf (specific latent heat of fusion of water) = 336 J/g
C (specific heat capacity of water) = 4.2 J/g°C
∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C
Lv (specific latent heat of vaporization of water) = 2,260 J/g
Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J
Answer:d
Explanation:I’m not 100% sure how to explain it but I’m almost for sure it’s d.
For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
To know more about hybrisisation:
brainly.com/question/23038117
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