The number of lone pairs that are most likely found on the central atom is zero. There are no lone pairs found on the central atom.
Answer:
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Explanation:
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Answer:
<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
Explanation:
given amount of salt at time t is A(t)
initial amount of salt =300 gm =0.3kg
=>A(0)=0.3
rate of salt inflow =5*0.4= 2 kg/min
rate of salt out flow =5*A/(200)=A/40
rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow
integrating factor
integrating factor 
multiply on both sides by
integrate on both sides
b)
after long period of time means t - > ∞
<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
Answer:
100.52
Explanation:
from the ideal gas equation PV=nRT
for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.
So n i.e no of moles will also be constant.
no of moles of Ar=3.224/40=0.0806
no of moles of unknown gas=0.0806
molecular wt of unknown gas=8.102/0.0806=100.52
Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
