An example can be perspiration or even rain.
Hope this helps.
Answer:
electron: orbiting around the nucleus
proton: inside the nucleus
neutron: inside the nucleus
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<span>Answer:
it shows that 1mol mCPHA provides the oxygens to 1 mol of propene, to make 1 mole of C3H6O
so: 1 mol C3H6 & 1 mol mCPHA --> 1 mol C3H6O
using molar masses, that equation becomes:
42.08grams C3H6 & 172.57grams mCPHA --> 58.08grams C3H6O
which is: 42.08 kg C3H6 & 172.57 kg mCPHA --> 58.08 kg C3H6O
to produce 1 kg of C3H6O, this becomes:
42.08 / 58.08 kg C3H6 & 172.57 / 58.08 kg mCPHA --> 58.08 /58.08 kg C3H6O
which is: 0.72452 kg C3H6 & 2.9712 kg mCPHA --> 1 kg C3H6O
but because the reaction gives only a 96% yield,
we scale up the reactants to get that desired 1 kg of C3H6O
(0.72452 kg ) (100/96) C3H6 & (2.9712 kg) (100/96) mCPHA --> 1 kg C3H6O
which is: 0.75471 kg C3H6 & 3.095 kg mCPHA --> 1 kg C3H6O
=========
costs per kg of C3H6O produced:
(0.75471 kg C3H6) ($10.97 per kg) = $8.279
(3.095 kg mCPHA) ($5.28 per kg) = $16.342
&
(0.75471 kg C3H6) / (0.0210 kg C3H6 / L dichloromethane) = 35.939 Litres dichloromethane
(35.939 Litres dichloromethane) ($2.12 per L) = $ 76.19
&
waste disposal is $5.00 per kilogram of propene oxide produced
total cost, disregarding labor,energy, & facility costs:
$8.279 & $16.342 & $ $ 76.19 & $5.00 = $105.81 per kg C3H6O produced
==========
profit: ($258.25 / kg C3H6O) - ($105.81 cost per kg) = $152.44 profit /kg
“Calculate the profit from producing 75.00kg of propene oxide”
(75.00kg) ($152.44 /kg) = $11,433
that answer rounded off to four sig figs, is $11,430</span>
Answer:
AgBr
Explanation:
The salt, AgBr has a very low solubility is pure water. However, it has a high solubility in 1 M NH3. The reason behind this higher solubility of AgBr in 1 M aqueous ammonia solution is the formation of a complex as shown below;
AgBr(s) + 2NH3(aq) ----> [Ag(NH3)2]^+(aq) + Br^-(aq)
The formation of this linear silver diammine complex accounts for the higher solubility of AgBr in 1 M aqueous ammonia solution.
