The answer is -1. 2 subtraction signs next to each other for an addition sign
Answer: D is the correct answer
Explanation: When there is no insulator surrounding a metal wire then there is nothing to stop the electrons from flowing outside of the wire, This means that anything the wire comes in contact with could get shocked or burnt when there is a current flowing through the wire.
Remains the same because temperature can't change energy
<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:
<h3>
C. 3</h3>
Explanation:
there are three electrons in a 2p sub-energy level of s neutral nitrogen atom.
