NH3 = 14 +3*1=17 g/mol
Mass = mol * molar mass = 0.687 * 17= 11.679 g
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Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
There's a slight error in your equation. I think you were trying to present it like this:
2C8H18 + 25O2 -> 16CO2 + 18H2O
Mole Ratio
O2 : H20
25 : 18
? moles : 18 moles
(18/18)×25 : 18 moles
25 moles : 18 moles
Final answer would be 25 moles of O2. :)
If you have any doubts that you want to clarify with me, please ask me! :)
I will do my utmost best to help you.
Answer:
<h2>0.06 % </h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula
From the question
error = 500 - 499.7 = 0.3
actual volume = 500 mL
We have
We have the final answer as
<h3>0.06 % </h3>
Hope this helps you
