4, 2, 1, and 3 is the answer
Answer:
Hey there!
This is already rounded to four significant figures!
Zeroes after the decimal but before the 7 don't count, and 7, 0, 6, and 2 count as significant figures.
So, the answer would be 0.007062.
Let me know if this helps :)
Answer:
20 neutrons
Explanation:
(not really any just look at a periodic table)
<span>a. Use PV = nRT and solve for n = number of mols O2.
mols NO = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols O2 to mols NO2. Do the same for mols NO to mols NO2. It is likely that the two values will not be the same which means one is wrong; the correct value in LR (limiting reagent) problems is ALWAYS the smaller value and the reagent producing that value is the LR.
b.
Using the smaller value for mols NO2 from part a, substitute for n in PV = nRT, use the conditions listed in part b, and solve for V in liters. This will give you the theoretical yield (YY)in liters. The actual yield at these same conditions (AY) is 84.8 L.
</span>and % will be 60%.
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
