Ok so this is what we know :
2KClO3 -> 2KCl + 3O2 (Always check if equation is balanced - in this case it is)
4.26moles
So we know that we have 4.26 moles of oxygen (O2). Now lets look at the ratio between KClO3 and O2.
We see that the ratio is 2:3 meaning that we need 2KClO3 in order to produce 3O2.
Therefore divide 4.26 by 3 and then multiply by 2.
4.26/3 = 1.42
1.42 * 2 = 2.84
Now we know that the molarity of KClO3 is 2.84 moles.
Multiply by R.M.M to find how many grams of KClO3 we have.
R.M.M of KClO3
K- 39
Cl- 35.5
3O- 3 * 16 -> 48
---------------------------
<span>122.5
</span>2.84 * 122.5 = 347.9 grams therefore the answer is (a)
348 grams needed of KClO3 to produce 4.26 moles of O2.
Hope this helps :).
Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten.
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
F = ma = (kg)(m/s2) = kg ´ m/s2 N
hope this helps :D
Answer:
30.6 g of C is formed.
Explanation:
2A + B → C
Average rate of reaction = 2[A]/Δt = [B]/Δt = [C]/Δt
Average rate of reaction = [C]/Δt
Average rate of reaction = 15 g / 9 min
Average rate of reaction = 1.7 g of C / min
Average rate of reaction = [C]/Δt
[C] = Average rate of reaction x Δt
[C] = 1.7 g of C / min x 18 min
[C] = 30.6 g of C
Since the Carbon C is 17.39% by mass hence the Fluorine F
is 82.61% by mass. Divide each mass % by the respective molar masses, that is:
C = 17.39 / 12 = 1.45
F = 82.61 / 19 = 4.35
Divide the two by the smaller number, so divide by 1.45
C = 1.45 / 1.45 = 1
F = 4.35 / 1.45 = 3
So the empirical formula is:
CF3
