K:
m=155g
M=39g/mol
n = 155g / 39g/mol ≈ 3,97mol
KNO₃:
m=122g
M=101g/mol
n = 122g/101g/mol = 1,21mol
2K + 10KNO₃ ⇒ 6K₂O + N₂
2mol : 10mol
3,97mol : 1,21mol
limiting reagent
KNO₃ is limiting reagent
Hi :)
The bottom number tells you how many protons you have in the nucleus of this element.
In Krypton-84, this means that you have 84 nucleons, where 36 of these are protons, and the remaining 48 are neutrons.
Hope this helps :)
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
Answer: permanent removal of hair by energy or heat
Explanation:
