Answer:
Explanation:
In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases, and a decrease in pressure (increase in volume) favors the net reaction that increases the total number of moles of gases.
Δn= b - a
Δn= moles of gaseous products - moles of gaseous reactants
Therefore, <u>after the increase in volume</u>:
- If Δn= −1 ⇒ there are more moles of gaseous reactants than gaseous products. The equilibrium will be shifted towards the products, that is, from left to right, and K>Q.
- If Δn= 0 ⇒ there is the same amount of gaseous moles, both in products and reactants. The system is at equilibrium and K=Q.
- Δn= +1 ⇒ there are more moles of gaseous products than gaseous reactants. The equilibrium will be shifted towards the reactants, that is, from right to left, and K<Q.
Answer:
The bombarding particle is a Proton
Explanation:
A Nuclear transmutation reaction occurs when radioactive element decay, usually converting them from one element/isotope into another element. Transmutation is the process which causes decay, generally, alpha or beta.
¹⁶₈O(P,alpha) ¹³₇N, can be written as
¹⁶₈O + x goes to ¹³₇N + ⁴₂He
Where x can be anything, balancing the equation in order to give us the correct amount of proton number and nucleus number
16 + x = 13 + 4
x = 17 – 16 = 1, Hence we can say that x = ¹₁P
<u>¹⁶₈O + ¹₁P goes to ¹³₇N + ⁴₂He</u>
Here we can clearly see the bombarding particle is ¹₁P (proton). The ejected particle being ⁴₂He which is also known as an alpha particle
32.6 grams divided by the molar mass of C2H6, which is 18.0584g/mol = 1.8 moles of C2H6.
As there are two carbon atoms per C2H6, we must multiply the number of moles of C2H6 by 2 to get the number of moles of Carbon which is 3.6 moles.
The answer is 3.6 moles.
Hope this helps.
(Sorry for previously incorrect answer)
Answer:
V = 134.5 L
Explanation:
Given data:
Number of moles of KClO₃ = 4 mol
Litters of oxygen produced at STP = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Now we will compare the moles of KClO₃ with oxygen.
KClO₃ : O₂
2 : 3
4 ; 3/2×4 = 6 mol
Litters of oxygen at STP:
PV = nRT
V = nRT/P
V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm
V = 134.5 L / 1
V = 134.5 L
