You can not see between opaque materials because light cannot pass through them.
STRUCTURE OF BROMOUS ACID: H–O–Br=O
<span>In this structure, all the elements have a formal charge of
zero. The formal charge of each element is calculated below: </span><span>
H: 1 – 1/2(2) – 0 = 0
O: 6 – 1/2(4) – 4 = 0
Br: 7 – 1/2(6) – 4 = 0
<span>O: 6 – 1/2(4) – 4 = 0</span></span>
The correct answers are A and C.
Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
The strength of an Arrhenius base determines percentage of ionization of base and the number of OH⁻ ions formed.
Strong base completely ionize in water and gives a lot of hydroxide ions (OH⁻), for example sodium
hydroxide: NaOH(aq) → Na⁺(aq)
+ OH⁻(aq).
Weak base partially ionize in water and gives a few hydroxide ions (OH⁻), for example ammonia: NH₃ + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq).
