The Ka : 1.671 x 10⁻⁷
<h3>Further explanation</h3>
Given
Reaction
HA (aq) + H2O (l) ←→ A- (aq) + H3O+ (aq).
0.3 M HA
pH = 3.65
Required
Ka
Solution
pH = - log [H3O+]
![\tt [H_3O^+]=10^{-3.65}=2.239\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20%5BH_3O%5E%2B%5D%3D10%5E%7B-3.65%7D%3D2.239%5Ctimes%2010%5E%7B-4%7D)
ICE method :
HA (aq) ←→ A- (aq) + H3O+ (aq).
0.3 0 0
2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴
0.3-2.239.10⁻⁴ 2.239.10⁻⁴ 2.239.10⁻⁴
![\tt Ka=\dfrac{[H_3O^+][A^-]}{[HA]}\\\\Ka=\dfrac{(2.239.10^{-4}){^2}}{0.3-2.239.10^{-4}}\\\\Ka=1.671\times 10^{-7}](https://tex.z-dn.net/?f=%5Ctt%20Ka%3D%5Cdfrac%7B%5BH_3O%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%5C%5C%5C%5CKa%3D%5Cdfrac%7B%282.239.10%5E%7B-4%7D%29%7B%5E2%7D%7D%7B0.3-2.239.10%5E%7B-4%7D%7D%5C%5C%5C%5CKa%3D1.671%5Ctimes%2010%5E%7B-7%7D)
It’s 34 I had this question
Answer:
119
Explanation:
Tin is present in group sixteen. Its atomic number is 50 and atomic weight is 119 amu.
The atomic number is the number of electrons or number of protons. The atomic mass is equal to the sum of number of protons and number neutrons.
Tin have 50 protons and 69 neutrons.The atomic weight will be 50+69 = 119 amu.
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons.
Answer:
i think A.
Explanation:
bc B. makes the molecules hight temp = move faster = more collisions = higher rate, C. it's use is to make reactions rates increase, D. somthung abt more surface area and easier collisions
The balanced chemical reaction is written as:
2H2+O2 = 2H2O
We are given the amount of H2 gas to be used in the reaction. This will be used as the starting point for the calculations. We calculate as follows:
.050 L H2 ( 1 mol H2 / 22.4 L H2 ) ( 1 mol O2 / 2 mol H2 ) ( 22.4 L / 1 mol ) = 0.025 L O2 or 25 mL O2