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)
La materia no se crea ni se destruye solo se transforma
Explanation:
«En toda reacción química la masa se conserva, es decir, la masa total de los reactivos es igual a la masa total de los productos».
Los átomos ni de crean ni se destruyen,durante una reacción química.Por lo tanto una ecuación química ha de tener el mismo numero de atomos de cada elemento a ambos lados de la flecha. Se dice entonces que la ecuación esta balanceada
Espero y te sirva!!
Answer:
Explanation:
Here's where all that equation balancing is going to come into use. Since the main object of the question is not the equation, I'm just going to balance it and use it.
4Fe + 3O2 ====> 2Fe2O3
Step One
Find the number of mols of O2 in 24.9 grams of O2
1 mol O2 = 2*16 = 32 grams
x mol O2 = 24.9 grams Cross multiply
32x = 24.9 * 1 Divide by 32
x = 24.9/32
x = 0.778 moles of O2
Step Two
Type the findings under the balanced equations parts. Solve for the number of moles of Fe
4Fe + 3O2 ====> 2Fe2O3
x 0.778
Step Three
Set up the proportion
4/x = 3/0.778 Cross multiply
Step Four
Solve the proportion moles of Fe
4*0.778 = 3x
3.112 = 3x Divide by 3
3.112/3 = 3x/3
x = 1.037 moles of Fe
Step Five
Find the mass of Fe
1 mol Fe = 56 grams
1.037 mol Fe = x Cross Multiply
x = 56*1.037
x = 58.1 grams
Corrected question:
Bromine-88 is radioactive and has a half life of 16.3seconds. What percentage of a sample would be left after 35seconds? Round your answer to 2 significant digits.
Answer:
23% (3 significant digits)
Explanation:
where 
is mass remaining
is original mass
t1/2= 16.3s , t=35s
n =35/16.3 =2.147
= 
0.2258*100% =22.5%
≈23%
