Answer: the answer is C oxidizing
I believe the answer is <span>can be elements or compounds
In this case, elements in the decomposition reaction is the substance that cannot be separated into simpler substances.
Compounds, technically act as a reactant in the decomposition reaction, but since the reaction breakdown one substance into two or more, sometimes it exists in the product.</span>
Answer:
Explanation:
We will need a balanced chemical equation with masses, moles, and molar masses.
1. Gather all the information in one place:
Mᵣ: 18.02
2Na + H₂O ⟶ 2NaOH + H₂
m/g: 72.0
2. Moles of H₂O
3. Moles of Na
The molar ratio is 2 mol Na/1 mol H₂O.
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
Answer:
4.1x10⁻⁵
Explanation:
The dissociation of an acid is a reversible reaction, and, because of that, it has an equilibrium constant, Ka. For a generic acid (HA), the dissociation happens by:
HA ⇄ H⁺ + A⁻
So, if x moles of the acid dissociates, x moles of H⁺ and x moles of A⁻ is formed. the percent of dissociation of the acid is:
% = (dissociated/total)*100%
4.4% = (x/[HA])*100%
But x = [A⁻], so:
[A⁻]/[HA] = 0.044
The pH of the acid can be calcualted by the Handersson-Halsebach equation:
pH = pKa + log[A⁻]/[HA]
3.03 = pKa + log 0.044
pKa = 3.03 - log 0.044
pKa = 4.39
pKa = -logKa
logKa = -pKa
Ka = 
Ka = 
Ka = 4.1x10⁻⁵
