The common neutralization reaction that involve NaOH reacting with HNO3 produces
NaNO3 and H2O
The equation for reaction is as folows
NaOH + HNO3 = NaNO3 + H2O
that is 1 mole of NaOH reacted with 1 mole of HNO3 to form 1 mole of NaNO3 and 1 mole of H2O
Reactants- Water, Light, Carbon dioxide
Products- Oxygen and Glucose
I a chemical reaction, we have the reactants in the left part,followed by an arrow and then the products.
So, the reactants are the two that are in the left part of the equation:
Sodium hydroxide and Hydrochloric acid
1. Answer:
1.0 × 10–9 M OH–
Explanation:
pH = -Log[H+]
pOH = -Log[OH-]
But;
pH + pOH = 14
Therefore;
[H+] + [OH-] = 1.0 × 10^-14 M
Therefore;
[OH-] = 1.0 × 10^-14 M - (1.0 × 10^–5 M)
= 1.0 × 10^-9 M OH–
2. Answer;
pH = 7.28
Explanation;
pH = -Log[H3O+]
Given;
[H3O+] = 5.2 × 10^–8 M
Therefore;
pH = - log [5.2 × 10^–8 M]
= 7.28
The pH is 7.28
This question is quite vague, as the initial concentration of ethanol is not provided. However, from experience I can tell you that most laboratory work is done with 98% ethanol, and not absolute ethanol (100%). So in order to calculate the final concentration, we need to take the given values, which includes the initial concentration (98%), the initial volume (50.0mL) and the final volume (100.0mL). We apply the following equation to calculate the final concentration:
C1V1 = C2V2
C1 = Initial concentration
C2 = Final concentration
V1 = Initial volume
V2 = Final volume
(98%)(50.0mL) = (C2)(100.0mL)
Therefore, the final concentration (C2) = 49%
