In an acidic solution, the concentration of H+ is greater than the concentration of OH-. The pH will be less than 7.
In a basic solution, the concentration of OH- is greater than the concentration of H+. The pH will be greater than 7.
In a neutral solution, the concentration of H+ ions to OH-ions will be equal, and will therefore have a pH of 7. (This is due to water autoionization, which we usually ignore because it is small in other circumstances.)
LBr is ionic compound because k for potassium is metal which means it’s on the left side of the periodic table and Br which is bromine is a non metal which means is on the right side of the periodic table.
In Conclusion when a non metal and metal come together the are called ionic compound
The enthalpy of this reaction is -5315 KJ/mol.
The equation of the reaction is;
2C4H10(g) + 13O2(g) -----> 8CO2 (g) + 10H2O(g)
We know that the enthalpy of reaction can be obtained from the enthalpy of formation of the reactants and products as follows;
ΔHrxn = ΔHf(products) - ΔHf(reactants)
We have the following information from the question;
ΔHf C4H10 = - 125. 6 kJ/mol
ΔHf CO2 = - 393. 5 kJ/mol
ΔHf H2O = - 241. 82 kJ/mol
ΔHf O2 = 0 KJ/mol
Hence;
[(8 × (- 393. 5 )) + (10 × (- 241. 82))] - [2( - 125. 6))]
= -5315 KJ/mol
Learn more: brainly.com/question/13164491
Answer:
The equation for the rate of this reaction is R = [NO] + {O2}
Explanation:
The rate-determining step of a reaction is the slowest step of a chemical reaction which determines the rate (speed) at which the overall reaction would take place.
Reaction mechanism:
The slow and fast reactions both have NO3 which is cancelled out on both sides, in order to get the overall reaction.
The rate law for this reaction would be that for the rate determining step:
R = [NO] + {O2}
Answer:
C
Explanation:
When the Kb is given, the Henderson-Hasselbalch equation can be used to calculate the pOH of a buffer solution:
pOH = pKb + log ([A⁻] / [HA]) = -log(Kb) + log ([BH+] / [B])
Here, moles can be used in place of the concentration since the pairs listed are both dissolved in 5L, which cancel due to the fraction in the logarithm.
a) pOH = -log(1.8 x 10⁻⁵) + log(1.5/1.0) = 4.92
pH = 14 - pOH = 14 - 4.92 = 9.08
b) pOH = -log(1.8 x 10⁻⁵) + log(1.0/1.5) = 4.57
pH = 14 - pOH = 14 - 4.57 = 9.43
c) pOH = -log(1.7 x 10⁻⁹) + log(1.5/1.0) = 8.95
pH = 14 - pOH = 14 - 8.95 = 5.05
d) pOH = -log(1.7 x 10⁻⁹) + log(1.0/1.5) = 8.59
pH = 14 - pOH = 14 - = 5.41
