Answer:
pH = 9.48
Explanation:
We have first to realize that NH₃ is a weak base:
NH₃ + H₂O ⇔ NH₄⁺ + OH⁻ Kb = 1.8 x 10⁻⁵
and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.
Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:
pOH = pKb + log ( [ conjugate acid ] / [ weak base ]
mol NH₃ = 0.139 L x 0.39 M = 0.054 mol
mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol
Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)
pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52
pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48
The solution is basic which agrees with NH₃ being a weak base.
<h3>During physical changes, the composition od the original substance is not altered, but the properties of the original substance are altered. During a chemical change the composition od the original substance is not altered and the change is irreversible. Melting of butter and wax is an example of chemical changes.</h3>
Answer:
golfer b had more space giving him the opportunity to gain momentum and strength to strike the ball
Answer:
7.35atm
Explanation:
Data obtained from the question include:
V1 = 28L
T1 = 42°C = 42 + 273 = 315K
P1 =?
V2 = 49L
T2 = 27°C = 27 + 273 = 300K
P2 = 4atm
Using P1V1/T1 = P2V2/T2, the original pressure can be obtained as follows:
P1V1/T1 = P2V2/T2
P1 x 28/315 = 4 x 49/300
Cross multiply to express in linear form
P1 x 28 x 300 = 315 x 4 x 49
Divide both side by 28 x 300
P1 = (315 x 4 x 49) /(28 x 300)
P1 = 7.35atm
Therefore, the original pressure is 7.35atm
