Answer:
0.144M
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
HNO3 + KOH —> KNO3 + H20
From the equation,
nA = 1
nB = 1
From the question given, we obtained the following:
Ma =?
Va = 30.00mL
Mb = 0.1000M
Vb = 43.13 mL
MaVa / MbVb = nA/nB
Ma x 30 / 0.1 x 43.13 = 1
Cross multiply to express in linear form
Ma x 30 = 0.1 x 43.13
Divide both side by 30
Ma = (0.1 x 43.13) /30 = 0.144M
The molarity of the nitric acid is 0.144M
Answer: YOU MAY FIND THE ANSWER IF YOU SEARCH THE WEB.
Explanation:
Answer:
SAMPLE A - pure substance.
SAMPLE B - homogeneous mixture.
SAMPLE C - heterogeneous mixture.
Explanation:
the answer to your question is
M1V1=M2V2
V2=750-150=600 ml
0.75M*750 ml = M2*600
M2=0.75*750/600 ≈ 9.38 M
Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
