I believe it is b. I hope this helps you
Answer:
semiconducting,tellurium
Explanation:
just completed the assignment
Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Answer:
0.01144L or 1.144x10^-2L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 20.352 mL
P1 (initial pressure) = 680mmHg
P2 (final pressure) = 1210mmHg
V2 (final volume) =.?
Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:
P1V1 = P2V2
680 x 20.352 = 1210 x V2
Divide both side by 1210
V2 = (680 x 20.352)/1210
V2 = 11.44mL
Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:
1000mL = 1 L
11.44mL = 11.44/1000 = 0.01144L
Therefore the volume of the container is 0.01144L or 1.144x10^-2L
Answer:
It has denatured
Explanation:
When the temperature get high the enzymes tend to change shape and denaturing occurs.