Decreased, this is due to the roots of the plants holding the soil together
It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.
Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.
Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.
Using the formula;
k = 0.693/t1/2
k = 0.693/21 hours = 0.033 hr-1
Using;
N=Noe^-kt
N = mass of radioactive sample at time t
No = mass of radioactive sample initially present
k = decay constant
t = time taken
Substituting values;
2.5/320= e^- 0.033 t
0.0078 = e^- 0.033 t
ln (0.0078) = 0.033 t
t = ln (0.0078)/-0.033
t = 147 hours
Learn more: brainly.com/question/6111443
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:
There are many errors possible while titrating the acid of an unknown concentration with a base like NaOH.
Main error that leads to the error in results is misreading of the end point volume .
End point is when the reaction between the analyte and solution of known concentration has stopped .
Sometimes Burette is not straight enough to read the volume of the end point. One way to misread the volume of burette is by looking at the burette volume at an angle .
From above , volume seems to be higher. Indicators are used to indicate the color change of the reaction. In Acid-Base titrations , indicators first lighten up then changes its color.
So, error may have occurred in wrongly judging of the end point by color change of the indicator .
Answer: C) middle 50 percent of the data
The interquartile range (IQR) spans from the first quartile Q1 to the third quartile Q3.
25% of the data is below Q1 and 75% of the data is below Q3. The gap between the two endpoints consists of 75-25 = 50 percent of the data, or half of the data.
