I believe that the answers are 3 and 4
For this problem we can use half-life formula and radioactive decay formula.
Half-life formula,
t1/2 = ln 2 / λ
where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days
Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days
Radioactive decay law,
Nt = No e∧(-λt)
where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.
Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days
By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg
Hence, mass of remaining sample after 13.0 days = 0.499 mg
The answer is "e"
When we have the balanced equation for this reaction:
AB3 ↔ A+3 + 3B-
So we can get Ksp:
when Ksp = [A+3][B-]^3
when [A+3] = 0.047 mol and from the balanced equation when
1 mol [A+3] → 3 mol [B-]
0.047 [A+3] → ??
[B-] = 3*0.047 = 0.141
so by substitution in Ksp formula:
∴Ksp = 0.047 * 0.141^3
= 1.32x10^-4
Answer:
indicator
Explanation:
Indicators are the weak organic dyes which shows different colors in acidic and basic mediums.
Due to the fact that a noticeable pH change occurs near the equivalence point of acid-base titrations, an indicator can be used to signal the end point of a titration.
Therefore, choosing the proper indicator, scientists can minimize the difference in these two numbers, allowing more accurate measurements in the lab.
Answer:
1.98 M
Explanation:
Given data
- Initial volume (V₁): 93.2 mL
- Initial concentration (C₁): 2.03 M
- Volume of water added: 3.92 L
Step 1: Convert V₁ to liters
We will use the relationship 1 L = 1000 mL.
Step 2: Calculate the final volume (V₂)
The final volume is the sum of the initial volume and the volume of water.
Step 3: Calculate the final concentration (C₂)
We will use the dilution rule.
