We can calculate how long the decay by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span><span>From the half-life data, we can calculate for k.
</span>
1/2(Ao) = Ao e^-k(30)
<span>k = 0.023
</span>
0.04Ao = Ao e^0.023(t)
<span>t = 140 sec</span>
Initial concentration of magnesium nitrate M1 = 2.13 M
Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L
Final concentration of MgNO3, M2 = 1.60 M
Let the final volume of MgNO3 upon dilution be V2
Formula to use:
M1*V1 = M2*V2
V2 = M1*V1/M2
= 2.13 M * 1.24 L/1.60 M = 1.65 L
Thus, the final volume of magnesium nitrate solution upon dilution is 1.65 L
The free energy change of the reaction; Fe (s) + Au3+ (aq) -> Fe3+ (aq) + Au (s) is calculated to be -443.83KJ/mol.
For the reaction shown in question 7, we can divide it into half equations as follows;
Oxidation half equation;
6 Al (s) -------> 6Al^3+(aq) + 18e
Reduction half equation;
3Cr2O7^2-(aq) + 42H^+(aq) + 18e -----> 6Cr^3+(aq) + 21H2O(l)
The balanced reaction equation is;
6Al(s) + 3Cr2O7^2-(aq) + 42H^+(aq) -----> 6Al^3+(aq) + 6Cr^3+(aq) + 21H2O(l)
The E° of this reaction is obtained from;
E° anode = -0.04 V
E°cathode = +1.50 V
E° cell = +1.50 V - (-0.04 V) = 1.54 V
Given that;
ΔG° = -nFE°cell
n = 3, F = 96500, E°cell = 1.54 V
ΔG° = -(3 × 96500 × 1.54)
ΔG° = -443.83KJ/mol
Learn more: brainly.com/question/967776
Answer:
50.0mL 0.10M NaOH
Explanation:
The chemical equation of H₂SO₄ with NaOH to reach the first equivalence point is:
H₂SO₄ + NaOH → HSO₄⁻ + Na⁺ + H₂O
<em>Where 1 mole of the H₂SO₄ reacts per mole of NaOH</em>
<em />
The initial moles of H₂SO₄ are:
50.0mL = 0.0500L × (0.10 mol / L) = 0.0050 moles of H₂SO₄
As 1 mole of the acid reacts per mole of NaOH, to reach the first equivalence point we need to add 0.0050 moles of NaOH. As molarity of NaOH is 0.10M, the volume that we need to add to reach 1st equivalence point is:
0.0050 moles NaOH ₓ (1L / 0.10 moles NaOH) = 0.050L NaOH 0.10M =
<h3>50.0mL 0.10M NaOH</h3>
