Text me I don’t understand your question?
Answer:
The answer would be this: an independent research organization test the product and record the results.
Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Answer:
a. C: +3 ; b. N: +5 ; c. S:+6 ; d. C: +4; e. Mn: +7 ; f. Cr: +6.
Explanation:
Global charges in molecules is 0
You sum all the oxidation states to determine the oxidation state for the compound.
Na₂C₂O₄ → Sodium oxalate → Global charge: 0
Oxidation state for C: +3
HNO₃ → Nitric acid → Global charge: 0
Oxidation state for N: +5
H₂SO₄ → Sulfuric acid → Global charge: 0
Oxidation state for S: +6
HCO₃⁻ → Bicarbonate → Global charge: -1, this is an anion
Oxidation state for C: +4
KMnO₄ → Potassium permanganate → Global charge: 0
Oxidation state for Mn: +7
Cr₂O₇⁻ → Anion dichromate → Global charge: -2
Oxidation state for Cr: +6
