Answer:
6 days
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Half life (t½) =?
Next, we shall determine the decay constant. This can be obtained as follow:
Original amount (N₀) = 100 mg
Amount remaining (N) = 6. 25 mg
Time (t) = 24 days
Decay constant (K) =?
Log (N₀/N) = kt / 2.303
Log (100/6.25) = k × 24 / 2.303
Log 16 = k × 24 / 2.303
1.2041 = k × 24 / 2.303
Cross multiply
k × 24 = 1.2041 × 2.303
Divide both side by 24
K = (1.2041 × 2.303) / 24
K = 0.1155 /day
Finally, we shall determine the half-life of the isotope as follow:
Decay constant (K) = 0.1155 /day
Half life (t½) =?
t½ = 0.693 / K
t½ = 0.693 / 0.1155
t½ = 6 days
Therefore, the half-life of the isotope is 6 days
Answer:
Subscript, or B
Explanation:
<em>AgS is the equation for the chemic reaction of Silver + Sulfur, since the subscript of 2 goes by the g in the equation, this would equal subscript.</em>
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<em>Remember subscripts are located at the BOTTOM of a coefficient and superscripts are located at the TOP of a coefficient.</em>
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Answer:
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g)
Explanation:
Which ONE of the following is an oxidation–reduction reaction?
A) PbCO₃(s) + 2 HNO₃(aq) ⇒ Pb(NO₃)₂(aq) + CO₂(g) + H₂O(l). NO. All the elements keep the same oxidation numbers.
B) Na₂O(s) + H₂O(l) ⇒ 2 NaOH(aq). NO. All the elements keep the same oxidation numbers.
C) SO₃(g) + H₂O(l) ⇒ H₂SO₄(aq). NO. All the elements keep the same oxidation numbers.
D) CO₂(g) + H₂O(l) ⇒ H₂CO₃(aq). NO. All the elements keep the same oxidation numbers.
E) C₂H₄(g) + H₂(g) ⇒ C₂H₆(g). YES. <u>C is reduced</u> and <u>H is oxidized</u>.
