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
Electronegativity is related to covalent bonding because when two atoms have the same electronegativity bond, they will form a pure covalent bond.
Answer:
I think its b
Explanation:
but I wouldn't depend on this answer
Hey there mate ;), Im Benjemin and lets solve your question.
★ (Alkanes) : forms single bonds between carbon atoms.
The first four elements are gases and others are liquid in state.
★(Alkenes) : forms double bonds between carbon atoms.
The first three alkenes are gases and rest are liquid.
★ (Alkynes) : forms triple bonds between carbon atoms.
First three are gases and the last one is liquid.
According to boiling point :
The larger structure of the hydrocarbons, the higher the boiling points they have.
In the 3 tables, we can see that the boiling point increases.
We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
