Answer:
D. a program on a computer
Explanation:
The half-life gets longer as the initial concentration increases in zero-order reaction.
The amount of time it takes for the concentration of a given reactant to reach 50% of its initial concentration is known as the half-life of a chemical reaction (i.e. the time taken for the reactant concentration to reach half of its initial value).
For zero order reaction:
The half-life is given as:
where k is the rate constant of the reaction and is the initial concentration.
As we can see that the half-life is directly proportional to the initial concentration. Therefore, when the initial concentration increases the half-life gets longer.
For the first-order reaction,
The half-life is given as:
A first-order reaction's half-life is independent of the initial concentration.
For a second-order reaction,
The half-life is:
The initial concentration is inversely proportional to the half-life, so when the initial concentration increases the half-life will get shorter.
Learn more about half life here:
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Answer:
The answer to your question is 242 ml
Explanation:
Data
HI 0.211 M Volume = x
KMnO₄ 0.354 M Volume = 24 ml
Balanced Chemical reaction
12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O
Process
1.- Calculate the moles of KMnO₄ 0.354 M in 24 ml
Molarity = moles / volume (L)
moles = Molarity x volume (L)
moles = 0.354 x 0.024
moles = 0.0085
2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,
12 moles of HI --------------- 2 moles of KMnO₄
x --------------- 0.0085 moles of KMnO₄
x = (0.0085 x 12)/2
x = 0.051 moles of HI
3.- Calculate the milliliters of HI 0.211 M
Molarity = moles/volume
Volume = moles/molarity
Volume = 0.051/0.211
Volume = 0.242 L or Volume = 242 ml
2Al+6HCl ->2AlCl3+3H2 .nAl=m/M=13,5/27=0,5(mol) =>nH2=0,5.3/2=0,75(mol)....VH2=n.22,4=0,75.22,4=16,8(lít)...
Answer:
Explanation:
A group of two or more smaller molecules is the correct answer