Answer:
0.01228s⁻¹ = rate constant
Half-life = 56.4s
Explanation:
The first order reaction follows the equation:
ln[A] = -kt + ln[A]₀
<em>Where [A] is amount of reactant after time t = 45.0%, k is rate constante and [A]₀ initial amount of reactant = 100%</em>
ln[45%] = -k*65s + ln[100%]
-0.7985 = -k*65s
0.01228s⁻¹ = rate constant
Half-life is:
Half-life = ln2 / k
<h3>Half-life = 56.4s</h3>
<em />
Answer is 14.5 g L⁻¹.
<em>Explanation;</em>
Here, the question says reduce the units as one.
The presented units are g/L. To reduce the units to one, what we can do is take L to the upper side.
This can be done according to the rules of indices;
1 / aˣ = a⁻ˣ
Like that, we can write 1 / L as L⁻¹.
Hence, the reduced unit is g L⁻¹.
But remember to keep a space between when writing two different units.
Actually, this is an unit for density.
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.
Astronomy deals with the study of outer space. The study focuses on planets and other celestial bodies. Generally telescopes are used to study outer space. The particular type of telescope that is used depends on the need.
In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.
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