Answer:
I do not know the Answer I'm just trying to get my point
Explanation:
Thank you
Answer:
Answers are in the explanation.
Explanation:
- The half‑life of A increases as the initial concentration of A decreases. order: <em>2. </em>In the half-life of second-order reactions, the half-life is inversely proportional to initial concentration.
- A three‑fold increase in the initial concentration of A leads to a nine‑fold increase in the initial rate. order: <em>2. </em>The rate law of second-order is: rate = k[A]²
- A three‑fold increase in the initial concentration of A leads to a 1.73‑fold increase in the initial rate. order: <em>1/2. </em>The rate law for this reaction is: rate = k √[A]
- The time required for [A] to decrease from [A]₀ to [A]₀/2 is equal to the time required for [A] to decrease from [A]₀/2 to [A]₀/4. order: <em>1. </em>The concentration-time equation for first-order reaction is: ln[A] = ln[A]₀ - kt. That means the [A] decreasing logarithmically.
- The rate of decrease of [A] is a constant. order: <em>0. </em>The rate law is: rate = k -<em>where k is a constant-</em>
Answer:
1040%
Explanation:
To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:
Percent yield = Actual yield (5.40g) / Theoretical yield * 100
<em>Moles Fe -Molar mass: 55.845g/mol-:</em>
10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.
For a complete reaction of these moles there are necessaries:
0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.
As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>
The moles of H2 produced are:
0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2
The mass is:
0.277 moles H2 * (2.016g/mol) = 0.558g H2
Percent yield is:
5.40g / 0.558g * 100 = 1040%
It is possible the experiment wasn't performed correctly
Answer:
Volume is the amount of space taken up an object.
The instrument is a Geiger counter and is used to measure radioactive level around people's bodies.