Answer:
Higher K at very high temperatures.
Explanation:
It is inferred is K is significantly greater at very high temperatures, allowing high production at low cost.
Answer:
25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.
Explanation:
The balanced reaction is:
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O
By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:
- C₂H₂: 2 moles
- O₂: 5 moles
- CO₂: 4 moles
- H₂O: 2 moles
The molar mass of each compound is:
- C₂H₂: 26 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- C₂H₂: 2 moles* 26 g/mole= 52 g
- O₂: 5 moles* 32 g/mole= 160 g
- CO₂: 4 moles* 44 g/mole= 176 g
- H₂O: 2 moles* 18 g/mole= 36 g
Then you can apply the following rule of three: if by stoichiometry 52 grams of C₂H₂ react with 160 grams of O₂, 8.15 grams of C₂H₂ react with how much mass of O₂?
mass of O₂= 25.08 grams
<u><em>25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.</em></u>
Monkey is good and that is all their it to it
Answer:
This has some interesting History in it.
Peat moss was once pounded into a sort of brick; sometimes with a little coal, these bricks were put to dry, and used for fuel.
The “Historical” part is in the name. I won’t go into why, ‘cause I type so slow and it is “extra”, but the peat, or coal, being TRULY organic, (that is made from plants) and burned as fuel is called “carbonixation”.
Explanation: Sort of a “Carbon + Oxidation”; many things in Chemistry have strange names that seem at first to make no sense; like “essence of Hartshorn”, is Ammonia (or ammonium hydroxide). A Hart is/was a type of deer in England and that area, and amines in the horn, when distilled, would yield Ammonia, “The Essence of Hartshorn”. Even the term “aromatic” has a long History, as do many compounds and nomenclature.
I hope that will do it.
4.14x10^-3 per minute
First, calculate how many atoms of Cu-61 we initially started with by
multiplying the number of moles by Avogadro's number.
7.85x10^-5 * 6.0221409x10^23 = 4.7273806065x10^19
Now calculate how many atoms are left after 90.0 minutes by subtracting the
number of decays (as indicated by the positron emission) from the original
count.
4.7273806065x10^19 - 1.47x10^19 = 3.2573806065x10^19
Determine the percentage of Cu-61 left.
3.2573806065x10^19/4.7273806065x10^19 = 0.6890455577
The formula for decay is:
N = N0 e^(-λt)
where
N = amount left after time t
N0 = amount starting with at time 0
λ = decay constant
t = time
Solving for λ:
N = N0 e^(-λt)
N/N0 = e^(-λt)
ln(N/N0) = -λt
-ln(N/N0)/t = λ
Now substitute the known values and solve:
-ln(N/N0)/t = λ
-ln(0.6890455577)/90m = λ
0.372447889/90m = λ
0.372447889/90m = λ
0.00413830987 1/m = λ
Rounding to 3 significant figures gives 4.14x10^-3 per minute as the decay
constant.
