Answer:
See explanation
Explanation:
The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;
238/92U + 4/2 He -------> 241/94Pu + 1/0 n
238/92U + 4/2 He ------> 241/94Pu + 1/0 n
14/7N + 4/2 He------> 17/8O + 1/1 p
56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Answer:
1.Ecologists often estimate the size and density of. populations using quadrats and the mark-recapture method.
2. The two key components of ecosystem stability are resilience and resistance.
3.The availability of abiotic factors (such as water, oxygen, and space) and biotic factors (such as food) affect population
4. When the climate is changing in the colder direction it effects the planet in the opposite way than it does when it is changing in the warmer direction. Ad nauseam the contemporary, it’ll keep getting hotter and at a greater rate so the ecosystem is going to die.
Answer:
16.5 dm³
Explanation:
Data Given:
no. moles of O₂ = 0.735 moles
volume of O₂ = ?
Solution:
Now
we have to find volume of O₂ gas
Formula used for this purpose
No. of moles = Volume / molar volume
where
molar volume at STP for Oxygen (O₂) = 22.4 dm³/mol
No. of moles O₂ = Volume of O₂ / 22.4 dm³/mol . . . . . .(1)
Put values in equation 1
0.735 = Volume of O₂ / 22.4 dm³/ mol
rearrange above equation
Volume of O₂ = 0.735 x 22.4 dm³/ mol
Volume of O₂ = 16.5 dm³
So,
the volume of O₂ at STP is 16.5 dm³
