Answer:
12.09 L
Explanation:
Step 1: Convert 826.1 mmHg to atm
We will use the conversion factor 760 mmHg = 1 atm.
826.1 mmHg × 1 atm/760 mmHg = 1.087 atm
Step 2: Convert 427.8 J to L.atm
We will use the conversion factor 101.3 J = 1 L.atm.
427.8 J × 1 L.atm/101.3 J = 4.223 L.atm
Step 3: Calculate the change in the volume
Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:
w = P × ΔV
ΔV = w/P
ΔV = 4.223 L.atm/1.087 atm = 3.885 L
Step 4: Calculate the final volume
V₂ = V₁ + ΔV
V₂ = 8.20 L + 3.885 L = 12.09 L
Probably when u close your eyes you’d see black or whenever you about to past out
Answer and Explanation:
As the temperature of the substances increases, the average energy of the molecules increases, and average energy of attraction between the molecules decreases consequently intermolecular spacing between the molecules increases. As a result, a substance change in succession from gas to liquid to solid.
Great question, but I believe you are mixing up atomic number with mass number. Assuming you are, 12.011 amu is the average mass of a carbon atom. For carbon, it can come in three forms: carbon-12, carbon-13, carbon-14. The number following carbon is the mass number of that particular carbon "isotope". The reason the average is so close to 12 is because carbon-12 is by far the most common, so the average should be (and is) very close to 12. Therefore, 12.011 is a weighted average of all carbon molecules, and carbon-14 is a particular carbon molecule that weighs 14 amu.
El número de Avogadro es 6,022 x 10^23, y es el número de átomos que hay en un mol de dicho elemento. A su vez un mol es la cantidad de un elemento cuya masa en gramos coincide con el peso atómico.
Por tanto, 6,022 x 10^23 átomos del elemento tienen una masa en gramos igual a su peso atómico. Hacemos una regla de tres:
1 gramo -------- 1,5 x 10^22 átomos
x ------------------ 6,022 x 10^23 átomos
=> x = 40,1 gramos por mol del elemento.
De modo que su peso atómico es 40,1. Se trata del calcio.
Saludos.
