Answer:
2 moles
Explanation:
In one mole of O2 there are 16 grams. So in 2 moles there are 32 grams
A. iron(III) oxide + Carbon-monoxide -> Iron + Carbon-di-oxide.
b.56*2 +16*3
c.Moles=mass(g) / Mr which is 3200/160
d. 2 moles
e.2 moles
f. Moles x Mr = mass so, 20 x 112=2240g
Answer:
"ite"
Explanation:
Se sabe que los elementos del grupo 16 de la tabla prioritaria muestran estados de oxidación o valencias de 2,4 y 6 respectivamente, dependiendo del compuesto formado.
Por ejemplo, el azufre forma los siguientes compuestos;
sulfato de sodio (el azufre tiene una valencia de 6)
Sulfito de sodio (el azufre tiene una valencia de 4)
Sulfuro de sodio (el azufre tiene una valencia de 2)
Por lo tanto, en compuestos en los que exhiben una valencia de 4, la terminación común es "ite"
Answer: NaCl (s) → NaCl (aq)
Explanation:
Entropy is often associated with the disorder or randomness of a system. Therefore, in each reaction, it is necessary to evaluate if the disorder increases or decreases to understand what happens to the entropy:
1) KCl (aq) + AgNO₃ (aq) → KNO₃ (aq) + AgCl (s) - In this reaction, we have only aqueous reactants in the beginning and in the product we have a precipitate. The solid state is more organised than the liquid, consequently, the entropy decreases.
2) NaCl (s) → NaCl (aq) - In this case, oposite to the first one, we go from a solid state to an aqueous state. The solvation of the ions Na⁺ and Cl⁻ is random while the solid state is very organised. Therefore, in this reaction the entropy increases.
3) 2NaOH (aq) + CO₂ (g) → Na₂CO₃ (aq) + H₂O (l) - In this reaction, the reactants have higher entropy because of the gas CO₂. Therefore, the entropy decreases.
4) C₂H₅OH (g) → C₂H₅OH (l) - In this reaction, the reactant is a gas and the product a liquid. Therefore, the entropy decreases.
Answer:
See below
Explanation:
Use Ideal Gas Law
PV = n RT using R = .082057366 l-atm/k-mol
T must be in Kelvin
solve for 'n'
.92 * 1.6 = n * .082057366 * 287
n = .0625 moles
then the mole weight: .0625 * x = .314
x = mole weight = 5.025 gm
"Systems with either very <u>low pressures or high temperatures</u> enable real gases to be estimated as “ideal.” "
