2SO2(g)+O2(g)→2 SO3(g), here reaction entropy decreases as the number of gas moles decreases from reactions to products.
HCL(g)+NH3(g)→NH4CL(s), entropy decreases as molecules of gas are converted into solid.
CO2(s)→CO2(g), entropy increases as gas is formed from a solid.
Cao(s)+CO2(g)→Caco3(s), entropy increases as gas is converted into a solid.
Answer:
Option (2) At equilibrium, there is a much higher concentration of products than reactants.
Explanation:
The equilibrium constant for a reaction is simply the ratio of the concentration products raised to their mole ratio divided by the concentration of the reactants raised to their mole ratio.
If the equilibrium constant is close to 1 or 1, it means the concentration of the reactants and products are almost the same. But if the equilibrium constant is large as in the case of the question given above, it means that at equilibrium, the concentration of the products are higher than that of the reactants.
Answer:
80.1 grams
Explanation:
Find the molar mass of CH3OH first by using the periodic table values.
12.011 g/mol C + (1.008*3 g/mol H) + 15.999g/mol O + 1.008 g/mol H
=32.042 so that is the molar mass
Now that you have 2.50 moles of CH3OH, you can calculate the mass in g
2.50molCH3OH * (32.042g CH3OH / 1 mol CH3OH) = 80.105
32.042g / 1 mol is the same as 32.042 g/mol
Since there are 3 sig figs in the problem (2.50 has 3 sig figs), you round to 80.1 g CH3OH
Iodine (I) Plz mark brainliest. HOPE THIS HELPS!
1mol ------------ 3×6,02×10²³ oxygen atoms
x ------------ 2,55×10²⁴
x = 2,5×10²⁴ × 1mol / 3×6,02×10²³ ≈ 0,138*10 mol = 1,38mol
