Answer:
The empirical formula is Ag2O.
The empirical formula is Ag2O.Explanation:
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to 2O.
do the steps ...
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
—————————————————−———mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
Answer:
The pressure is higher than the atmospheric one, therefore the temperature is less.
Explanation:
When it is closed permanently, the pressure of the pot inside it increases, generating that the atoms and particles of the water are closer together, increasing their kinetic energy, if intermolecular friction and therefore the boiling point is lower, because the water reaches a boil or boil at a lower temperature.
The equilibrium constant of the reaction is 282. Option D
<h3>What is equilibrium constant?</h3>
The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.
Concentration of bromine = 0.600 mol /1.000-L = 0.600 M
Concentration of iodine = 1.600 mol/1.000-L = 1.600M
In this case, we must set up the ICE table as shown;
Br2(g) + I2(g) ↔ 2IBr(g)
I 0.6 1.6 0
C -x -x +2x
E 0.6 - x 1.6 - x 1.190
If 2x = 1.190
x = 1.190/2
x = 0.595
The concentrations at equilibrium are;
[Br2] = 0.6 - 0.595 = 0.005
[I2] = 1.6 - 0.595 = 1.005
Hence;
Kc = [IBr]^2/[Br2] [I2]
Kc = ( 1.190)^2/(0.005) (1.005)
Kc = 282
Learn more about equilibrium constant:brainly.com/question/15118952
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