PbO2
You have to take the mass of lead in the problem, and divide by the molar mass.
When you do the same with oxygen, you get a number about twice as large as when you divide the mass of lead by the molar mass of lead. This means that the simplest formula would be PbO2
Answer : The temperature of the gas is, 245.9 K
Explanation :
To calculate the temperature of gas we are using ideal gas equation:

where,
P = pressure of gas = 2770.96 torr = 3.646 atm
Conversion used : (1 atm = 760 torr)
V = volume of gas = 88.84 L
T = temperature of gas = ?
R = gas constant = 0.0821 L.atm/mole.K
w = mass of gas = 609.64 g
M = molar mass of
gas = 38 g/mole
Now put all the given values in the ideal gas equation, we get:


Therefore, the temperature of the gas is, 245.9 K
<u>Answer:</u> The number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
<u>Explanation:</u>
In a chemical equation, the chemical species are termed as reactants or products.
Reactants are defined as the species which react in the reaction and are written on the left side of the reaction arrow.
Products are defined as the species which are produced in the reaction and are written on the right side of the reaction arrow.
For the given chemical equation:

On the reactant side:
Number of carbon atoms = (6 × 2) = 12
Number of hydrogen atoms = (14 × 2) = 28
Number of oxygen atoms = (2 × 19) = 38
Hence, the number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
<u>Answer:</u> The chemical equations and equilibrium constant expression for each ionization steps is written below.
<u>Explanation:</u>
The chemical formula of carbonic acid is
. It is a diprotic weak acid which means that it will release two hydrogen ions when dissolved in water
The chemical equation for the first dissociation of carbonic acid follows:

The expression of first equilibrium constant equation follows:
![Ka_1=\frac{[H^+][HCO_3^{-}]}{[H_2CO_3]}](https://tex.z-dn.net/?f=Ka_1%3D%5Cfrac%7B%5BH%5E%2B%5D%5BHCO_3%5E%7B-%7D%5D%7D%7B%5BH_2CO_3%5D%7D)
The chemical equation for the second dissociation of carbonic acid follows:

The expression of second equilibrium constant equation follows:
![Ka_2=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}](https://tex.z-dn.net/?f=Ka_2%3D%5Cfrac%7B%5BH%5E%2B%5D%5BCO_3%5E%7B2-%7D%5D%7D%7B%5BHCO_3%5E-%5D%7D)
Hence, the chemical equations and equilibrium constant expression for each ionization steps is written above.