Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol
The answer is 43 L if I am correct.
Answer:
Explanation:
1 x 10²³ molecules of methane
6.02 x 10²³ molecules of methane = 1 mole of methane
10²³ molecules of methane = 1 / 6.02 moles of methane
= .166 moles .
Answer 1:
Equilibrium constant (K) mathematically expressed as the ratio of the concentration of products to concentration of reactant. In case of gaseous system, partial pressure is used, instead to concentration.
In present case, following reaction is involved:
2NO2 ↔ 2NO + O2
Here, K =
![\frac{[PNO]^2[O2]}{[PNO2]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BPNO%5D%5E2%5BO2%5D%7D%7B%5BPNO2%5D%5E2%7D%20)
Given: At equilibrium, <span>PNO2= 0.247 atm, PNO = 0.0022atm, and PO2 = 0.0011 atm
</span>
Hence, K =
![\frac{[0.0022]^2[0.0011]}{[0.247]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.0022%5D%5E2%5B0.0011%5D%7D%7B%5B0.247%5D%5E2%7D%20)
= 8.727 X 10^-8
Thus, equilibrium constant of reaction = 8.727 X 10^-8
.......................................................................................................................
Answer 2:
Given: <span>PNO2= 0.192 atm, PNO = 0.021 atm, and PO2 = 0.037 atm.
Therefore, Reaction quotient = </span>
![\frac{[PNO]^2[O2]}{[PNO2]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BPNO%5D%5E2%5BO2%5D%7D%7B%5BPNO2%5D%5E2%7D%20)
=
![\frac{[0.021]^2[0.037]}{[0.192]^2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5B0.021%5D%5E2%5B0.037%5D%7D%7B%5B0.192%5D%5E2%7D%20)
= 4.426 X 10^-4.
Here, Reaction quotient > Equilibrium constant.
Hence, <span>the reaction need to go to
reverse direction to reattain equilibrium </span>
In a reduction reaction,
some electrons are gained by the substance being reduced. The balanced
half-reaction to this would be:
<span>MnO2(s) + 4 H+ (aq)
+ 2e ---> Mn^2+ (aq) + 2 H2O
(aq)</span>
<span>It is called balanced reaction
since the number of each element in the left side is equal to the number of
each element on the right side.</span>
Molality is defined as the number of moles of solute dissolved in 1 kg of solvent.
To calculate molality, we need to calculate the number of moles of CaCl₂.
Mass of CaCl₂ - 5.0 g
Molar mass of CaCl₂ - 111 g/mol
The number of moles of CaCl₂ - 5.0 g / 111 g/mol = 0.045 mol
we need to then calculate the number of moles in 1 kg solvent.
number of CaCl₂ moles in 500 g water - 0.045 mol
Therefore number of moles in 1 kg water - 0.045 mol / 500g x 1000 g = 0.090 mol
Molality of CaCl₂ - 0.090 mol/kg