Combined gas law is
PV/T = K (constant)
P = Pressure
V = Volume
T = Temperature in Kelvin
For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂
P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL
hence the volume of gas at stp is 2012.6 mL
The original mass of krypton 81 that is present in the ice is 6.70 grams.
<h3>How do we calculate original mass?</h3>
Original mass of any substance will be calculated as below for the decomposition reaction is:
N = N₀(1/2)ⁿ, where
N = remaining mass of krypton-81 = 1.675g
N₀ = original mass of krypton-81 = ?
n will be calculated as:
n = T/t, where
T = total time period = 458,000 years
t = half life time = 229,000 years
n = 458,000/229,000 = 2
Now putting all these values on the above equation, we get
N₀ = 1.675 / (1/2)²
N₀ = 6.70 g
Hence required mass is 6.70 g.
To know more about half life time, visit the below link:
brainly.com/question/2320811
<u>Answer:</u> The standard free energy change of formation of
is 92.094 kJ/mol
<u>Explanation:</u>
We are given:

Relation between standard Gibbs free energy and equilibrium constant follows:

where,
= standard Gibbs free energy = ?
R = Gas constant = 
T = temperature = ![25^oC=[273+25]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5DK%3D298K)
K = equilibrium constant or solubility product = 
Putting values in above equation, we get:

For the given chemical equation:

The equation used to calculate Gibbs free change is of a reaction is:
![\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the Gibbs free energy change of the above reaction is:
![\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag%5E%2B%28aq.%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28Ag_2S%28s%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol](https://tex.z-dn.net/?f=285.794%3D%5B%282%5Ctimes%2077.1%29%2B%281%5Ctimes%20%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%7D%29%5D-%5B%281%5Ctimes%20%28-39.5%29%29%5D%5C%5C%5C%5C%5CDelta%20G%5Eo_f_%7B%28S%5E%7B2-%7D%28aq.%29%29%3D92.094J%2Fmol)
Hence, the standard free energy change of formation of
is 92.094 kJ/mol
You have to move the decimal point forward 2 times to make the number 3.4,which makes the scientific notation 3.4×10^-2