17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.
<h3>How to calculate mass?</h3>
The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.
However, the number of moles of the gas must be calculated first as follows:
PV = nRT
Where;
- P = pressure = 1.0796941atm
- V = volume = 12.3L
- n = number of moles
- T = temperature = 288.4K
- R = gas law constant = 0.0821 Latm/molK
1.079 × 12.3 = n × 0.0821 × 288.4
13.27 = 23.68n
n = 13.27/23.68
n = 0.56mol
Mass = 0.56 × 32
mass of oxygen gas = 17.93g
Therefore, 17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C.
Learn more about mass at: brainly.com/question/19694949
I believe the correct answer from the choices listed above is option B. Stainless steel is an example of a solid-solid solution. It is an alloy which is made up of different metals <span>such as </span>carbon<span>, </span>manganese<span>, phosphorus, sulfur, nickel, chromium and others. Hope this answers the question.</span>
Exothermic reaction is where there is release of energy during a reaction
The enthalpy of exothermic reaction is negative
The relation between energy of products, reactants and enthalpy of reaction is
Enthalpy of reaction = sum of enthalpy of formation of products - sum of enthalpy of formation of reactants
.
As enthalpy of reaction is negative, it means the enthalpy of products is less than the enthalpy of reactants so answer is
:
In an exothermic reaction the energy of the product is less than the energy of the reactants.
Answer:
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)
Explanation:
Write the balance chemical equation ,

initial concenration of 
lets assume that degree of dissociation=
concenration of each component at equilibrium:
![[SO_2Cl_2] = 0.1-0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha)
![[SO_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BSO_2%5D%20%3D%200.1%5Calpha)
![[Cl_2] = 0.1\alpha](https://tex.z-dn.net/?f=%5BCl_2%5D%20%3D%200.1%5Calpha)


as
is very small then we can neglect 
therefore ,



Eqilibrium concenration of ![[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.1-0.1%5Calpha%20%3D%200.1-0.1%5Ctimes%200.00173)
![[SO_2Cl_2] = 0.09983 M](https://tex.z-dn.net/?f=%5BSO_2Cl_2%5D%20%3D%200.09983%20M)