Answer:
See below
Explanation:
propane mole weight = 44 gm / mole
100 gm / 44 gm / mole = 2.27 moles
From the equation, 5 times as many moles of OXYGEN (O2)are required
= 11.36 moles of oxygen
at <u>STP</u> this is 254.55 liters of O2 (because 22.4 L = one mole) and
Using oxygen as 21 percent of air means that
.21 x = 254.55 = x = <u>1212.12 liters of air required </u>
Answer : The value of
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)

conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction
.

![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)

Now we have to calculate the Gibbs free energy of reaction
.
As we know that,

At room temperature, the temperature is 500 K.


Therefore, the value of
for the reaction is -959.1 kJ
Adding an atom will increase the repulsion between existing atoms and lone pairs. Added atom will result in bond pair-bond pair and bond pair-lone pair repulsion. The magnitude of the lone pair-bond pair repulsion is greater than the bond pair-bond pair repulsion. The added atom will change the electron geometry and bring about a distortion in the molecular geometry.

Inside the stomach, Hydrochloric acid kills micro-organisms in the food. Stomach juices begin to break down <u>proteins</u> to amino acids.
✤ So, Fill the blank with proteins.
<h3>
<u>Explanation:-</u></h3>
- Inside the stomach, the digestion of proteins starts due to the action of pepsin enzyme.
- But this enzyme remains inactive and is activated by the Hydrochloric acid(HCl).
- The Hydrochloric acid also helps in killing the germs and microbes which entered along with food.
- The mucous lines the wall of stomach to protect it from the harm caused by HCl because HCl is a strong acid.
- In stomach, The partial digestion of proteins occur and rest is digested in the small intestine by Trypsin(Pancreatic enzyme) and Intestinal juices.
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer: 
Explanation:
A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
Spectator ions are defined as the ions which does not get involved in a chemical equation or they are ions which are found on both the sides of the chemical reaction present in ionic form.
The given chemical equation is:

The complete ionic equation is:

The ions which are present on both the sides of the equation are lithium and chlorate ions. and hence are not involved in net ionic equation.
Thus the net ionic equation is:
