Answer:
it is D
Explanation:
NaOH and chlorine gas
Electrolysis of Aqueous NaCl Since water can be both oxidized and reduced, it competes with the dissolved Na+ and Cl– ions. Rather than producing sodium, hydrogen is produced. ... The net process is the electrolysis of an aqueous solution of NaCl into industrially useful products sodium hydroxide (NaOH) and chlorine gas.
" There will be a net movement of oxygen from outside the cell to inside the cell " Statement is True.
Explanation:
The partial pressure for oxygen in alveoli is greater under normal circumstances, and oxygen moves neatly into the blood. In addition, the partial carbon dioxide pressure throughout the blood usually is higher, such that carbon dioxide migrate clearly into the alveoli.
The few common molecules which can traverse the cell membrane by absorption (or diffusion of a sort recognized as osmosis) are water, carbon dioxide and oxygen. Metabolism is typically oxygen-needed, which is lowest in the cell within the animal and plant, so that net oxygen flows to the cell.
Answer:
H₂ is excess reactant and O₂ the limiting reactant
Explanation:
Based on the chemical reaction:
2H₂(g) + O₂(g) → 2H₂O
<em>2 moles of H₂ react per mole of O₂</em>
<em />
To find limiting reactant we need to convert the mass of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
10g H₂ * (1mol / 2.016g) = 4.96 moles
<em>Moles O₂ -Molar mass: 32g/mol-:</em>
22g O₂ * (1mol / 32g) = 0.69 moles
For a complete reaction of 0.69 moles of O₂ are needed:
0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂
As there are 4.96 moles,
<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>
Answer:
a). P = 688 atm
b). P = 1083.04 atm
c).Δ G = 16.188 J/mol
Explanation:
a). Fugacity 'f' can be calculated from the following equations :
where, P = pressure , Z = compressibility
Now, the virial equation is :
........(1)
Also, PV=ZRT for real gases .......(2)
∴ 
So from the fugacity equation ,
Putting the value of P = 500 atm in the above equation, we get,
f = 688 atm
b). Given f = 2P
∴ P = 1083.04 atm
c). dG = Vdp -S dt at constant temperature, dT = 0
Therefore, dG = V dp
![$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$](https://tex.z-dn.net/?f=%24%5CDelta%20G%3DR%5B%5Cln%5Cfrac%7BP_2%7D%7BP_1%7D%2B6.4%20%5Ctimes%2010%5E%7B-4%7D%28P_2-P_1%29%5D%24)
![$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$](https://tex.z-dn.net/?f=%24%5CDelta%20G%3D8.314%5Ctimes%20298%5B%5Cln%5Cfrac%7B500%7D%7B1%7D%2B6.4%20%5Ctimes%2010%5E%7B-4%7D%28500-1%29%5D%24)
Δ G = 16.188 J/mol
