1747.2 ml of HNO3 is required to reach methyl orange endpoint.
Explanation:
Number of moles of NaOH solution having 17.5 gram NaOH in 350ml of water.
Atomic weight of NaOH = 40 g/mol
thus number of moles= 17.5/40
= 0.437 moles
Now the molarity of the solution is calculated as
M=n/C
= 0.437/O.35 LITRES
= 1.248 M solution of the base
the molarity of acid is given as 0.25 M of HNO3
From the formula
M1V1=M2V2 we can calculate the volume of HNO3 required to make solution acidic.
1.248*350= 0.25*v2
v2= 1.248*350/0.25
= 1747.2 ml
Thus 1747.2 ml of HNO3 is required.
Answer:
B
Explanation:
When stem cells differentiate, they are different (almost like mutation).
Brhrhrhfhfhrhr r tbrhrnrbd
Answer:
Explanation:
We must use the Nernst equation
The equation for the cell reaction is is
2Cl⁻(0.384 mol·L⁻¹) + 2Co³⁺(0.324 mol·L⁻¹) ⇌ Cl₂(5.80 atm) + 2Co²⁺(0.158 mol/L)
Data:
E° = 0.483 V
R = 8.314 J·K⁻¹mol⁻¹
T = 25 °C
n = 2
F = 96 485 C/mol
Calculation:
T = 25 + 273.15 = 298.15 K
![Q = \dfrac{\text{[Cl}^{-}]^{2}[\text{Co}^{3+}]^{2}}{p_{\text{Cl}_{2}}^{2}\text{[Co}^{3+}]^{2}} = \dfrac{0.384^{2} \times 0.324^{2}}{5.80 \times 0.158^{2}} =0.1069\\\\E = 0.483 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(0.1069)\\\\=0.483 -0.01285 \times (-2.236) = 0.483 + 0.02872 = \textbf{0.512 V}\\\text{The cell potential is } \large\boxed{\textbf{0.512 V}}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BCl%7D%5E%7B-%7D%5D%5E%7B2%7D%5B%5Ctext%7BCo%7D%5E%7B3%2B%7D%5D%5E%7B2%7D%7D%7Bp_%7B%5Ctext%7BCl%7D_%7B2%7D%7D%5E%7B2%7D%5Ctext%7B%5BCo%7D%5E%7B3%2B%7D%5D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B0.384%5E%7B2%7D%20%5Ctimes%200.324%5E%7B2%7D%7D%7B5.80%20%5Ctimes%200.158%5E%7B2%7D%7D%20%3D0.1069%5C%5C%5C%5CE%20%3D%200.483%20-%20%5Cleft%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298.15%20%7D%7B2%20%5Ctimes%2096485%7D%5Cright%20%29%20%5Cln%280.1069%29%5C%5C%5C%5C%3D0.483%20-0.01285%20%5Ctimes%20%28-2.236%29%20%3D%200.483%20%2B%200.02872%20%3D%20%5Ctextbf%7B0.512%20V%7D%5C%5C%5Ctext%7BThe%20cell%20potential%20is%20%7D%20%5Clarge%5Cboxed%7B%5Ctextbf%7B0.512%20V%7D%7D)
All of the above is the answer
