It oxidized cu since it gains electons.
![\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D%3D1.50%5Ctimes%2010%5E%7B-6%7D)
- dt remains same for reaction
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cdfrac%7B3%7D%7B2%7D%5Cdfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cdfrac%7B3%7D%7B2%7D%281.5%5Ctimes%2010%5E%7B-6%7D%29)
![\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Cleadsto%20%5Cdfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D2.25%5Ctimes%2010%5E%7B-6%7DMs%5E%7B-1%7D)
M is molarity here not metre
<h3><u>Answer;</u></h3>
= 78 g of NaOH
<h3><u>Explanation;</u></h3>
Concentration = Moles of solute/Volume of solution
Thus;
Moles of the solute = Volume × Concentration
= 7.80 Moles/L × 0.250 L
= 1.95 moles
But; 1 mole of NaOH = 40.0 g
Thus;
Mass of NaOH = moles × molar mass
= 1.95 moles × 40 g/mole
<u> = 78 g of NaOH</u>
Answer:
pH = 8.477
Explanation:
∴ ni 0.3 - -
nf - 0.3 0.3
∴ ni 0.50 - 0.3
nf 0.5 - X X 0.3 + X
∴ Ka = 2.0 E-9 = ([H3O+]*[OBr-]) / [HOBr]
⇒ Ka = 2.0 E-9 = ((X/1L)*(0.3 + X)/1 L) / ((0.5 - X)/1L)
⇒ 2.0 E-9 = ( 0.3X + X² ) / (0.5 - X)
⇒ X² + 0.3X - 1 E-9 = 0
⇒ X = 4.333 E-9 M
according Henderson-Hauselbach:
- pH = pk + Log [A-] / [HA]
∴ [OBr-] = 0.3 mol/ 1 L + 4.333 E-9 M = 0.300 M
∴ [HOBr] = 0.5 mol / 1 L - 4.33 E-9 M = 0.500 M
∴ pKa = - log Ka = - Log ( 2.0 E-9 ) = 8.6989
⇒ pH = 8.6989 + Log ( 0.300/0.500 )
⇒ pH = 8.477
