im found only in plant cells im as green as can be i make food for the plant using the suns energy what am i?
ANS:-Chloroplast
Weak base: [OH⁻] = √Kb.C
pKb = 4.2

c = concentration
MM Amphetamine (C9H13N) = 135.21 g/mol
c = 215 mg/L = (0.215 g : 135,21 g/mol) / L = 0.00159 mol/L = 1.59 x 10⁻³ mol/L
![\tt [OH^-]=\sqrt{10^{-4.2}\times 1.59\times 10^{-3}}=3.17\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20%5BOH%5E-%5D%3D%5Csqrt%7B10%5E%7B-4.2%7D%5Ctimes%201.59%5Ctimes%2010%5E%7B-3%7D%7D%3D3.17%5Ctimes%2010%5E%7B-4%7D)
pOH = 4 - log 3.17
pH = 14 - (4 - log 3.17)
pH = 10 + log 3.17 = 10.50
Answer:
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Explanation:
Let's consider the oxidation and reduction half-reactions and the global reaction.
Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻
Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)
Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red,cat - E°red,an
E° = 0.771 V - 0.154 V = 0.617 V
The Nernst equation allows us to calculate the cell potential (E) under the given conditions.
![E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V](https://tex.z-dn.net/?f=E%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20logQ%5C%5CE%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20log%5Cfrac%7B%5BSn%5E%7B%2B4%7D%5D.%5BFe%5E%7B2%2B%7D%5D%7D%7B%5BSn%5E%7B2%2B%7D%5D.%5BFe%5E%7B3%2B%7D%20%5D%7D%20%5C%5CE%3D0.617V-%5Cfrac%7B0.05916%7D%7B2%7D%20log%5Cfrac%7B%280.13%29.%280.0037%29%7D%7B%280.0023%29.%280.11%29%7D%20%5C%5CE%3D0.609V)
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
The chemical equation represents the reaction describes is;
4NH3 + 5O2 = 4NO + 6H2O
Therefore 4 moles of NH3 reacts with 5 moles of O2.
1 mole of O2 (molar mass) = 2 * 16 = 32g.
5 moles of O2 = 5 * 32 = 160g
4 moles of NH3 = 4 (14 + 3*1) = 68g
Therefore, 68g of NH3 reacts with 160g of O2.
But, we have only 4.5 g of oxygen.
68g reacts with 160g
Xg reacts with 4.5
X = 68*4.5 / 160 = 1.9125g