Answer:
%age Yield = 51.45 %
Solution:
Step 1: Convert Kg into g
68.5 Kg CO = 68500 g CO
8.60 Kg H₂ = 8600 g
Step 2: Find out Limiting reactant;
The Balance Chemical Equation is as follow;
CO + 2 H₂ → CH₃OH
According to Equation,
28 g (1 mol) CO reacts with = 4 g (2 mol) of H₂
So,
68500 g CO will react with = X g of H₂
Solving for X,
X = (68500 g × 4 g) ÷ 28 g
X = 9785 g of H₂
It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.
Step 3: Calculate Theoretical Yield
According to equation,
4 g (2 mol) H₂ reacts to produce = 32 g (1 mol) Methanol
So,
8600 g H₂ will produce = X g of CH₃OH
Solving for X,
X = (8600 g × 32 g) ÷ 4 g
X = 68800 g of CH₃OH
Step 4: Calculate %age Yield
%age Yield = Actual Yield ÷ Theoretical Yield × 100
Putting Values,
%age Yield = 3.54 × 10⁴ g ÷ 68800 g × 100
%age Yield = 51.45 %
4.06x20^24/6.02x10^23 = 6.744 moles x 55.845 g/mole = 376.61868grams
Answer:
Ka = 
Explanation:
Initial concentration of weak acid = 
pH = 6.87
![pH = -log[H^+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%2B%5D)
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
![[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6.87%7D%3D1.35%20%5Ctimes%2010%5E%7B-7%7D%5C%20M)
HA dissociated as:
(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = 
![Ka = \frac{[H^+][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
0.000000135 <<< 0.00045
The answer is phototropism, which is the growth of an organism responding to the presence of light.
Answer:
A) One that occurs on its own
