Answer : The pH of a 0.1 M phosphate buffer is, 6.86
Explanation : Given,
Concentration of acid = 0.1 M
Concentration of conjugate base (salt) = 0.1 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of a 0.1 M phosphate buffer is, 6.86
Answer:
The answer to your question is 245.9 g of Fe₂O₃
Explanation:
Data
grams of Fe₂O₃ = ?
grams of Fe = 172 g
Balanced chemical reaction
Fe₂O₃ + 3H₂ ⇒ 2Fe + 3H₂O
Process
1.- Calculate the molar mass of of Iron Oxide and Iron
Fe₂O₃ = (2 x 56) + (3 x 16) = 112 + 48 = 160 g
2Fe = 2 x 56 = 112 g
2.- Use proportions and cross multiplication to solve this problem
160 g of Fe₂O₃ --------------- 112 g of Fe
x ---------------- 172 g of Fe
x = (172 x 160)/112
x = 27520/112
x = 245.9 g of Fe₂O₃
I think its because its more accurate because it shows you the numbers rather than you reading the approximate temperature on a liquid thermometer
Answer:
Across a period, effective nuclear charge increases as electron shielding remains constant. A higher effective nuclear charge causes greater attractions to the electrons, pulling the electron cloud closer to the nucleus which results in a smaller atomic radius. ... This results in a larger atomic radius.
Explanation:
Now lets d8
Answer:
Metal more reactive than non metal
