Answer:
.0556 L
Explanation:
First, convert the 1.35 M to 1.35 mol/L in order for the units to correctly cancel out.
Then, multiply (0.0725 moles Na2CO3/1) times (L/ 1.35 mol).
Finally, the answer will be .0556 L.
<h3 />
Nothing, he shouldn’t be able to move it. Think about it like this say you try really hard to push something that is 5,000 pounds and you push as hard as you can. Well you can’t move it bc it weighs more than you can push. I’m sure their is a equation you can use to see how much you can push (body weight=force?)
Answer:
2C₂H₆ + [7]O₂ → [4]CO₂ + [6]H₂O
Explanation:
Chemical equation:
C₂H₆ + O₂ → CO₂ + H₂O
Balanced chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Step 1:
2C₂H₆ + O₂ → CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 1
H = 12 H = 2
O = 2 O = 3
Step 2:
2C₂H₆ + O₂ → 4CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 2
O = 2 O = 9
Step 3:
2C₂H₆ + O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 2 O = 14
Step 4:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 14 O = 14
Answer:
pKa = 4.89.
Explanation:
We can solve this problem by using the <em>Henderson-Hasselbach equation</em>, which states:
pH = pKa + log ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.
We <u>input the given data</u>:
4.63 = pKa + log 
And <u>solve for pKa</u>:
pKa = 4.89
