Answer:
10−8 M.
Explanation:
In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.
Answer:
0.665 moles of CO₂
Explanation:
The balance chemical equation for the combustion of Ethane is as follow:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
Step 1: <u>Calculate moles of C₂H₆ as;</u>
Moles = Mass / M.Mass
Putting values,
Moles = 10.0 g / 30.07 g/mol
Moles = 0.3325 moles
Step 2: <u>Calculate Moles of CO₂ as;</u>
According to balance chemical equation,
2 moles of C₂H₆ produced = 4 moles of CO₂
So,
0.3325 moles of C₂H₆ will produce = X moles of CO₂
Solving for X,
X = 0.3325 mol × 4 mol ÷ 2 mol
X = 0.665 moles of CO₂
D:
When electrons are gained, the charge of the atom decreases.
When you are given an atom with a charge, the oxidation of that atom is the charge. So by going from a Cr^3+ (Oxidation Number = 3) to a Cr^2+ (Oxidation Number = 2), the Oxidation Number thus decreases.
