Answer:
The suitable equation for this reaction is
2CO + O₂ -----> 2CO₂
Here, we are given that we have 2 grams of O₂
From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂
Moles of O₂:
2/32 = 1/16 moles
Therefore, the number of moles of CO₂ is twice the moles of O₂
Moles of CO₂ = 2 * 1/16
Moles of CO₂ formed = 1/8 moles
Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂
Mass of CO₂ formed = 44 * 1/8
Mass of CO₂ formed = 5.5 grams
Hence, option B is correct
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<u>Answer:</u> The pH of the solution is 13.0
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of KOH = 716. mg = 0.716 g (Conversion factor: 1 g = 1000 mg)
Molar mass of KOH = 56 g/mol
Volume of solution = 130 mL
Putting values in above equation, we get:

1 mole of KOH produces 1 mole of hydroxide ions and 1 mole of potassium ions
- To calculate hydroxide ion concentration of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
[tex[[OH^-]=0.098M[/tex]
Putting values in above equation, we get:

To calculate the pH of the solution, we use the equation:
pOH + pH = 14
So, pH = 14 - 1.00 = 13.0
Hence, the pH of the solution is 13.0
Answer:
Exothermic
q = -8.03 kJ/g
Explanation:
<em>When 2 moles of Na(s) react with H₂O(l) to form NaOH(aq) and H₂(g) according to the following equation, 369 kJ of energy are evolved.</em>
The thermochemical equation is:
2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) ΔH = -369 kJ
Since the energy is evolved, the reaction is exothermic, which is why the enthalpy of the reaction (ΔH) is negative.
The heat released (q) per gram of Na(s) is (molar mass 22.98 g/mol):

Divide the number by 6.02 x 10^23 so the measurement is in moles and multiply by 22.4 to convert it into liters