Answer:
D 4,3,2
Explanation:
4 Co + 3 O2 ----> 2 Co2O3
Answer:
0.800 mol of O2
Explanation:
<em>Calculate the moles of oxygen produced by the reaction of 0.800mol of carbon dioxide.</em>
The balanced equation for the reaction is given as;
6CO2 + 6H2O → C6H12O6 + 6O2
From the reaction;
6 mol of CO2 produces 6 mol of O2
0.0800 mol of CO2 would produce x mol of O2
6 = 6
0.0800 = x
Solving for x;
x = 6 * 0.800 / 6
x = 0.800 mol
<u>Answer:</u> The pH of resulting solution is 8.7
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of TRIS acid solution = 0.1 M
Volume of solution = 50 mL
Putting values in above equation, we get:
Molarity of TRIS base solution = 0.2 M
Volume of solution = 60 mL
Putting values in above equation, we get:
Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)
- To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[salt]}{[acid]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D%29)
![pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7BTRIS%20base%7D%5D%7D%7B%5B%5Ctext%7BTRIS%20acid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of TRIS acid = 8.3
![[\text{TRIS acid}]=\frac{0.005}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20acid%7D%5D%3D%5Cfrac%7B0.005%7D%7B0.11%7D)
![[\text{TRIS base}]=\frac{0.012}{0.11}](https://tex.z-dn.net/?f=%5B%5Ctext%7BTRIS%20base%7D%5D%3D%5Cfrac%7B0.012%7D%7B0.11%7D)
pH = ?
Putting values in above equation, we get:
Hence, the pH of resulting solution is 8.7
Answer:
Given: 42 g of N2
Solve for O2 mass that contains the same number of molecules to 42 g of N2.
Solve for the number of moles in 42 g of N2
1 mole of N2 = (14 * 2) g = 28 g so the number of moles in 42 g of N2 is equal to 42 g / 28 g per mole = 1.5 moles
Solve for mass of 1 mole of oxygen
1 mole of O2 = 16 g * 2 = 32 g per mole
Solve for the mass of 1.5 moles of oxygen
mass of 1.5 moles of O2 = 32 g per mole * 1.5 moles
mass of 1.5 moles of O2 = 48 g
So 48 g of O2 contains the same number of molecules as 42 g of N2
