Answer:
30.8 g of water are produced
Explanation:
First of all we need the equation for the production of water:
2H₂ + O₂ → 2H₂O
2 moles of hydrogen react with 1 mol of oxygen in order to produce 2 moles of water.
As we assume, the oxygen in excess, we determine the moles of H₂.
1.03ₓ10²⁴ molecules . 1 mol/ 6.02ₓ10²³ molecules = 1.71 moles
Ratio is 2:2, so 1.71 moles will produce 1.71 moles of water
Let's convert the moles to mass: 1.71 mol . 18g / 1mol = 30.8 g of water are produced
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
The correct answer for the question that is being presented above is this one: "<span>16.728 g."</span>
Given that
ΔHsolid = -5.66 kJ/mol.
This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies
When 5.57 kJ of heat is released
amount of NH3 solidifies = 5.57/5.66 = 0.984 moles
<span>molar mass of NH3 = 17 g/mole </span>
<span>1 mole of NH3 = 17 g </span>
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g
Answer:
group 8 of the periodic table
Explanation:
