![pH=-\log_{10} [H_3O^+] \Rightarrow [H_3O^+]=10^{-pH} \\ \\ pH=2.8 \\ \ [H_3O^+]=10^{-2.8} \approx 1.58 \times 10^{-3}](https://tex.z-dn.net/?f=pH%3D-%5Clog_%7B10%7D%20%5BH_3O%5E%2B%5D%20%5CRightarrow%20%5BH_3O%5E%2B%5D%3D10%5E%7B-pH%7D%20%5C%5C%20%5C%5C%0ApH%3D2.8%20%5C%5C%0A%5C%20%5BH_3O%5E%2B%5D%3D10%5E%7B-2.8%7D%20%5Capprox%201.58%20%5Ctimes%2010%5E%7B-3%7D)
The [H₃O⁺] of the solution is approximately 1.58 × 10⁻³ M.
False. carbon-carbon bonds that share 2 pairs of electrons are double bonds. An unsaturated hydrocarbon isnt necessary to only have double bonds. they can also have single bonds or triple bonds.
Answer:
Mass = 2.12 g
Explanation:
Given data:
Volume of KMnO₄ = 255 mL (255/1000 = 0.255 L)
Molarity = 0.0525 M
Mass in gram = ?
Solution:
First of all we will calculate the number of moles.
<em>Molarity = number of moles of solute / volume in litter</em>
0.0525 M = number of moles of solute / 0.255 L
Number of moles of solute = 0.0525 M ×0.255 L
Number of moles of solute = 0.0134 mol
Mass in gram:
<em>Number of moles = mass/ molar mass</em>
Mass = moles × molar mass
Mass = 0.0134 mol × 158.04 g/mol
Mass = 2.12 g
Mass of N₂ will be formed : 50.12 g
<h3>Further explanation</h3>
Reaction
2 NH₃(g) + 3 CuO(s) → N₂(g) + 3 Cu(s) + 3 H₂O(g)
moles NH₃ = 3.58
ratio mol NH₃ : mol N₂ = 2 : 1
so mol N₂ :
mass N₂ (MW=28 g/mol) :
