Answer:
See below
Explanation:
From periodic table, find Mole Weight of the element in question
this is the grams of the element in Avagadro's Number (6.022 x 10^23) of atoms
(5.50 x 10^22) / (6.022 x 10^23) x mole weight = grams
Hello!
datos:
Molarity = 
ps: The ionization constant of the nitric acid is strong (100% ionized in water) or completely dissociates in water, so the pH will be:
![pH = - log\:[H_3O^+]](https://tex.z-dn.net/?f=%20pH%20%3D%20-%20log%5C%3A%5BH_3O%5E%2B%5D%20)
![pH = - log\:[2*10^{-4}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-%20log%5C%3A%5B2%2A10%5E%7B-4%7D%5D%20)
Note:. The pH <7, then we have an acidic solution.
I Hope this helps, greetings ... DexteR!
Answer:
Explanation:
a) The mass of the reactants is 2.36 grams, and the mass of the products is 1.57 grams plus the mass of the carbonic acid. Thus, using the law of conservation of mass, we get the mass of the carbonic acid is 2.36 - 1.57 = 0.79 grams.
b) The gram-formula mass of sodium bicarbonate is 84.006 g/mol, meaning that 2.36/84.006 = 0.028 moles were consumed. Thus, this means that in theory, 0.014 moles of carbonic acid should have been produced, which would have a mass of (0.014)(62.024)=0.868 grams. Thus, the percentage yield is (0.79)/(0.868) * 100 = 91%
39.25 g of water (H₂O)
Explanation:
We have the following chemical reaction:
2 H₂ + O₂ → 2 H₂O
Now we calculate the number of moles of each reactant:
number of moles = mass / molar weight
number of moles of H₂ = 14.8 / 2 = 7.4 moles
number of moles of O₂ = 34.8 / 32 = 1.09 moles
We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:
if 1 moles of O₂ produces 2 moles of H₂O
then 1.09 moles of O₂ produces X moles of H₂O
X = (1.09 × 2) / 1 = 2.18 moles of H₂O
mass = number of moles × molar weight
mass of H₂O = 2.18 × 18 = 39.25 g
Learn more about:
limiting reactant
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The anserr is c that what i think
