The balanced chemical reaction is:
<span>2HC2H3O2(aq) + Ba(OH)2(aq) + ----> 2H2O(l) + Ba(C2H3O2)2(aq)
We are given the amount of </span><span>HC2H3O2 to be used in the reaction. This will be the starting point for the calculation.
</span> 0.461 mol HC2H3O2 ( 1 mol Ba(OH)2 / 2 mol HC2H3O2<span> ) = 0.231 mol Ba(OH)2</span>
Answer: The three end products of the oxidation-reduction reactions involved in metabolism are WATER, CARBON DIOXIDE and ATP.
Metabolism refers to the sum of all chemical reactions occurring in the body. It includes both catabolism and anabolism. Every metabolic reaction is characterized by oxidation-reduction reactions. The metabolic pathways include respiration, photosynthesis, etc. In all these reactions water is produced. In respiration CO2 is also a product. In general. in every metabolic reaction either an organic molecule is broken or synthesized producing CO2, water and energy.
Answer:
Explanation:
Arrhenius defined an acid as a substance that interacts with water to produce excess hydrogen ions in aqueous solution.
A base is a substance which interacts with water to yield excess hydroxide ions, in an aqueous solution according to Arrhenius.
Bronsted-Lowry theory defined an acid as a proton donor while a base is a proton acceptor.
Answer:
The correct option is: E. No precipitate will form.
Explanation:
A solubility chart refers to the list of solubility of various ionic compounds. It shows the solubility of the various compounds in water at room temperature and 1 atm pressure.
Also, according to the solubility rules, the salts of chlorides, bromides and iodides are generally soluble and mostly all salts of sulfate are soluble.
Since, all the compounds formed in this double replacement reaction are soluble in water. Therefore, no precipitate will be formed.
ZnSO₄ (aq) + MgCl₂ (aq) → ZnCl₂ (aq) + MgSO₄ (aq)
<h3><u>Answer;</u></h3>
321.8 g CaF2
321.5 g Al2(CO3)3
<h3><u>Explanation;</u></h3>
The equation for the reaction is;
3 CaCO3 + 2 AlF3 → 3 CaF2 + Al2(CO3)3
Number of moles of CaCO3 will be;
=(412.5 g CaCO3) / (100.0875 g CaCO3/mol)
= 4.12139 mol CaCO3
Number of moles of AlF3 will be;
= (521.9 g AlF3) / ( 83.9767 g AlF3/mol)
= 6.21482 mol AlF3
But;
4.12139 moles of CaCO3 would react completely with 4.12139 x (2/3) = 2.74759 moles of AlF3.
Thus; there is more AlF3 present than that, so AlF3 is in excess, and CaCO3 is the limiting reactant.
Therefore;
Mass of CaF2 will be;
(4.12139 mol CaCO3) x (3/3) x (78.0752 g CaF2/mol) = 321.8 g CaF2
Mass of Al2(CO3)3 on the other hand will be;
(4.12139 mol CaCO3) x (1/3) x (233.9903 g Al2(CO3)3/mol) = 321.5 g Al2(CO3)3
