The balanced reaction is:
<span>4Cr(s)+3O2 (g )= Cr2O3 (s)
Since we are not given the amount of any of the reactants, we assume we have one gram of chromium. Calculations are as follows:
1 g Cr ( 1 mol Cr / 52 g Cr ) ( 1 mol Cr2O3 / 4 mol Cr ) = <span>0.005 mol Cr2O3
</span></span>0.005 mol Cr2O3 (151.99 g Cr2O3 / 1 mol Cr2O3 ) = 0.7307 g <span>Cr2O3
</span>
Therefore, the theoretical yield for 1 gram of Cr is 0.005 mol Cr2O3 or 0.7307 g Cr2O3.
<u>Answer:</u> The mass of water produced in the reaction is 97.2 grams
<u>Explanation:</u>
We are given:
Moles of calcium hydroxide = 2.70 moles
The chemical equation for the reaction of calcium hydroxide and HCl follows:

By Stoichiometry of the reaction:
1 mole of calcium hydroxide produces 2 moles of water
So, 2.70 moles of calcium hydroxide will produce =
of HCl
To calculate mass for given number of moles, we use the equation:
Molar mass of water = 18 g/mol
Moles of water = 5.40 moles
Putting values in above equation, we get:

Hence, the mass of water produced in the reaction is 97.2 grams
Answer and Explanation:
- <em>Computers ( The computers are very old and very slow, and we have to notice that almost everyone are using computers at the same time at the school, which makes it even harder for it to load up assignments. )</em>
- <em>The Rick Rolling ( Everyone keeps sending links to teachers and students saying that it is part of some assignment but then you have to listen to Rick Astley, they should really block these links. )</em>
- <em>The lockers ( The lockers are also very old and they are breaking down and rusting a lot from the moisture in the hallways. One of the lockers even broke down today!!! I hope they can fix this so no one else gets hit with a locker door. )</em>
<em>Hope this helps! ;)</em>
Answer:
answer is strong base hope it helps
Question: A optically active compound, C5H10O, exhibits IR absorption at 1730 cm-1.
Its carbon NMR shifts are given below. The number of hydrogen's at each carbon, determined by DEPT, is given in parentheses after the chemical shift.
13C NMR: δ 22.6 (3), 23.6 (1), 52.8 (2), 202.4 (1)
Draw the structure of this compound in the window below
Explanation:
3-methylbutanal is a butanal substituted by the methyl group at the 3rd position. It is a volatile constituent in the olive. Also, it is used as a flavoring agent and a plant metabolite, it is also a Saccharomyces cerevisiae metabolite. It is also called as the Isovaleraldehyde organic compound. The liquid is colorless at STP, and also found in very low concentrations. It is also seen to be produced commercially for different use. Mostly used compound reagent in the preparation of pharmaceuticals and pesticides.