A very disgusting type of lemonade
Answer:
Cr(OH)2(s), Na+(aq), and NO3−(aq)
Explanation:
Let is consider the molecular equation;
2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)
This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.
Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.
Answer:
Explanation:
2C₂H₅OH = C₄H₆ + 2H₂O + H₂
2 mole 1 mole
molecular weight of ethyl alcohol
mol weight of C₂H₅OH = 46 gm
mol weight of C₄H₆ 54 gm
540 gm of C₄H₆ = 10 mole
10 mole of C₄H₆ will require 20 mol of ethyl alcohol .
20 mole of ethyl alcohol = 20 x 46
= 920 gm
ethyl alcohol required = 920 gm .
Answer:
Explanation:
An electrophilic addition reaction occurs when an electrophile attacks a substrate, with the end result being the inclusion of one or many comparatively straightforward molecules along with multiple bonds.
In the given question, the hydrogen bromide provides the electrophile while the bromide is the nucleophile. The mechanism proceeds with the attack of the electrophile on the carbon, followed by deprotonation. This process is continued with a formation of carbocation and the bromide(nucleophile) finally bonds to the carbocation to form a stable product.
The first diagram showcases the possible various starting molecules for the synthesis while the second diagram illustrates their mechanism.
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
