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kow [346]
2 years ago
14

Consider the reaction 2AI(OH)3 + 3H2SO4 Right arrow. X + 6Y. What are X and Y? X = AI2(SO4)3; Y = H2O X = AI2(SO4)3; Y = H2 X =

AI2(SO3)3; Y = H2O X = AI2(SO3)3; Y = H2
Chemistry
1 answer:
Rom4ik [11]2 years ago
4 0

Answer :

X = Al_2(SO_4)_3

Y = H_2O

Explanation :

Balanced chemical equation : It is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

The given chemical equation is:

2Al(OH)_3+3H_2SO_4\rightarrow X+6Y

As we know that when aluminum hydroxide react with sulfuric acid then it gives aluminum sulfate and water as product.

The balanced chemical reaction will be:

2Al(OH)_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+6H_2O

Thus, X is Al_2(SO_4)_3 and Y is H_2O

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H2 is known to exist. For dihydrogen, H2, we can identify the frontier molecular orbitals (FMOs). The highest occupied molecular orbital (or HOMO) is the σ (sigma) 1s MO. The lowest unoccupied MO (LUMO) is the σ* (sigma star) 1s MO which is antibonding.


3 0
3 years ago
A lab technician mixed a 550 ml solution of water and alcohol. if​ 3% of the solution is​ alcohol, how many milliliters of alcoh
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The solution 550 ml total and first we will find the amount of alcohol. 3% = 0.03 550 ml x .03 = 16.5 ml alcohol
 Then to find the amount of water used, we just have to subtract the amount of alcohol from the total volume  
550 ml total - 16.5 ml alcohol = 533.5 ml water
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12. Describe the results of a chemical change. List four<br> indicators of chemical change.
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8 0
2 years ago
A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt
Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample  has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1:  1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams  ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr  

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0
3 years ago
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