Given the balanced equation representing a reaction:Al2(SO4)3 +6NaOH ==> 2Al(OH)3 + 3Na2SO4

The mole ratio of NaOH to Al(OH)3 is

(1) 1:1 (2) 3:1

(3) 1:3 (4) 3:7

Answers

Answer 1
Answer: The answer is number 2. 3:1. This is the simplified value of 6:2 which came from 6NaOH:2Al(OH)3. The original ratio is 6:2 but it can be simplified through dividing it each by 2. So 6 divided by 2 equals to 3 and 2 divided by 2 is 1. Therefore the simplified ratio is 3:1.

Related Questions

Can a tank of oxygen gas ever be half empty? explain
How can you demonstrate that helium takes up space?
Can someone explain #3
Which of the classifications for humans means that we have a hollow nerve chord?A) Class B) Kingdom C) Order D) Phylum
Explain how a gene pool and biological evolution are related

I’m general,what was would a scientist do after forming a testable question

Answers

Develop a hypothesis.

How could you tell whether a liquid is
honey or vinegar?

Answers

By it's color, texture, smell and the way it taste...

The _____ is the organelle that recycles unneeded molecules is the answer vacuole, mitochondria, or lysosome

Answers

Answer: Lysosomes

Explanation: The lysosome is an organelle that contains digestive enzymes and acts as the organelle-recycling facility of an animal cell. It breaks down old and unnecessary structures so their molecules can be reused. Lysosomes are part of the endomembrane system, and some vesicles that leave the Golgi are bound for the lysosome.

the answer is lysosome

Why does split petrol form a stick on top of water?

Answers

Because oil or petrol is less dense than water therefore, instead of being absorbed by the water, it separates leaving the petrol on top of the water instead of in it or underneath it

Please help me!!!30 ptsShow all work and box in your answers.
1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of
sulfur. Calculate the empirical formula of this compound.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen
indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?


3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 %
oxygen. Find the empirical formula of this compound.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that
4.00 grams of calcium are present. What is the empirical formula of the compound formed?

7. A 60.00 gram sample of tetraethyl lead, a gasoline additive, is found to contain 38.43grams of
lead, 17.83 grams of carbon, and 3.74 grams of hydrogen. Find its empirical formula.

8. Determine the molecular formula for the compound with empirical formula P2O5 and molar mass of
284 g/mol.

9. Determine the molecular formula for the compound with empirical formula OCNCl and molar mass
of 232.41 g/mol.

10. A compound is found to be 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. Its molar mass is
60.0 g/mol. What is its molecular formula?

11. A compound is 64.9%carbon, 13.5% hydrogen and 21.6% oxygen. Its molar mass is 74.0 g/mol.
What is its molecular formula?

12. A compound is 54.5% carbon, 9.1% hydrogen and 36.4% oxygen. Its molar mass is 88.0 g/mol.
What is its molecular formula?

Answers

Answer:

Question 7 to 12 are given in attached file because character limit is only 5000

Explanation:

1.  A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of  sulfur. Calculate the empirical formula of this compound.

Given data:

Mass of sample = 15 g

Mass of sodium = 8.83 g

Mass of sulfur = 6.17 g

Empirical formula = ?

Solution:

Number of gram atoms of Na = 8.83 / 23 = 0.4

Number of gram atoms of S = 6.17 / 32 = 0.2

Atomic ratio:

            Na               :               S          

           0.4/0.2         :            0.2/0.2  

            2                  :               1        

Na : S  = 2 :  1

Empirical formula is Na₂S.

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen  indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Given data:

Mass of phosphorus = 4.433 g

Mass of oxygen = 10.150 g - 4.433 g = 5.717 g

Empirical formula = ?

Solution:

Number of gram atoms of P = 4.433 / 30.9738 = 0.1431

Number of gram atoms of O = 5.717/ 15.999 = 0.3573

Atomic ratio:

            P                        :               O          

        0.1431/0.1431         :            0.3573/0.1431

            1                         :                  2.5        

P : O  = 2(1 : 2.5)

Empirical formula is P₂O₅.

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Given data:

Percentage of sodium = 36.48%

Percentage of sulfur = 25.41%

Percentage of oxygen = 38.11%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 36.48 / 23 = 1.6

Number of gram atoms of S = 25.41/ 32 = 0.8

Number of gram atoms of O = 38.11/ 16 = 2.4

Atomic ratio:

            Na              :               S              :      O

        1.6/0.8            :            0.8/0.8       :     2.4/0.8

            2                  :                1              :       3

Na: S : O  = 2 :  1 : 3

Empirical formula is Na₂SO₃.

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Given data:

Percentage of iron = 63.52%

Percentage of sulfur = 36.48%

Empirical formula = ?

Solution:

Number of gram atoms of Fe = 63.52 / 55.845 = 1.14

Number of gram atoms of S = 36.48 / 32 = 1.14

Atomic ratio:

            Fe                :                   S            

        1.14/1.14            :               1.14/1.14    

            1                  :                   1            

Fe : S  = 1 :  1

Empirical formula is FeS.

5. Qualitative analysis shows that a compound contains 32.38%sodium, 22.65% sulfur and 44.99 %  oxygen. Find the empirical formula of this compound.

Given data:

Percentage of sodium = 32.38%

Percentage of sulfur = 22.65%

Percentage of oxygen = 44.99%

Empirical formula = ?

Solution:

Number of gram atoms of Na = 32.38 / 23 = 1.4

Number of gram atoms of S = 22.65/ 32 = 0.7

Number of gram atoms of O = 44.99/ 16 = 2.8

Atomic ratio:

            Na              :               S               :            O

        1.4/0.7            :            0.7/0.7        :           2.8/0.7

            2                  :                1             :             4

Na: S : O  = 2 :  1 : 4

Empirical formula is Na₂SO₄.

6. Analysis of a 20.0 gram sample of a compound containing only calcium and bromine indicates that  4.00 grams of calcium are present. What is the empirical formula of the compound formed?

Given data:

Mass of sample = 20g

Mass of bromine = 20 g - 4 g = 16 g

Mass of calcium = 4 g

Empirical formula = ?

Solution:

Number of gram atoms of bromine = 16 / 80= 0.2

Number of gram atoms of calcium =  4/ 40= 0.1

Atomic ratio:

            Ca                :               Br      

        0.1/0.1              :            0.2/0.1

            1                   :                2      

Ca: Br  = 1 :  2

Empirical formula is CaBr₂.

Answer:

1)Na2S

2)P2O5

3)Na2SO3

4)FeS

5)Na2SO4

6)CaBr2

7)C8H20Pb

8)P4O10

9)C3Cl3N3O3

10)C2H4O2

11)C4H10O

12)C4H8O2

Explanation:

1. A 15.0 gram sample of a compound is found to contain 8.83 grams of sodium and 6.17 grams of  sulfur. Calculate the empirical formula of this compound.

Moles Na = 8.83 grams / 22.98 g/mol = 0.384 moles

Moles S = 6.17 grams / 32.065 g/mol = 0.192 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 0.384  / 0.192  = 2

S: 0.192 /0.192 = 1

For each mol S we have 2 mol Na

The empirical formula is Na2S

2. Analysis of a 10.150 gram sample of a compound known to contain only phosphorus and oxygen  indicates a phosphorus content of 4.433 grams. What is the empirical formula of this compound?

Mass of oxygen = 10.150 - 4.433 = 5.717 grams

Moles P = 4.433 grams / 30.97 g/mol

Moles P = 0.143 moles

Moles O = 5.717 grams / 16.0 g/mol = 0.357 moles

To find the mol ratio we divide by the smallest amount of moles

P: 0.143 moles / 0.143 moles = 1

O = 0. 357 moles / 0.143 moles = 2.5

For each P atom we have 2.5 O atoms

The empirical formula is P2O5

3. A compound is found to contain 36.48%Na, 25.41% S, and 38.11% O. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Na = 36.48 grams / 22.98 g/mol = 1.587 moles

Moles S = 25.41 grams / 32.065 g/mol = 0.792 moles

Moles O = 38.11 grams / 16.0 g/mol = 2.382 moles

To find the mol ratio we divide by the smallest amount of moles

Na: 1.587 moles / 0.792 moles = 2

S: 0.792 moles / 0.792 moles = 1

O: 2.382 moles / 0.792 = 3

The empirical formula = Na2SO3

4. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

Suppose the mass of the compound = 100 grams

Moles Fe = 63.52 grams / 55.845 g/mol = 1.137 moles

Moles S = 36.48 grams / 32.065 g/mol = 1.137 moles

The empirical formula is FeS

An air mass that forms over Texas will be a ____?Question 1 options:


continental polar air mass


continental tropical air mass


maritime polar air mass


maritime tropical air mass

Answers

Answer: continental tropical air mass

Explanation:

The continental tropical mass of air forms over the Texas. It forms in the interior of the subtropical continents at about 150 to 350 North to South latitude. They can be characterized by the dry and hot air masses due to the temperature and moisture present in their source region. The tropical air masses originate in the tropical regions (warm regions) these typically have low air pressure.

Answer:

2

Explanation:

continental tropical air mass