A bicycle rider travels 50.0km in 2.5 hours.what is the cyclist average speed

All substances are matter but all matters are not substance. Matter is generally a loose term used in respect to a substance. Matter and substance are sometimes used for the same context, this is certainly not correct. Various examples have already proved that a matter will not always be a substance depending on its physical nature, but a substance is always a matter.

(cc/ to Taskmasters , I just switched it up a little bit.)

Answer:

1. Temperature

2. Surface area

3. Catalyst

Explanation:

Answer:

Heat, molecular size, and surface area

Explanation:

Heat: higher temperatures are directly correlated to higher kinetic energy. With more kinetic energy, the solute molecules move faster and the bonds are more likely to break.

Molecule size: usually at the same temperature and pressure, the solute with smaller molecular size dissolves faster. A large molecule usually has a heavier weight and size, which makes it more difficult for the solvent to surround it and help break its bonds.

Surface area: increasing surface area generally increases the solubility rate. This is because with a higher area, more molecules are exposed to the solvent, so their bonds are more likely to break.

Answer: Humans might extract more water from the ground.

Explanation:

Humans may face the situations of unavailability of water supply. Majority of the human population is dependent upon the source of water like oceans, lakes, rivers, ponds and wells for their water supply.

But in the scarcity of water from the drainage supply by the water body human population should search for a relevant alternative such as groundwater. Groundwater is the underground reservoir of water. The water in the groundwater reservoir accumulates from the sources such as rainfall, snow melting, river and oceanic water.

Humans should search for groundwater reservoir for compensating the scarcity of water.

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