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IceJOKER [234]
3 years ago
7

A volcanic ash layer between sedimentary rock layers is used by geologists to

Chemistry
1 answer:
lozanna [386]3 years ago
8 0
I believe the answer is - Correlate widely separated sedimentary rock layers
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Compare and contrast a pure substance with a mixture and give an example.
Inessa [10]

Of any substance? What about two powders such as 1/2 cup of white flour, then mixing 1/2 cup of salt together with the flour. This would then form a more coarse mixture being that the two particles differ in shape.

3 0
3 years ago
Give two uses of compound under each of the following, in medicine and in industry​
NemiM [27]

Explanation:

#Medicine

Compounds used as medicines are most often organic compounds, which are often divided into the broad classes of small organic molecules (e.g., atorvastatin, fluticasone, clopidogrel) and "biologics" (infliximab, erythropoietin, insulin glargine),

#Industry

Polymers and plastics such as polyethylene, polypropylene, polyvinyl chloride, polyethylene terephthalate, polystyrene and polycarbonate

7 0
3 years ago
the development and use of the ______ a common and important piece of laboratory equipment, led scientist to the discovery of se
Liula [17]

The answer is the Microscope.

6 0
3 years ago
WS Percent yield don’t understand how to do would appreciate the help
Grace [21]

Answer:

1. Theoretical yield of NaOH is 22.72 g

2. Percentage yield of NaOH = 22.14%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaHCO₃ —> NaOH + CO₂

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 1 mole (i.e 40 g) of NaOH and 1 mole (i.e 44.01 g) of CO₂.

Next, we shall determine the number of mole of NaHCO₃ that will decompose to produce 25 g of CO₂. This can be obtained as follow:

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 44.01 g of CO₂.

Therefore, Xmol of NaHCO₃ will decompose to 25 g of CO₂ i.e

Xmol of NaHCO₃ = 25 / 44.01

Xmol of NaHCO₃ = 0.568 mole

1. Determination of the theoretical yield of NaOH.

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 40 g of NaOH.

Therefore, 0.568 mole of NaHCO₃ will decompose to produce = 0.568 × 40 = 22.72 g of NaOH.

Thus, the theoretical yield of NaOH is 22.72 g

2. Determination of the percentage yield of NaOH.

Theoretical yield of NaOH = 22.72 g

Actual yield of NaOH = 5.03 g

Percentage yield of NaOH =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.03 / 22.72 × 100

Percentage yield of NaOH = 22.14%

4 0
2 years ago
The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of
Olin [163]

Answer:

            Empirical Formula  =  C₃H₆O₁

Solution:

Data Given:

                      Mass of Ethyl Butyrate  =  3.61 mg  =  0.00361 g

                      Mass of CO₂  =  8.22 mg  =  0.00822 g

                      Mass of H₂O  =  3.35 mg  =  0.00335 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100

                      %C  =  (2.277) × (12 ÷ 44) × 100

                      %C  =  2.277 × 0.2727 × 100

                      %C  =  62.09 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.9279) × (2.02 ÷ 18.02) × 100

                      %H  =  0.9279 × 0.1120 × 100

                     %H  =  10.39 %


                      %O  =  100% - (%C + %H)

                      %O  =  100% - (62.09% + 10.39%)

                      %O  =  100% - 72.48%

                      %O  =  27.52 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 62.09 ÷ 12.01

                      Moles of C  =  5.169 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 10.39 ÷ 1.01

                      Moles of H  =  10.287 mol


                      Moles of O  =  %O ÷ At.Mass of O

                      Moles of O  = 27.52 ÷ 16.0

                     Moles of O  =  1.720 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     O

             5.169                                10.287                              1.720

       5.169/1.720                       10.287/1.720                     1.720/1.720

               3.00                                   5.98                                   1

                  3                                      ≈ 6                                     1

Result:

         Empirical Formula  =  C₃H₆O₁

8 0
3 years ago
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