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Studentka2010 [4]
3 years ago
9

What ingredients are used to make chloroform. :p

Chemistry
2 answers:
fenix001 [56]3 years ago
6 0
Bleach, rubbing alcohol or acetone and bleach
mr Goodwill [35]3 years ago
4 0

Answer: i dont no

Explanation:

ddd

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What phenomenon in nature do scientists use to define the length of a meter? Why is this a better definition than “the length of
sergey [27]

Answer:

The natural phenomenon used to describe the length of a meter is the speed of light. The length of a meter is the length a light path travels in 1/(299792458) seconds through a vacuum.

The definition is better due to the uncertainty involved in the use of the length of a standard meter stick because the length of the meter stick could change due to atmospheric conditions from place to place

Explanation:

8 0
3 years ago
A thermometer reads an outside air temperature of 35°c. What is the temperature in degrees Fahrenheit
Hoochie [10]

35°c is equal to 95°f

To do this multiply 35 and 1.8

35 x 1.8=63

Now add 32

Resulting in the answer 95

(The equation for to solve for c and f is c1.8+32=f

3 0
3 years ago
Which of the situations below describes an ethical dilemma related
fredd [130]

Answer: C

Explanation: Pesticides protect our crops, but some studies show that they cause consumers to gain weight:)

8 0
3 years ago
PLEASEEEEEEEEEEEE HELPPPPPPPP I BEGGGGG FOR HELPPPPP
Elza [17]

Answer: There are 21.08\times 10^{23} molecules in 63.00 g of H_2O

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles

1 mole of H_2O contains =  6.023\times 10^{23} molecules

Thus 3.5 moles of H_2O contains =  \frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23} molecules.

There are 21.08\times 10^{23} molecules in 63.00 g of H_2O

3 0
3 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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