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Olenka [21]
4 years ago
14

Give the name of the following molecule. A compound with a total of seven carbons with a double bond and the rest single bonds.

There is a chain of four carbons along the bottom with a double bond between the third and fourth carbon when counting from left to right. There is a propyl group branching off of the third carbon from the left.

Chemistry
2 answers:
victus00 [196]4 years ago
8 0
(1) compound with seven carbons with double bond and the rest single bond :
                                      HEPTENE
(2) four carbons with a double bond between the third and the fourth carbon:
                                     1-BUTENE
(3) Propyl group branching from the third carbon from the left:
                                     3-Propyl heptene
Sunny_sXe [5.5K]4 years ago
7 0

Answer:

2-ethyl-pentene.

Explanation:

Hello,

In this case, following the given indications, the molecule is shown of the attached picture. Thus, the double bond is placed as well as the propyl radical. In such a way, the resulting molecule is 2-ethyl-pentene since the longest chain has five carbons and at the second carbon there is an ethyl radical. Moreover, the double bond makes the molecule an alkene, finishing with "ene".

Best regards.

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Explanation:

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Number of moles = mass / molar mass

Number of moles = 442.8 / 147.6

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7 0
3 years ago
A student uses a solution of 1.2 molar sodium hydroxide (NaOH) to calculate the concentration of a solution of sulfuric acid (H2
Mila [183]

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CB = 1.2 M

VA =  50 mL

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8 0
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What is the de Broglie wavelength (in meters) of a 45-g golf ball traveling at 72 m/s?
Cerrena [4.2K]

Answer: The de broglie wavelength is 2.037 \times 10^{-34} m.

Explanation:

Calculate  \lambda = \frac{h}{p}as follows.

          \lambda = \frac{h}{p}

where,

          h = plank's constant = 6.6 \times 10^{-34} m^{2} kg/s

         p = momentum = mass \times velocity

Putting the values in the formula as follows.

        \lambda = \frac{h}{mass \times velocity}

                               =  \frac {6.6 \times 10^{-34} m^{2} kg/s}{0.045 kg \times 72 m/s}                        

                               =  2.037 \times 10^{-34} m

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6 0
3 years ago
Read 2 more answers
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
4 years ago
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