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Bess [88]
3 years ago
15

Write the mechanism for the two steps of the hydroboration/oxidation reaction of indene

Chemistry
1 answer:
iVinArrow [24]3 years ago
5 0
It's difficult to write it down, but I'll attach you a good example of hydroboration of indene. I hope you'll find it helpful.

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A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is it’s density
vichka [17]

Answer:

1.6g/mL

Explanation:

Density equation is D=m/v

Density = g/mL

m=mass of sample in grams

v = volume of sample in mL

The volume of a square can be calculated by V=l*w*h.

In this case it is 5cm*5cm*5cm = 125cm^3

Since we know that 1cm^3 ~ 1mL we can convert the volume to mL as so:

125cm^3 (1mL/(1cm^3)) = 125mL

Then simply plug into the density equation:

D=200g/125mL = 1.6g/mL

3 0
3 years ago
What is a chemical bond?
nadezda [96]
The answer is A. the attraction between atoms that enables the formation of chemical compounds.
8 0
3 years ago
a sample of solid is decomposed and found to contain 6.52g of potassium, 4.34 g of chromium and 5.34 of oxygen, what is the empi
Fynjy0 [20]

Answer:

K₂CrO₅

Explanation:

The empirical formula is the simplest formula of a compound. To find the empirical formula, we follow the procedure below:

Elements                         Potassium                 Chromium         Oxygen

Mass                                  6.52                             4.34                  5.34

Molar mass                          39                               60                      16

Number of moles             6.52/39                     4.34/60             5.34/16

                                             0.167                          0.072              0.333

Divide through by

the smallest                      0.167/0.072             0.072/0.072          0.333/0.072

                                            2.3                               1                                4.6

                                             2                                 1                                  5

Empirical formula K₂CrO₅

5 0
3 years ago
Please help me with this: Create 20 bullet points specifically about energy exchanges in Earth's systems. Also, it doesn't have
raketka [301]

The below is about the energy exchanges in earth systems.                                                                                                          

<u>Explanation</u>:

  • Energy exchanges in earth systems are of many types.  The earth systems are atmosphere, geosphere, stratosphere, hydrosphere, and biosphere. All these earth systems exchange energy with each other.
  • The earth gains energy reflected from the sky. It converts that energy back to space. That energy is equally given to all the planets in the sky.
  • Each planet will absorb that energy and radiate heat. This heat is absorbed by all the places on the earth. So this is the energy exchange in the earth systems.                                                                                
7 0
3 years ago
Read 2 more answers
Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

     = 13.09

4 0
3 years ago
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