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notka56 [123]
3 years ago
13

Draw the correct structure for (3S,4S)-3,4-dimethylhexane. Show stereochemistry clearly. To ensure proper grading, explicitly dr

aw all four groups, including wedge/dash bonds, around a chirality center. Indicate whether the compound could exist in an optically active form.

Chemistry
1 answer:
Neko [114]3 years ago
4 0

Find the figure in the attachment

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0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
What are the radicals?Write their types with four examples.​
polet [3.4K]

Answer:

<h3>A<em><u> group of atoms behaving as a unit in a number of compounds.</u></em><em><u> </u></em></h3>

Explanation:

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4 0
3 years ago
Read 2 more answers
Qué es la hidrólisis​
gregori [183]

Answer:

Explanation:    es una reacción química entre una molécula de agua y otra macromolécula, en la cual la molécula de agua se divide y rompe uno o más enlaces químicos y sus átomos pasan a formar unión de otra especie química.

8 0
3 years ago
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If a pork roast must absorb
Black_prince [1.1K]
Let us  assume propane was the fuel
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g) = 2217kJ
1 mole ofpropane produces 3 moles of CO2 
heat absorbed by pork = 0.11 x 2217
                                     = 243.87 kJ/mol
number of moles of propane = 1700kJ / 243.87 kJ/mol
                                              = 6.971 moles
1 mole of C3H8 = 3 moles ofCO2
6.971 moles of C3H8 = ?
3 x 6.971 = 20.913 moles of CO2
Convert to grams
mass = MW x mole
          = 44 x 20.913
          = 920.172g of CO2 emitted 
7 0
3 years ago
The head loss in a turbulent flow in a pipe varies Approximant as square of velocity • Direct as the velocity • Invers as square
Eddi Din [679]

Answer:

Head loss in turbulent flow is varying as square of velocity.

Explanation:

As we know that head loss in turbulent flow given as

h_F=\dfrac{FLV^2}{2gD}

Where

F is the friction factor.

L is the length of pipe

V is the flow velocity

D is the diameter of pipe.

So from above equation we can say that

h_F\alpha V^2

It means that head loss in turbulent flow is varying as square of velocity.

We know that loss in flow are of two types

1.Major loss :Due to surface property of pipe

2.Minor loss :Due to change in momentum of fluid.

5 0
4 years ago
Read 2 more answers
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