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3 years ago

Please help me I will mark you as BRAINLIST

Chemistry

1 answer:

sashaice [31]3 years ago

4 0

31: Z, 32:BB, 33:K, 34:L, 35: M, 36: O, 37: D, 38: i, 39: G

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When (R)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When

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Answer:

See explanation

Explanation:

The nucleophile here is CH3OH. We know that CH3OH is a good nucleophile that promotes SN2 reanction. However, (R)-6-bromo-2,6-dimethylnonane is a tertiary alkyl halide so the reaction proceeds by SN1 mechanism. This means that a racemic mixture is obtained at the end of the reaction because the attack occurs at the stereogenic carbon atom (6R) hence the product is optically inactive.

On the other hand, when (5R)-2-bromo-2,5-dimethylnonane is reacted with CH3OH, an optically active product is obtained because; though a tertiary alkyl halide and reaction occurs by SN1 mechanism, the attack does not occur at the stereogenic carbon atom (5R). Therefore, an optically active product is obtained in this case.

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3 years ago

What factors influence the efficiency of a solar oven?

Anna007 [38]

The more sunlight that is present, the better the oven will be able to produce electricity. This means that a solar oven will be more efficient on a sunny day than on a cloudy day. Also, depending on the time of day, if the solar panel is in shadow, the oven will greatly decrease in efficiency.

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3 years ago

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

Troyanec [42]

$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the <em>change</em> in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.