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3 years ago

Bond Energy

Chemistry

1 answer:

Vlad1618 [11]3 years ago

7 0

The required energy to break a C=0 bond is 749kj/mol and the energy to break an H-O bond is 428kj/mol, so in order to form those bonds we have to add a - for each of those values.

For bond making It takes roughly 100 kcal of energy to break 1 mol of C–H bonds, so we speak of the bond energy of a C–H bond as being about 100 kcal/mol.

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Which of the following is not true about bias help pls

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The second one, vocab and definition doesn't add up

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3 years ago

Hello can someone help me please and thanksss

Svetlanka [38]

the equation of the parabola is given as $y = 1/4 (x-6)^2 +1$ . D

<h3>How to find the parabola</h3>

Given the focus(6,2) and the directrix, y = 0

Use the formula

$\sqrt{(x-6)^{2} + (y-2)^{2} } }$ = (y - 0)

Find the square of both sides, square root is removed from the side with it and square added on the other side

$(x-6)^2 + (y-2)^2 = (y- 0)^{2}$

Expand the expression and bring all terms to one side

$x^{2} - 6x - 6x + 36 +y^{2} -2y- 2y +4$ = $y^{2} - 0$

Collect like terms

$x^{2} - 12x -4y + 40 = 0$

Make 'y' the subject of the formula

$4y = x^{2} - 12x + 40$

Divide through by 4 and substrate 4 from 40 to give a perfect quadratic equation

$y = 1/4 x^{2} - 12x + 36 + 4$

Divide factor by 4 to give 1

$y = 1/4( x^{2} -6x -6x +36) + 1$

Simplify the expanded quadratic equation

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Then insert in place into previous equation

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Learn more about a parabola here:

brainly.com/question/4061870

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2 years ago

I have an science question and i don’t know how to do it, Which math function can be used to find the time an object moved if yo

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3 years ago

For this question, the "entropy term" refers to "-TΔS". Addition reactions are generally favorable at low temperatures because _

nata0808 [166]

Answer:

Lowering the temperature typically reduces the significance of the decrease in entropy. That makes the Gibbs Free energy of the reaction more negative. As a result, the reaction becomes more favorable overall.

Explanation:

In an addition reaction there's a decrease in the number of particles. Consider the hydrogenation of ethene as an example.

$\rm H_2C\text{=}CH_2\; (g) + H_2\; (g) \stackrel{\text{Ni}^\ast}{\to} H_3C\text{-}CH_3\; (g)$ .

When $\rm H_2$ is added to $\rm H_2C\text{=}CH_2$ (ethene) under heat and with the presence of a catalyst, $\rm H_3C\text{-}CH3$ (ethane) would be produced.

Note that on the left-hand side of the equation, there are two gaseous molecules. However, on the right-hand side there's only one gaseous molecule. That's a significant decrease in entropy. In other words, $\Delta S < 0$ .

The equation for the change in Gibbs Free Energy for a particular reaction is:

$\Delta G = \Delta H + (\underbrace{- T \, \Delta S}_{\text{entropy}\atop \text{term}})$ .

For a particular reaction, the more negative $\Delta G$ is, the more spontaneous ("favorable") the reaction would be.

Since typically $\Delta S < 0$ for addition reactions, the "entropy term" of it would be positive. That's not very helpful if the reaction needs to be favorable.

$T$ (absolute temperature) is always nonnegative. However, lowering the temperature could help bring the value of

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3 years ago

A chemistry student weighs out 0.09666 g of phosphoric acid (H3PO4), a triprotic acid, into a 250.volumetric flask and dilutes t

Ber [7]

Answer: 14.62 ml

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$V_s$ = volume of solution in ml = 250 ml

${\text {moles of solute}=\frac{\text {given mass}}{\text {molar mass}}=\frac{0.09666g}{98g/mol}=9.9\times 10^{-4}$

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{9.9\times 10^{-4}\times 1000}{250ml}$

$Molarity=3.9\times 10^{-3}M$

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_3PO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=3.9\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.2000M\\V_2=?$

Putting values in above equation, we get:

$3\times 3.9\times 10^{-3}\times 250=1\times 0.2000\times V_2\\\\V_2=14.62ml$

Thus the volume of NaOH solution the student will need to add to reach the final equivalence point is 14.62 ml

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3 years ago