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3 years ago

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The standard heats of combustion (δh∘) per mole of 1,3-butadiene, c4h6(g); butane, c4h10(g); and h2(g) are −2540.2,−2877.6, and

−285.8kj, respectively. Use these data to calculate the heat of hydrogenation of 1,3-butadiene to butane. C4h6(g)+2h2(g)→c4h10(g)

1 answer:

ryzh [129]3 years ago

5 0

solution:

Hydration is the addition of water; hydrogenation is the addition of hydrogen.

desire rxn: _C4H6(g) + 2 H2(g)-----> C4H10(g)___dHhy = ??

knowns:

__________C4H6 + 11/2 O2 --------> 4CO2 + 3H2O______dHox = -2540.2 kJ/mole

__________4CO2 + 5H2O -----------> C4H10 + 13/2 O2___-dHox = 2877.6 kJ/mole

___________2(1/2 O2 + H2 -------------> H2O)___________2*dHox = 2(-285.8 kJ/mole)

Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem

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3 years ago

What is the area of triangle ABC with vertices A(x¹,y¹), B(x²,y²)and C (x³,y³)??????????<br>

AlexFokin [52]

Answer:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

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Explanation:

Given

$A = (x_1,y_1)$

$B = (x_2,y_2)$

$C = (x_3,y_3)$

Required

Determine the area

The area of a triangle is :

$Area = \frac{1}{2}|A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|$

By substituting values for the x and y coordinates of A, B and C;

We have:

$Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|$

So:

For instance

$A = (0,3)$

$B= (2,1)$

$C = (2,3)$

The area is:

$Area = \frac{1}{2}|0(1-3) + 2(3-3) + 2(3-1)|$

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2 years ago

A 25 liter balloon holding 1.5 moles of carbon dioxide leaks. If we are able to determine that 0.9 moles of carbon dioxide escap

ad-work [718]

Answer:

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Given that ::

v1 = 25L ; n1 = 1.5 mole ; v2 =? ; n2 = (1.5-0.9) = 0.6 mole

Using the relation :

(n2 * v1) / n1 = (n2 * v2) / n2

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v2 = (0.6 mole * 25 Litre) / 1.5 mole

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2 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

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calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

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3 years ago

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kirill [66]

Answer:

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Explanation:

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2 years ago