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2 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]2 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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Answer:

a. 123.9°C

b.

c.

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2 years ago

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

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T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

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$n=3.84mol$

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From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

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2 years ago

Find the de Broglie wavelength lambda for an electron moving at a speed of 1.00 \times 10^6 \; {\rm m/s}. (Note that this speed

masya89 [10]

(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

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So, its momentum is

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(B) $1.16\cdot 10^{-34}m$

In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

So, the momentum of the ball is

$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

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(D) $7.35\cdot 10^{-26}kg m/s$

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$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

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1 year ago

If the patient has to be administered a dosage of 2 tablets every 8 hours for 7 days, what is the number of tablets required for

Lilit [14]

If the patient has to take 2 tablets every 8 hours for 7 days.
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1 year ago

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nalin [4]

Answer:

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Explanation:

correct answer is option A

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1 year ago

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