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2 years ago

7

What happens to the reaction 2 N O 2 in equilibrium with N 2 O 4 plus 57.3 kilojoules when the temperature of the reaction is in

creased?
more product forms

more reactant forms

the reaction stops

no change occurs

2 answers:

Veseljchak [2.6K]2 years ago

6 0

Answer:

more reactant forms

Explanation:

Burka [1]2 years ago

3 0

Answer:

More reactant forms.

Explanation:

Given reaction is,

$2NO_{2}$ ⇒ $N_{2} O_{4}+57.3 KiloJoules$ per mole

This is an Exothermic Reaction,(ΔE=-57.3KJper mole)

We know the equilibrium point of all Exothermic reactions moves leftward and more reactant is formed at the equilibrium.

<u>Reason:</u>

As heat is being produced in the reaction the additional heat(57.3KJpermole) can be <u>thought of as a product</u> of the reaction.

So,if you increase the temperature ,you provide heat energy,

(in other words heat energy is given) and hence the concentration of the products increases.

So, with respect to LeChateliers Principle,

As the concentration of products is increased by external means,more of the reactants are produced at the equilibrium of the reaction.

Therefore amount of reactants increases as <u>more reactant forms.</u>

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An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b

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Explanation:

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3 years ago

Determine the velocity of a car that travels 28km in 2.2 hours

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Answer:

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2 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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Explanation:

The answer I think

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