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3 years ago

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xIn the reaction A B + C + heat, ______. there is a net input of energy the potential energy of the products is greater than tha

t of the reactant the potential energy of the products is the same as that of the reactant the potential energy of the products is less than that of the reactant entropy has decreased

1 answer:

kow [346]3 years ago

6 0

Answer: Option (d) is the correct answer.

Explanation:

A chemical reaction in which heat energy is liberated is known as an exothermic reaction.

For example, $A \rightarrow B + C + Heat$

In an exothermic reaction, energy of reactants is more than the energy of products.

This means that potential energy of products is less than the potential energy of reactants.

Thus, we can conclude that in the reaction $A \rightarrow B + C + heat$ , the potential energy of the products is less than that of the reactant.

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3 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

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$hc = 1242 eV-nm$

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$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

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3 years ago

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3 years ago

How long does it take electrons to get from

OlgaM077 [116]

We need to find the time it takes an electron to move in the given circuit.

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

I = Current = 134 A

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

A = Area = $38.9\ \text{mm}^2$

L = Length = 92.2 cm

$\rho$ = Density of copper = $8960\ \text{kg/m}^3$

M = Molar mass of copper = 63.5 g/mol

$n_v$ = Number of valence electrons of copper = 1

e = Charge of electron = $1.6\times 10^{-19}\ \text{C}$

Number of charge carriers per unit volume is given by

$n=\dfrac{\rho N_An_v}{M}\\\Rightarrow n=\dfrac{8960\times 6.022\times 10^{23}\times 1}{63.5\times 10^{-3}}\\\Rightarrow n=8.497\times 10^{28}\ \text{m}^{-3}$

Time taken is given by

$t=\dfrac{LAne}{I}\\\Rightarrow t=\dfrac{92.2\times 10^{-2}\times 38.9\times 10^{-6}\times 8.497\times 10^{28}\times 1.6\times 10^{-19}}{134}=3638.83\ \text{s}\\\Rightarrow t=\dfrac{3638.83}{60}=60.65\ \text{minutes}$

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

Learn more:

brainly.com/question/1426683

brainly.com/question/170663

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