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3 years ago

How many moles of nitrogen are needed to completely convert 6.34 mol of hydrogen?

Physics

2 answers:

Sergeu [11.5K]3 years ago

5 0

Answer:

n = 2.11 moles

Explanation:

As we know by the reaction of hydrogen and nitrogen

$N_2 + 3H_2 --> 2NH_3$

so here from above equation we know that for complete conversion of 3 moles of hydrogen we need 1 mole of nitrogen

so here we know that 6.34 moles of hydrogen is available

so we require 1/3 times of moles of nitrogen for complete reaction

so number of moles of hydrogen will be

$n = \frac{1}{3} \times 6.34$

$n = 2.11 moles$

so here we can say it requires 2.11 moles of nitrogen for complete reaction of 6.34 moles of hydrogen

____ [38]3 years ago

4 0

If we use the equation:
N2 + 3H2 --> 2NH3
Then
1 mol of Nitrogen required 3 moles of Hydrogen
x mols : 6.34mols
X = 6.34/3
X = 2.11 moles of Nitrogen are required.

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When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g

mr Goodwill [35]

The distance an object falls from rest through gravity is

   D = (1/2) (g) (t²)

Distance = (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected
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So I'm going to start by rearranging the equation
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D = (1/2) (g) (t²)

Multiply each side by 2 :    2 D =      g t²

Divide each side by ' g ' : 2 D/g = t²

Square root each side: t = √ (2D/g)

Looking at the equation now, we can see what happens
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-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
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-- They don't change by the same factor, because 1/g is inside
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Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.

So we expect ' t ' to increase by √6 = 2.45 times.

It would take the same bottle (2.45 x 4.95) = 12.12 seconds
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4 0

3 years ago

Read 2 more answers

A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.

charle [14.2K]

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