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MatroZZZ [7]
2 years ago
15

Under appropriate conditions, nitrogen and hydrogen undergo a combination reaction to yield ammonia: N2 (g) + 3H2 (g) → 2NH3 (g)

A 9.3-g sample of hydrogen requires ________ g of N2 for a complete reaction.
Chemistry
1 answer:
kherson [118]2 years ago
6 0

Answer:

43 g N₂

Explanation:

To find the mass of N₂ required, you need to (1) convert grams H₂ to moles H₂ (via molar mass), then (2) convert moles H₂ to moles N₂ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles N₂ to grams N₂ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 2 sig figs to match the amount of sig figs of the given value.

Molar Mass (H₂): 2(1.008 g/mol)

Molar Mass (H₂): 2.016 g/mol

Molar Mass (N₂): 2(14.009 g/mol)

Molar Mass (N₂): 28.018 g/mol

1 N₂(g) + 3 H₂(g) -----> 2 NH₃(g)

9.3 g H₂           1 mole             1 mole N₂             28.018 g
--------------  x  ----------------  x  --------------------  x  ------------------  =  43 g N₂
                        2.016 g            3 moles H₂            1 mole

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When a metal was exposed to light at a frequency of 4.07× 1015 s–1, electrons were emitted with a kinetic energy of 3.30× 10–19
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Answer :  The maximum number of electrons released = 1.432\times 10^{12}electrons

Explanation : Given,

Frequency = 4.07\times 10^{15}s^{-1}

Kinetic energy = 3.30\times 10^{-19}J

Total energy = 3.39\times 10^{-7}J

First we have to calculate the work function of the metal.

Formula used :

K.E=h\nu -w

where,

K.E = kinetic energy

h = Planck's constant = 6.626\times 10^{-34}J/s

\nu = frequency

w = work function

Now put all the given values in this formula, we get the work function of the metal.

3.30\times 10^{-19}J=(6.626\times 10^{-34}J/s\times 4.07\times 10^{15}s^{-1})-w

By rearranging the terms, we get

w=2.367\times 10^{-18}J

Therefore, the works function of the metal is, 2.367\times 10^{-18}J

Now we have to calculate the maximum number of electrons released.

The maximum number of electrons released = \frac{\text{ Total energy}}{\text{ work function}}

The maximum number of electrons released = \frac{3.39\times 10^{-7}J}{2.367\times 10^{-19}J}=1.432\times 10^{12}electrons

Therefore, the maximum number of electrons released is 1.432\times 10^{12}electrons

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Answer : The atom with the same number of neutrons as ^{12}C is, ^{11}B

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Number of neutrons = Atomic mass number - Atomic number = 14 - 7 = 7

Therefore, the atom with the same number of neutrons as ^{12}C is, ^{11}B

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