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3 years ago

One method to produce nitrogen in the lab is to react ammonia with copper (II) oxide: NH3(

Chemistry

2 answers:

7nadin3 [17]3 years ago

7 0

Answer:

47.05% is the percent yield of nitrogen in the reaction.

Explanation:

$2NH_3(g)+3CuO(s)\rightarrow 3Cu(s)+3H_2O(l) +N_2(g$

Theoretical yield of nitrogen gas = x

Moles of ammonia = $\frac{40.0 g}{17 g/mol}=2.3529 mol$

According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.

Then 2.3529 mol of ammonia will give:

$\frac{1}{2}\times 2.3529 mol=1.1764 mol$ of nitrogen gas

Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g

Experiential yield of nitrogen gas = 15.5 g

Percentage yield:

$\% yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$\% yield=\frac{15.5 g}{32.94 g}\times 100=47.05\%$

47.05% is the percent yield of nitrogen in the reaction.

Tpy6a [65]3 years ago

6 0

40g(1mol NH3/17.04g per mol), N2's molar mass=28.02g
3CuO+2NH3=>3Cu+N2+3H2O
(2.3473mol NH3)(1 mol N2/2 mol NH3)(28.02g/1mol)=32.8873g N2
15.5/32.8873=47.1306% yield

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A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay fo

Ksenya-84 [330]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:

$T = x*P$

$80 = x*20$

$80 = 20\:x$

$20\:x = 80$

$x = \dfrac{80}{20}$

$\boxed{x = 4}$

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

$m = \dfrac{m_o}{2^x}$

$m = \dfrac{43}{2^{4}}$

$m = \dfrac{43}{16}$

$\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

8 0

4 years ago

Quiz. Formula Weight and % Composition

astraxan [27]

Answer:

#1- the percentage is 2.77%

#2 - 218.1336g

#3- 7.89%

#4- 63.8489g

#5- 136.1406g

#6- 340.3515g

#7- 2.387238

#8-

Explanation:

7 0

4 years ago

Iron is extracted from fe 2o 3 by a reaction with carbon (c) at very high temperature. if you take 2.86 x 10 4 g of fe 2o 3 to r

umka21 [38]

Balanced equation:
2 Fe₂O₃ + 3 C → 4 Fe + 3 CO₂
number of moles of Fe₂O₃ = weight (g) / molar mass = (2.86 x 10⁴ g) / 159.69 g/mol = 179 moles
number of moles of C = (9.05 x 10³) / 12 = 754.16 moles
2 moles of Fe₂O₃ reacts with 3 moles of C
so 179 moles of Fe₂O₃ need 268.5 moles of C only
so carbon present in excess (754.16) and Fe₂O₃ is the limiting reactant so the number of moles used from Fe₂O₃ will be all the 179 moles

6 0

3 years ago

Read 2 more answers

4. Which of the following statements is true?

Aleksandr [31]

4) The answer is: Chemical reaction rates vary with the conditions of the reaction, but nuclear decay rates do not.

Differences between nuclear and chemical reactions:

1) The electrons of an atom participates in a chemical reaction and protons in nuclear reaction.

2) Chemical reactions involve the rearrangement of electrons, because there is transfer, loss, gain and sharing of electrons in chemical reactions.

3) Nuclear reactions can produce new elements, because number of protons and neutrons is changed, chemical reactions donot produce new element.

4) Chemical reactions require heat or an input of energy to occur, this is activation energy.

5) Isotopes of the same element behave differently in a nuclear reaction, because isotopes have different number of neutrons, so products of a nuclear reaction is different.

6) The answer is: nuclear change.

Transmutation is a process that involves a change in the nucleus of an atom. It can be natural or artificial.

For example: 14/7 N + 4/2 He --> 17/8 O + 1/1 H.

This example is the first artificial nuclear transmutation described by Rutherford in 1919.

Artificial transmutation can be expressed by nuclear reaction:

Target Nuclide + Bombardment Particle → New Nuclide + Ejected Particle.

5 0

3 years ago

How many grams of NaOH are produced with the reaction of 5.00 moles of water? *

Andre45 [30]

Answer:

200 grams of NaOH are produced with the reaction of 5.00 moles of water

Explanation:

First of all you apply a rule of three to know the amount of moles of NaOH as follows: if 2 moles of water produce 2 moles of NaOH, 5 moles of water how many moles would they produce?

$moles of NaOH=\frac{5 moles of water*2 moles of NaOH}{2 moles of water}$

moles of NaOH= 5

Being:

Na: 23 g/mole
O: 16 g/mole
H: 1 g/mole

the molar mass of NaOH is: 23 g/mole + 16 g/mole + 1 g/mole= 40 g/mole

Then a rule of three applies as follows: if in 1 mole there are 40 g of NaOH, in 5 moles how much mass is there?

$mass=\frac{5 moles*40g}{1 mole}$

mass= 200 g

<u><em>200 grams of NaOH are produced with the reaction of 5.00 moles of water</em></u>

4 0

4 years ago