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3 years ago

Ammonia (NH3) can be produced by the reaction of hydrogen gas with nitrogen gas:

Chemistry

1 answer:

klio [65]3 years ago

3 0

Answer:

10%

Explanation:

Given data:

Actual yield = 0.13 moles

Percent yield = ?

Solution:

Balanced chemical equation.

3H₂ + N₂ → 2NH₃

First of all we will calculate the mass of given moles of ammonia.

Number of moles = mass / molar mass

Mass = 0.13 × 17 g/mol

Mass = 2.21 g

2.21 g is experimental yield of ammonia.

Hydrogen is limiting reactant because only two moles of hydrogen present.

we will compare the moles of hydrogen and ammonia from balance chemical equation.

H₂ : NH₃

3 : 2

2 : 2/3×2 = 1.33 moles

Mass of ammonia

Mass = number of moles × molar mass

Mass = 1.33 mol × 17 g/mol

Mass = 22.61 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 2.21 g / 22.61 g

Percent yield =9.8% which is almost 10%

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Calculate the oxidation numbers of the elements in this equation 2 Al2O3 → 4 Al + 3 O2

riadik2000 [5.3K]

Explanation:

Answer

Open in answr app

The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O ) = 0

2× ( Oxidation number of Al) +3(−2)=0

2× ( oxidation number of Al) +6

∴ Oxidation number of Al =+3

5 0

3 years ago

A sample of neon initially has a volume of 2.80 L at 23 degrees Celsius. What is final temperature, in degrees Celsius, is neede

finlep [7]

The relation between the volume and the temperature of the gas is given by Charles's law. The final temperature of the gas at 0.75 liters is -193.8°C.

<h3>What is Charles's law?</h3>

Charles's law was derived from the ideal gas equation and is used to state the relationship between the temperature and the volume of the gas. With a decrease in volume the temperature decreases.

If the pressure is kept constant then with an increase in temperature the volume of the gas expands. The law is given as,

V₁ ÷ T₁ = V₂ ÷ T₂

Given,

Initial volume (V₁) = 2.80 L

Initial temperature (T₁) = 23 °C = 296.15 K

Final volume (V₂) = 0.75 L

Final temperature = T₂

Substituting the values above as:

T₂ = (V₂ × T₁) ÷ V₁

= 0.75 × 296.15 ÷ 2.80

= 79.325 K

Kelvin is converted as, 79.325K − 273.15 = -193.8°C

Therefore, the final temperature is -193.8°C.

Learn more about Charle's law, here:

brainly.com/question/16927784

#SPJ1

7 0

1 year ago

The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con

bonufazy [111]

Answer: The concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=3.82days$

Putting values in above equation, we get:

$k=\frac{0.693}{3.82}=0.181days^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.181days^{-1}$

t = time taken for decay process = 3.00 days

$[A_o]$ = initial amount of the reactant = $1.45\times 10^{-6}mol/L$

[A] = amount left after decay process = ?

Putting values in above equation, we get:

$0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}$

$[A]=3.83\times 10^{-30}mol/L$

Hence, the concentration of radon after the given time is $3.83\times 10^{-30}mol/L$

7 0

3 years ago

If 25.4 grams of water react to form 2.8 grams of hydrogen, how many grams of oxygen must simultaneously be formed?

aalyn [17]

The mass of oxygen and hydrogen must be equal to the mass of the substance they create the water. So if the hydrogen is 2.8 g the oxygen must account for the rest of the mass. Basically just subtract 25.4-2.8=mass of oxygen

5 0

3 years ago

Read 2 more answers

Vanyuwa [196]

Answer:

If the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

Explanation:

Boyle's law that says "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

P * V = k

To obtain the proportionality factor k you must make the quotient:

$k=\frac{V2}{V1} =\frac{6 L}{2 L}$

k= 3

This means that if the volume of a gas increased from 2 to 6 L while the temperature was held constant, the pressure of the gas decreased by a factor of 3.

4 0

3 years ago