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3 years ago

2C(s) + 2H2(g) → C2H4(g): ΔH = +52.4 kJ/mol Which statement about this reaction is correct?

Chemistry

1 answer:

artcher [175]3 years ago

6 0

<h3><u>Answer;</u></h3>

26.2 kJ of energy are absorbed for every mole of carbon reacted

<h3><u>Explanation;</u></h3>

ΔH is the enthalpy change which is given by subtracting the energy of reactants from the energy of products.
That is; ΔH = energy of products - energy of reactants
<u>In this case, it means that C2H4 has more energy than the the reactants. 52.4 J of energy to be precise, this energy is absorbed during the reaction.</u>
<u>However, Two moles of Carbon require that much energy, thus one mole will require only 26.2 to react. </u>

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The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:

bagirrra123 [75]

Answer:

$12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow 14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)$

Explanation:

The steps of the Ostwald process:

$4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)$

$2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g)$

$3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g)$

Combinning the equations:

$4 NH_3 (g) + 5 O_2 (g) \longrightarrow 4 NO (g) + 6 H_2O (g)$

$(2 NO (g) + O_2 (g) \longrightarrow 2 NO_2 (g))*2$

$(3 NO_2 (g) + H_2O (l) \longrightarrow 2 HNO_3 (g) + NO (g))*4/3$

$4 NH_3 (g)+ 4 NO (g)+ 7 O_2 (g) + 4 NO_2 (g) +4/3 H_2O (l) \longrightarrow 4 NO (g) + 6 H_2O (g) + 4 NO_2(g) + 8/3 HNO_3 (g) + 4/3 NO (g)$

Simplifying:

$4 NH_3 (g)+ 7 O_2 (g) + \longrightarrow 14/3 H_2O (l) + 8/3 HNO_3 (g) + 4/3 NO (g)$

$12 NH_3 (g)+ 21 O_2 (g) + \longrightarrow 14 H_2O (l) + 8 HNO_3 (g) + 4 NO (g)$

The overall reaction is endothermic becuase the formation of new chemical bonds requires energy consumption.

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3 years ago

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3 years ago

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What are the 3 principles of the Kinetic molecular theory?

Sever21 [200]

Answer:

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3 years ago

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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o

Andreas93 [3]

<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>

The reaction between NaOH and H₂SO₄ is a neutralization reaction.
The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>

Number of moles = Mass ÷ Molar mass
Molar mass of NaOH is 40.0 g/mol

Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

= 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>

From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
Thus, Moles of H₂SO₄ = moles of NaOH × 2

= 0.33 moles × 2

= 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>

When given the molarity of an acid and the number of moles we can calculate the volume of the acid.

That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

= 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

7 0

3 years ago

What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme

boyakko [2]

<u>Answer:</u> The longest wavelength of light is 656.5 nm

<u>Explanation:</u>

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.096776\times 10^7m^{-1}$

$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

4 0

3 years ago