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Nadusha1986 [10]
2 years ago
14

write a balanced chemical equation depicting the formation of one mole of pocl3(l) from its elements in their standard states.

Chemistry
2 answers:
drek231 [11]2 years ago
5 0
Thank you for posting your question here at brainly. Below is the answer:

At 25 C and 1 atm pressure (the standard state): P is a solid, O2 is a gas, and Cl2 is a gas. 

<span>P(s) + O2(g) + Cl2(g) ==> POCl3(l) </span>

<span>To make 1 mole of POCl3, we need to start with 1 mole of P, 1/2 mole of O2, and 3/2 mole of Cl2. </span>

<span>P(s) + 1/2O2(g) + 3/2Cl2(g) ==> POCl3(l) </span>

<span>NOTE: Some people write P as P4(s), in which case you would need 1/4 mole P4.</span>
7nadin3 [17]2 years ago
3 0

Answer:

Explanation:

First off, it's important to identify the constituent elements of the Compound.

From the compound, we can identify the following elements;

Phosphorus (P), Oxygen (O) and Chlorine (Cl)

So we have;

P + O + Cl --> POCl3

The standard states of elements are the phases that they adopt at a Temperature of 25°C and Pressure of 1 atm.

For POCl3, the standard form of P is P4 (s) while O2 (g) for O and Cl2 (g) for Cl. With this, we now have;

P4 + O2 + Cl2 --> POCl3

Upon balancing, we have;

P4 + 2O2 + 6Cl2 --> 4POCl3

The above equation shows 4 moles of POCl3, since the question stated one mole, we now have;

1/4 P₄ (s) +  1/2 O₂(g) +  3/2 Cl₂ (g) -----> POCl₃ (l)

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Alexxandr [17]

We are given:

Wavelength of Green light (λ) = 531 nm  OR   531 * 10⁻⁹ m

Speed of light (c) = 3* 10⁸ m/s

<u>Solving for the Frequency of the wave:</u>

We know the relation:

c = νλ    

<em>(where ν is the frequency, c is the speed of light and λ is the wavelength)</em>

3*10⁸ = ν*(531*10⁻⁹)

ν = 3*10⁸ / 531*10⁻⁹                  [dividing both sides by the wavelength]

ν = 3*10⁸*10⁹ / 531                   [ since a⁻¹ = 1/a]

ν = 3*10¹⁷ / 531                         [ since mᵃ * mᵇ = mᵃ⁺ᵇ]

ν = 10¹⁷ / 177

ν = 1000 * 10¹⁴ / 177                 [we split the numerator to simplify]

ν = 5.65 * 10¹⁴

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8 0
1 year ago
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How many grams of copper (I) chloride can be produced from the reaction of 73.5 g of copper (I) oxide with hydrochloric acid acc
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The balanced equation for the reaction is as follows

Cu₂O + 2HCl ---> 2CuCl + H₂O

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mass of Cu₂O reacted - 73.5 g

Number of moles of Cu₂O reacted - 73.5 g / 143 g/mol = 0.51 mol

According to the molar ratio,

when 1 mol of  Cu₂O reacts then 2 mol of CuCl is formed

therefore when 0.51 mol of Cu₂O reacts then - 2 x 0.51 mol of CuCl is formed

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Convert 1.21 kg to grams
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Answer:

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