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3 years ago

How many moles of PCl5 can be produced from 24.0 g of P4

Chemistry

1 answer:

posledela3 years ago

6 0

Hey there!

Given the reaction:

P4 + 10 Cl2 ------------------ 4 PCl5

Molar mass P4 = 124 g/mol

Number of moles P4:

n = mass of solute / molar mass

n = 24.0 / 124

n = 0.1935 moles of P4

Therefore:

1 mole P4 --------------- 4 moles PCl5

0.1935 moles P4 ------- moles PCl5

moles PCl5 = 0.1935 * 4

= 0.774 moles of PCl5

Hope that helps!

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Ancient Romans built often out of bricks and mortar. A key ingredient in their mortar was quicklime (calcium oxide), which they

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The question has missing information. At part 1 it is "Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO3) into solid calcium oxide and gaseous carbon dioxide."

Part 2. "Suppose 19.0 L of carbon dioxide gas are produced by this reaction, at a temperature of 290.0°C and pressure of exactly 1 atm. Calculate the mass of calcium carbonate that must have reacted (...)"

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Explanation:

1. Calcium oxide has molecular formula CaO and carbon dioxide CO₂, thus, the reaction will be:

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The equation is already balanced because there's the same number of each element on both sides.

2. First, let's calculate the number of moles of CO₂ produced by the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (290°C = 273 = 563 K).

1*19 = n*0.082*563

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1 mol of CaCO₃ ------ 1 mol of CO₂

x ----- 0.4116 mol

By a simple direct three rule:

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Read 2 more answers

Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:

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The order of frequency is d < a < c < b

$E_{n} = 13.6 * z^{2} / n^{2}$ eV

where z = atomic mass number

n = energy level

For hydrogen z = 1

Therefore, Energy for n = 1

$E_{n} = 13.6 * 1^{2} / 1^{2}$ eV

= -13.6 eV

for n = 2

$E_{n} = 13.6 * 1^{2} / 2^{2}$ eV

= -3.40 eV

for n = 3

$E_{n} = 13.6 * 1^{2} / 3^{2}$ eV

= -1.51 eV

for n = 4

$E_{n} = 13.6 * 1^{2} / 4^{2}$ eV

= -0.85 eV

for n = 5

$E_{n} = 13.6 * 1^{2} / 5^{2}$ eV

= -0.544 eV

n = 2 to n = 4 (absorption)

ΔE = E4 - E2 = -0.85 - (-3.40) = 2.55 eV

n = 2 to n = 1 (emission)

ΔE = E1 - E2 = -13.6 - (-3.40) = -10.2eV

The negative sign indicates that emission will take place.

n = 2 to n = 5 (absorption)

ΔE = E5 - E2 = -0.544 - (-3.40) = 2.856 eV

n = 4 to n = 3 (emission)

ΔE = E3 - E4 = -1.51 - (-0.85) = -0.66 eV

We know that

E = h * υ

Therefore, Energy is proportional to frequency.

So increasing the order of energy is

E4 < E1 < E3 < E2

order of frequency is

d < a < c < b

For more information click on the link below:

brainly.com/question/17058029

# SPJ4

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2 years ago