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3 years ago

What is the empirical formula for a compound whose molecular formula is P4O10

Chemistry

1 answer:

Bas_tet [7]3 years ago

6 0

Answer:

The empirical formula is P2O5 (option B)

Explanation:

An empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers.

The actual numbers of atoms of each element that occur in the smallest freely existing unit or molecule of the compound is expressed by the molecular formula of the compound.

The molecular formula of a compound may be the empirical formula, or it may be a multiple of the empirical formula.

If the molecular formula is P4O10, this means for each for P-atoms we have 10-O atoms this is a ratio 4:10 or 1: 2.5

To find the empirical formula we divide the molecular formula by 2 what will give us P2O5

For each 2 P atoms we have 5 O-atoms. This is a ratio 1: 2.5

This is the simpliest form for the compound P4O10.

The empirical formula is P2O5 (option B)

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It has 4 valance electrons.

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3 years ago

PLEASE HELP!!

uranmaximum [27]

Answer: - 25.8 kJ

The overall process of interest in the question is the following

Hg (g) (650 K) → Hg (l) (297 K)

However, for mercury to pass from a gaseous state in 650 K to a liquid state in 297 K, it must go through a series of steps:

Step 1. Gaseous mercury at 650 K should cool down to 629.88 K, temperature corresponding to the vaporization temperature of this substance.

Step 2. Gaseous mercury goes to liquid state at 629.88 K .

Step 3. The liquid mercury at 629.88 K is cooled until it reaches 297 K.

This series of steps can be represented through the following diagram:

Hg (g) (650 K) → Hg (g) (629.88 K) → Hg (l) (629.88 K) → Hg (l) (297 K)

(1) (2) (3)

Then the total heat involved in the process will be equal to the sum of the heats inherent to steps 1, 2 and 3. We proceed to calculate the heats for each of the steps.

Step 1:

The heat in step 1 will be given by

Q = n Cp ΔT

where n is the number of moles of mercury, Cp is the heat capacity and ΔT is equal to the difference between the temperatures at the end (T₂) and at the beginning of the process (T₁), that is to say

ΔT = T₂ - T₁

You should know that the heat capacity or thermal capacity is the energy needed to increase the temperature of a certain substance in a unit of temperature. The heat capacity of mercury is Cp = 27.983 J / mol K

Then the heat in step 1 will be,

Q₁ = 75.0 g x $\frac{1 mol}{200.59 g}$ x 27.983 $\frac{J}{mol K}$ x (629.88 K - 650 K)

→ Q₁ = - 210.5 J

Step 2:

In this step a change in the state of the mercury occurs, since it condenses from a gaseous state to a liquid state. In this case the heat involved in the process will be given by ,

Q = - n ΔHvap

where ΔHvap is the enthalpy of vaporization. The enthalpy of vaporization is the amount of energy necessary for the mass unit of a substance that is in equilibrium with its own vapor at a pressure of one atmosphere to pass completely from the liquid state to the gaseous state. Therefore, to determine the energy necessary for the mercury to pass from gaseous state to liquid, the negative of the enthalpy of vaporization must be taken, as it is done in the previous equation with the minus sign that is placed.

The enthalpy of vaporization of mercury is ΔHvap = 59.11 kJ/mol

Then the heat in step 2 will be,

Q₂ = - 75.0 g x $\frac{1 mol}{200.59 g}$ x 59.11 $\frac{kJ}{mol}$

→ Q₂ = 22.10 kJ → Q₂ = 22100 J

Step 3:

The heat in step 3 will be

Q₃ = n Cp ΔT = 75.0 g x $\frac{1 mol}{200.59 g}$ x 27.983 $\frac{J}{mol K}$ x (297 K - 629.88 K)

→ Q₃ = - 3483 J

Finally the heat involved in the overall process will be ,

Q = Q₁ + Q₂ + Q₃ = - 210.5 J - 22100 J - 3483 J = - 25794 J

→ Q = - 25.8 kJ

So, the heat lost when 75.0 g of mercury vapor at 650 K condenses to a liquid at 297 K is - 25.8 K

3 0

3 years ago

use the periodic table to determine the molar mass of each of the following elements. Use the correct number of significant figu

Yanka [14]

Answer: Potassium is 39.10 g/mol

Explanation: The numbers below the element are the molar mass.

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3 years ago

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Mamont248 [21]

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4 0

4 years ago

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti

ss7ja [257]

Answer:

1.89 of Sodium Carbonate

3.94 g of Silver Carbonate

2.43 g of Sodium Nitrate

Zero grams of Silver Nitrate

Explanation:

We have to start with the reaction:

$AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~NaNO_3$

Now, we can balance the reaction:

$2AgNO_3~+~Na_2CO_3~->~Ag_2CO_3~+~2NaNO_3$

Now, we have to calculate the limiting reagent and we have to follow a few steps:

1) Convert to moles (using the molar mass of each compound)

2) Divide by the coefficient of each reactive (given by the balanced reaction)

Convert to moles

$3.40~g~Na_2CO_3\frac{105.98~g~Na_2CO_3}{1~mol~Na_2CO_3}=0.032~mol~Na_2CO_3$

$4.86~g~AgNO_3\frac{169.8~g~AgNO_3}{1~mol~AgNO_3}=0.0286~mol~AgNO_3$

Divide by the coefficient

$\frac{0.032~mol~Na_2CO_3}{1}=0.032$

$\frac{0.0286~mol~AgNO_3}{2}=0.0143$

The smallest value is for $AgNO_3$ , therefore the 4.86 g of $AgNO_3$ .

Now we can calculate the amount of compounds produced is we follow a few steps:

1) Use the molar ratio

2) Convert to moles (using the molar mass of each compound)

Amount of Silver Carbonate

$0.0286~mol~AgNO_3\frac{1~mol~AgCO_3}{2~mol~AgNO_3}\frac{275.74~g~AgCO_3}{1~mol~AgCO_3}=3.94~g~AgCO_3$

Amount of Sodium Nitrate

$0.0286~mol~AgNO_3\frac{2~mol~NaNO_3}{2~mol~AgNO_3}\frac{84.99~g~NaNO_3}{1~mol~NaNO_3}=2.43~g~NaNO_3$

Amount of Sodium Carbonate (Excess reactive)

$0.0286~mol~AgNO_3\frac{1~mol~NaCO_3}{2~mol~AgNO_3}\frac{105.98~g~NaCO_3}{1~mol~NaCO_3}=1.51~g~NaCO_3$

$3.4~g~NaCO_3-1.51~g~NaCO_3=1.89~g~NaCO_3$

Amount of Silver Nitrate

All the silver nitrate would be consumed in the reaction

I hope it helps!

7 0

4 years ago