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2 years ago

How many moles of phosphoric acid can be formed from 3000 g of diphosphorus pentoxide?

Chemistry

1 answer:

Semenov [28]2 years ago

7 0

Answer:

42.2 moles of H3PO4

Explanation:

The equation of the reaction is:

P2O5(s) + 3 H2O(l) ⟶ 2 H3PO4.

First we must obtain the number of moles of P2O5 from

Number of moles of P2O5= reacting mass of P2O5/molar mass of P2O5

Molar mass of P2O5= 141.9445 g/mol

Number of moles= 3000g/141.9445 g/mol = 21.1 moles of P2O5

From the reaction equation;

1 mole of P2O5 yields 2 moles of H3PO4

21.1 moles of P2O5 will yield 21.1 ×2/ 1 = 42.2 moles of H3PO4

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PLEASE HELP!!

grandymaker [24]

Answer:

1. 136 °C.

2. 0.21 atm.

Explanation:

1. Determination of the new temperature in °C.

Initial volume (V1) = 1.35L

Final volume (V2) = 1.95L

Initial temperature (T1) = 283 K

Final temperature (T2) =...?

Using the Charles' law equation, the new temperature of the gas can be obtained as follow:

V1 /T1 = V2 /T2

1.35/283 = 1.95/T2

Cross multiply

1.35 × T2 = 283 × 1.95

1.35 × T2 = 551.85

Divide both side by 1.35

T2 = 551.85/1.35

T2 = 408.8 ≈ 409 K

Finally, we shall convert 409 K to °C. This can be obtained as follow:

T (°C) = T(K) – 273

T(K) = 409 K

T (°C) = 409 – 273

T (°C) = 136 °C

Therefore, the new temperature of the gas is 136 °C.

2. Determination of the new pressure.

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1.67 L

Final pressure (P2) =.?

Next, we shall convert 1.67 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

1.67 L = 1.67 L × 1000 mL / 1 L

1.67 L = 1670 mL

Therefore, 1.67 L is equivalent to 1670 mL.

Finally, we shall determine the new pressure of the gas as follow:

Initial pressure (P1) = 1.34 atm

Initial volume (V1) = 267 mL

Final volume (V2) = 1670 mL

Final pressure (P2) =.?

P1V1 = P2V2

1.34 × 267 = P2 × 1670

357.78 = P2 × 1670

Divide both side by 1670.

P2 = 357.78 / 1670

P2 = 0.21 atm.

Therefore, the new pressure of the gas is 0.21 atm.

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2 years ago

1. Interpret the following equation using moles, molecules, and volumes (assume STP). Compare the mass of the reactants to the m

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1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules. Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas. The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.

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3 years ago

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. the density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v). densit

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Density is intensive because it is the ratio of two extensive properties that is mass to volume.

<h3>What is Density ?</h3>

Density is defined as mass per unit volume. S.I unit if density is kg/m³.

It is expressed as

Density = $\frac{\text{Mass}}{\text{Volume}}$ or $d = \frac{m}{V}$

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Intensive property is the property which does not depend on the size of the system. Intensive property can be easily identified. Intensive property is the independent property. The size does not change in this.

Example of intensive property are Density, Freezing point, melting point, colour, Lustre, etc.

<h3>What is Extensive property ?</h3>

Extensive property is the property which depend on the substance. Extensive property cannot be easily identified.

Example of extensive property length, weight, volume, mass etc.

Thus from the above conclusion we can say that Density is intensive because it is the ratio of two extensive properties that is mass to volume.

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Disclaimer: The given question is incomplete on the portal. Here is the complete question.

Question: The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (v)

density = $\frac{\text{Mass}}{\text{Volume}}$ or $d = \frac{m}{V}$

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max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

ΔG = ΔH - TΔS

On this note, it follows that;

ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

ΔG = 14.6 - 71.52
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Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

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