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3 years ago

Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)

Chemistry

1 answer:

Vsevolod [243]3 years ago

6 0

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

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The atmospheric pressure in Francisco on ascertain day was 97.6kpa what was the pressure on mmhg

vodka [1.7K]

Answer:

732.0601 mmHg

Explanation:

Given data:

Pressure = 97.6 KPa

Given pressure in mmHg = ?

Solution:

Kilo pascal and millimeter mercury both are units of pressure.

Kilo pascal is denoted as "KPa"

Millimeter mercury is denoted as " mmHg"

Kilo pascal is measure of force per unit area while also define as newton per meter square.

It is manometric unit of pressure. It is the pressure generated by column of mercury one millimeter high.

Conversation of kilopascal to mmHg:

97.6 × 7.501 = 732.0601 mmHg

3 0

3 years ago

Which compound contains both covalent and ionic bonds? a. NCl3 b. NaF c. HCN d. NH4OH

Marina86 [1]

<h2> $NH_4OH$ contains both covalent and ionic bonds.</h2>

Explanation:

A covalent bond is formed when an element shares its valence electron with another element. This bond is formed between two non metals.

An ionic bond is formed when an element completely transfers its valence electron to another element. The element which donates the electron is known as electropositive element or the metal and the element which accepts the electrons is known as electronegative element or non metal.

a. $NCl_3$ contain covalent bonds as they are made up of non metals only.

b. $NaF$ contain ionic bonds as they are made up of sodium metal and fluorine non metal.

c. $HCN$ contain covalent bonds as they are made up of non metals only.

d. $NH_4OH$ contain ionic bonds between $NH_4$ and $OH$ and covalent between N and H in $NH_4^+$

Learn more about ionic and covalent bonds

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8 0

3 years ago

if the percent yield ofr the following reaction is 75% and 45.0g of NO2 are consumed with ercess water in the reaction. how many

NISA [10]

Answer:

30.8 grams of nitric acid are produced

Explanation:

Let's state the reaction:

3 NO₂ + H₂O → 2 HNO₃ + NO

If water is the excess reagent, then the limiting is the gas.

We convert the mass to moles:

45 g . 1 mol/ 46 g = 0.978 moles

Ratio is 3:2. 3 moles of gas can produce 2 moles of acid

Then, 0.978 moles may produce (0.978 . 2) /3 = 0.652 moles of acid

This is the 100% yield, but in this case, the percent yield is 75%

0.652 moles . 0.75 = 0.489 moles

Let's convert the moles to mass → 0.489 mol . 63g / 1mol = 30.8 g

6 0

3 years ago

What is the charge of a chlorine ion that has gained 1 electron?

nikdorinn [45]

schoOOOL SUcKS sO BADDDDDD

7 0

3 years ago

Read 2 more answers

Calculate the radius of tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6 g/cm, and an atomic wei

Ivahew [28]

Answer:

The radius of tantalum (Ta) atom is $R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

Explanation:

From the Body-centered cubic (BBC) crystal structure we know that a unit cell length a and atomic radius R are related through

$a=\frac{4R}{\sqrt{3} }$

So the volume of the unit cell $V_{c}$ is

$V_{c}= a^3=(\frac{4R}{\sqrt{3} } )^3=\frac{64\sqrt{3}R^3}{9}$

We can compute the theoretical density ρ through the following relationship

$\rho=\frac{nA}{V_{c}N_{a}}$

where

n = number of atoms associated with each unit cell

A = atomic weight

$V_{c}$ = volume of the unit cell

$N_{a}$ = Avogadro’s number ( $6.023 \times 10^{23}$ atoms/mol)

From the information given:

A = 180.9 g/mol

ρ = 16.6 g/cm^3

Since the crystal structure is BCC, n, the number of atoms per unit cell, is 2.

We can use the theoretical density ρ to find the radio R as follows:

$\rho=\frac{nA}{V_{c}N_{a}}\\\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}$

Solving for R

$\rho=\frac{nA}{(\frac{64\sqrt{3}R^3}{9})N_{a}}\\\frac{64\sqrt{3}R^3}{9}=\frac{nA}{\rho N_{a}}\\R^3=\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}} \\R=\sqrt[3]{\frac{nA}{\rho N_{a}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}$

Substitution for the various parameters into above equation yields

$R=\sqrt[3]{\frac{2\cdot 180.9}{16.6\cdot 6.023 \times 10^{23}}\cdot \frac{1}{\frac{64\sqrt{3}}{9}}}\\R = 1.43 \times 10^{-8} \:cm = 0.143 \:nm$

7 0

3 years ago