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3 years ago

The chemical equation below shows the reaction between carbon dioxide (CO2) and lithium hydroxide (LiOH).

Chemistry

2 answers:

likoan [24]3 years ago

8 0

Answer:

c on edge

Explanation:

mariarad [96]3 years ago

6 0

$M_{CO_{2}}=44,01\frac{g}{mol}\\
m=25,5g\\\\
n=\frac{m}{M}=\frac{25,5g}{44,01\frac{g}{mol}}\approx0,58mol$

CO₂ + 2LiOH ⇒ Li₂CO₃ + H₂O
1mol : 2mol
0,58mol : x

$x=\frac{0,58mol*2mol}{1mol}=\underline{1,16mol}$

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2ca(s)+o2(g) → 2cao(s) δh∘rxn= -1269.8 kj; δs∘rxn= -364.6 j/k

Alinara [238K]

Gibbs free energy change for the reaction at 29°C. is equal to -1378.93 KJ.

<h3>What is Gibbs's free energy?</h3>

Gibbs free energy can be described as the enthalpy of the system minus the product of the temperature and entropy.

If the chemical reaction can be carried out under constant temperature ΔT = 0:

ΔG = ΔH – TΔS

The above equation is known as the Gibbs-Helmholtz equation.

ΔG > 0 non-spontaneous and endergonic and ΔG < 0 spontaneous and exergonic, ΔG = 0 is representing equilibrium.

Given the ΔS = -364 J/K, ΔH = -1269.8 KJ, T = 29°C = 29 + 273 = 302 K

ΔG = - 1269 - 302 × 364

ΔG = -1269 KJ - 109.93 KJ

ΔG = - 1378.93 KJ

Learn more about Gibbs's free energy, here:

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Your question is incomplete, most probably the complete question was,

2Ca(s)+O₂(g) → 2CaO(s)

ΔH∘rxn= -1269.8 kJ; ΔS∘rxn= -364.6 J/K

For this problem, assume that all reactants and products are in their standard states.

Calculate the free energy change for the reaction at 29°C.

6 0

2 years ago

Sodium chloride (table salt,) whose formula is NaCl: <br> Ionic <br> Covalent

Vesna [10]

Since sodium chloride contains both a metal AND a nonmetal, the combination of those would result in an ionic bond.

7 0

3 years ago

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If I have 3.2 moles of gas at a pressure of 0.095 atm and at a temperature o 43 C what is the volume of the container that the g

andriy [413]

Answer

The volume of the container that the gas is in = 874.30 L

Explanation

Given:

Moles, n = 3.2 mol

Pressure, P = 0.095 atm

Temperature, T = 43 °C = (43 + 273.15 K) = 316.15 K

Molar gas constant, R = 0.0821 atm•L/mol•K

What to find:

The volume of the container that the gas occupied.

Step-by-step solution:

The volume of the container that the gas occupied can be calculated using the ideal gas equation.

$\begin{gathered} PV=nRT \\ \\ \Rightarrow V=\frac{nRT}{P}=\frac{3.2mol\times0.0821atm•L/mol•K\times316.15K}{0.095atm} \\ \\ V=\frac{83.058928\text{ }atm•L}{0.095\text{ }atm}=874.30\text{ }L \end{gathered}$

Therefore, the volume of the container that the gas occupied is 874.30 L

5 0

1 year ago

Assuming dopant atoms are uniformly distributed in a silicon crystal, how far apart are these atoms when the doping concentratio

enyata [817]

Answer:

d =~ 5.8μm

d =~ 0.13 μm

Explanation:

when the doping concentrations are 5 × 10^15 cm^-3

d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^15

d = 1/ 170997.5

d = 5.85 × 10 ^ -6

d =~ 5.8μm

when the doping concentrations are 5 × 10^20 cm^-3

d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^20

using the principle of surds and standard forms, we have

d = 1/ ∛0.5 × 10^21

d = 1/7937005.26

d = 1.26 × 10 ^ -7

d = 0.126 × 10 ^ -6

d =~ 0.13 μm

8 0

3 years ago

How to use molar mass

ruslelena [56]

Explanation:

the molar mass of a compound can be caucaleted by adding the standar atomic masses.

8 0

3 years ago

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