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3 years ago

What are the products of this reaction? (High temperature) Na2SO4 * 10H2O + C —> ????

Chemistry

1 answer:

Thepotemich [5.8K]3 years ago

6 0

Answer:

A chemical equation can be described as a symbolic representation of a chemical reaction. It can be written in the form of words or symbols.

The following chemical reaction can be completed as :

Na2SO4·10H2O + 2C = Na2S + 10H2O + 2CO2

Sodium Sulfate Decahydrate + Diamond = Sodium Sulfide + Water + Carbon Dioxide

The products of the reaction are:

Sodium Sulfide (Na2S)
Water (H2O)

Carbon dioxide (CO2)

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How many milligrams of a 20mg sample of cesium-137 remain after 60 years

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(1/4)*20mg=5mg left

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Which is the correct Lewis dot structure of NH2? A.A B.B C.C D.D E.E

Anarel [89]

I think D is your answer

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2 years ago

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LICUICY NICULUM

Evgen [1.6K]

Answer:

Missing component

Cl₂

H₂O

Explanation:

Reaction given:

? + 2NaBr → 2NaCl+Br

Mg+? → MgO+H2

missing component = ?

Solution:

Reaction 1

? + 2NaBr -----→ 2NaCl + Br

if we look at the reactants and products there is one element is in product side but not shown in the reactant side.

In products there are following atoms

Na (Sodium)

Cl (Chlorine)

Br (Bromine)

In Reactant there are following atoms

Na (Sodium)

Br (Bromine)

So only Chlorine is missing in the reactant side.

if we put chlorine in the reactant then this reaction will be complete

Cl + 2 NaBr -----→ 2NaCl + Br

So chlorine react with water and give sodium chloride and bromine molecule, this is a single displacement reaction and chlorine replace bromine in the above reaction

Now balance the equation

Cl₂ + 2 NaBr -----→ 2NaCl + Br₂

So, the missing component is

Cl₂

____________

Reaction 2

Mg + ? ---------→ MgO + H₂

if we look at the reactants and products there are two element in product side but not shown in the reactant side.

In products there are following atoms

Mg (Magnesium)

O (Oxygen)

H (Hydrogen)

In Reactant there is only one following atom

Mg (Magnesium)

So Oxygen and Hydrogen is missing in the reactant side.

if we put water molecule (H₂O) in the reactant then this reaction will be complete

Mg + H₂O ---------→ MgO + H₂

in this reaction magnesium (Mg) react with water (H₂O) and give magnesium oxide and hydrogen gas.

So, the missing component is

H₂O

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3 years ago

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What is water classified as

Montano1993 [528]

Water, or H2O, is a compound composed of the elements hydrogen (H) and oxygen (O).

3 0

3 years ago

A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm

Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys) =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

⇒ Vf = (1*0.08206 *300)/1

⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0

3 years ago