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1 year ago

What is the balanced chemical equation for the reaction used to calculate ΔH∘f of SrCO3(s)?

Chemistry

1 answer:

Igoryamba1 year ago

5 0

Answer :

Standard enthalpy of formation : It is defined as the enthalpy change for the reaction that forms one mole of compound from its elements. All the substances in their standard states.

The balanced chemical equation for the reaction is,

$Sr(s)+C(s)+\frac{3}{2}O_2(g)\rightarrow SrCO_3(s)$

In the balance reaction, Strontium (Sr), Carbon (C) and Strontium carbonate $SrCO_3$ are in solid state and oxygen is in gaseous state.

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Write the expression for the equilibrium constant Kp for the following reaction.Enclose pressures in parentheses and do NOT writ

maxonik [38]

Answer: The expression for $K_p$ is written below.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as $K_p$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_p$ is written as:

$K_p=\frac{P_{C}^c\times P_{D}^d}{P_{A}^a\times P_{B}^b}$

The partial pressure for solids and liquids are taken as 1.

For the given chemical equation:

$NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)$

The expression for $K_p$ for the following equation is:

$K_p=\frac{(P NH_3)\times (P H_2S)}{(P NH_4HS)}$

The partial pressure of $NH_4HS$ will be 1 because it is solid.

So, the expression for $K_p$ now becomes:

$K_p=\frac{(P NH_3)\times (P H_2S)}{1}$

Hence, the expression for $K_p$ is written above.

5 0

2 years ago

2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) [balanced] How many moles of N2H4 is required to produce 28.3 g of N2? Assume that all rea

JulijaS [17]

Answer: 0.67 moles of $N_2H_4$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{28.3}{28.02}=1mole$

$2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)$

According to stoichiometry:

3 moles of $N_2$ is produced by 2 moles of $N_2H_4$

Thus 1 mole of $N_2$ is produced by= $\frac{2}{3}\times 1=0.67moles$ of $N_2H_4$

Thus 0.67 moles of $N_2H_4$ are required to produce 28.3 g of $N_2$

6 0

2 years ago

Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2

Svet_ta [14]

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

4 0

1 year ago

In a hydrogen fuel cell, hydrogen gas and oxygen gas are combined to form water. Write the balanced chemical equation describing

xxTIMURxx [149]

Answer: The chemical reaction is given below.

Explanation:

A fuel cell is defined as the electrochemical cell which converts the chemical energy of a fuel (often used hydrogen) and an oxidizing agent (often used oxygen) into electrical energy via a pair of redox reactions.

The reactions which occur in hydrogen-oxygen fuel cell are:

At cathode: $H_2+2OH^-\rightarrow 2H_2O+2e^-$

At anode: $\frac{1}{2}O_2+H_2O+2e^-\rightarrow 2OH^-$

Net reaction: $H_2+\frac{1}{2}O_2\rightarrow H_2O$

Thus, the chemical reaction is given above.

5 0

2 years ago

Read 2 more answers

A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro

Liula [17]

Answer:

The specific heat of the metal is 0.335 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Initial temperature of the metal = 90.0 °C

Mass of the water = 25.0 grams

Initial temperature of water = 22.5 °C

Final temperature of water (and metal) = 25.0 °C

The specific heat of water = 4.18 J/g°C

Step 2: Calculate the specific heat of the metal

Qgained = -Qlost

Qwater = -Qmetal

Q= m*c*ΔT

m(metal) *c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒ mass of the metal = 12.0 grams

⇒ c(metal) = TO BE DETERMINED

⇒ ΔT(metal) = T2 - T1 = 25.0 - 90.0 °C = -65.0

⇒ mass of the water = 25.0 grams

⇒ c(water) = the specific heat of water = 4.18 J/g°C

⇒ ΔT(water) = T2 - T1 = 25.0 - 22.5 = 2.5°C

12.0 * c(metal) * -65.0 = -25.0 * 4.18 * 2.5

c(metal) = 0.335 J/g°C

The specific heat of the metal is 0.335 J/g°C

6 0

1 year ago