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3 years ago

Consider the reaction. c2h5oh(l)⟶ c2h6(g)+12o2(g). state what further data would be needed to obtain δrh∘ at 500∘c and 1 bar.

1 answer:

6 0

For this type of problem, it is essential for you to have a data on the standard heats of formation of the substances given. For elemental substances or diatomic gases, the standard heat of formation is 0. Standard means the temperature is at 0°C and pressure at 1 atm. Calculate the standard heat of reaction using:

ΔH°rxn = ∑(Stoichiometric coefficient×ΔHf of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)
Then, use this equation to find the reaction at T = 500°C and P = 1 bar:

ΔHrxn = ΔH°rxn + [∑(Stoichiometric coefficient×Cp of products) - ∑(Stoichiometric coefficient×ΔHf of reactants)]ΔT
So, you also need the Cp or specific heat capacities of the substances.

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Explain why photosynthesis and cellular respiration are the “real circle of life”

BabaBlast [244]

Explanation:Photosynthesis is important because the earth and all of the creatures in the sea and land need the plants to be able to feed themselves with the help of the sun and with cellular respiration, those plants are able to take in the carbon dioxide that we give and make into oxygen with we need. there fore without them we would not be able to exist.

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3 years ago

How many atoms are in 0.625 moles of ge (atomic mass = 72.59 amu)?

lord [1]

0.625 moles of ge X 6.02x10^23 atoms/ 1 mol of ge equal to 3.76x10^23 atoms of ge, just times with Avogadro.

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3 years ago

Read 2 more answers

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(

kkurt [141]

Answer: The $\Delta H^o_{rxn}$ for the reaction is 72 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)$ $\Delta H^o_{rxn}=?$