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svp [43]
3 years ago
7

Modern pennies are composed of zinc coated with copper. A student determines the mass of a penny to be 2.483 g and then makes se

veral scratches in the copper coating (to expose the underlying zinc). The student puts the scratched penny in hydrochloric acid, where the following reaction occurs between the zinc and the HCl (the copper remains undissolved): Zn(s)+2HCl(aq)→H2(g)+ZnCl2(aq) The student collects the hydrogen produced over water at 25 ∘C. The collected gas occupies a volume of 0.894 L at a total pressure of 794 mmHg . Part A Calculate the percent zinc in the penny. (Assume that all the Zn in the penny dissolves.)

Chemistry
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

97.1%

Explanation:

Using the ideal gas equation, the number of moles of hydrogen gas produced can be calculated from information provided about the volume of gas evolved at a given temperature and pressure.

The stoichiometry of the reaction is now used to obtain the number of moles of Zn that will produce a given number of moles of hydrogen from the balanced reaction equation as shown. This gives us the number of moles of zinc reacted hence the mass of zinc in the coin since it is assumed that all the zinc reacts.

This is now used to calculate the mass percentage of Zn as shown.

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What is the change in enthalpy when 11. 0 g of liquid mercury is heated by 15°c?
Vlad1618 [11]

Enthalpy change is the difference between energy used and energy gained. The change in enthalpy of the liquid mercury is 0.0231 kJ.

<h3>What is the enthalpy change?</h3>

Enthalpy change is the difference between the energy used to break chemical bonds and the energy gained by the products formed in a chemical reaction.

The enthalpy change is given by,

\rm \Delta H_{rxn} = \rm q_{rxn}

and,

\rm q = mc\Delta T

Given,

Mass of the liquid mercury (m) = 11.0 gm

The specific heat of mercury (c) = 0.14 J per g per degree Celsius

Temperature change = 15 degrees Celsius

Enthalpy change is calculated as:

\begin{aligned} \rm q &= \rm mc\Delta T\\\\&= 11 \times 0.14 \times 15\\\\&= 23.1 \;\rm J\end{aligned}

Therefore, 0.0231 kJ is the change in enthalpy.

Learn more about enthalpy change here:

brainly.com/question/10932978

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6 0
2 years ago
The Formation Of Lithium Iodide
IRISSAK [1]

<u>Lithium Iodide</u><u>:</u>

~formed by the reaction of hydroxide with hydroiodic acid

Hope this helped you, have a good day bro cya)

6 0
3 years ago
The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst
OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction  is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

<em>Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product? </em>

<em>     A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol. </em>

<em>     </em>Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

    <em>C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol</em>

    Not feasible. ΔG°rxn = ΔG°f(product) > 0.

    <em>D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol</em>

    Feasible. ΔG°rxn = ΔG°f(product) < 0.

3 0
4 years ago
Wofür steht die Abkürzung „PUR“?
Lunna [17]

Explanation:

polyurethane, a polymer

8 0
3 years ago
Why does one Oxygen atom bond with two Hydrogen atoms to form water? Why don't they bond in a different ratio?
Vladimir79 [104]

This is because oxygen (2.8.6) requires two electrons on its valence shell to attain stable configuration (2.8.8). Hydrogen (1) on the other hand requires one electron on its valence shell to attain stable configuration (2). Therefore in a covalent bond, it requires two hydrogen and one oxygen to share electrons and achieve stable configuration.  

4 0
3 years ago
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