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beks73
1 month ago
15

In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g h2o i

nitially at 23.2°c. the final temperature of the solution was 31.8°c. what was the change in enthalpy for the dissolution of this compound? give your answer in units of joules per gram of compound. assume that the specific heat of the solution is the same as that of pure water, 4.18 j ⁄ g ⋅ °c.
A) 2.7 x 10^2 J/g
B) -3.1 x 10^3 J/g
C) -3.1 x 10^2 J/g
D) 3.1 x 10^2 J/g
Chemistry
2 answers:
Alekssandra [3K]1 month ago
8 0
The enthalpy change for dissolving this compound is 2.70 x 10²J/g. The mass of the soluble ionic compound is 10.00g, the mass of water is 75.0g, the initial water temperature is 23.2⁰C (T1), and the final temperature is 31.8⁰C (T2). The specific heat of water is 4.18 J/g.⁰C. To compute the heat gained by the water, use the formula for heat transfer: Q = mcΔT. Calculating Q gives you 2696 J. The heat gained by water matches the heat lost by the ionic compound, expressed as q(ionic) = 2696 J. To find ΔH: ΔH = q(ionic)/m, leading to ΔH = 2696 J/10.00g, which results in 2.70 x 10²J/g. Hence, the enthalpy change for this compound's dissolution is 2.70 x 10²J/g.
VMariaS [2.9K]1 month ago
7 0
Given: Mass of the ionic compound = 10.00 g Mass of water = 75.0 g Initial temperature of water T1= 23.2 C Final temperature of water T2 = 31.8 C Specific heat of water c = 4.18 J/gC To determine: Enthalpy of dissolution of the ionic compound Heat gained by water equation: Q = mcΔT m = mass of water c = specific heat ΔT = change in temperature (T2-T1) Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution) Therefore, q(ionic) = 2696 J ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g Answer: A) enthalpy change = 2.7*10² J/g
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V(Fe) = 1 dL = 1 dL · 100 mL/1dL.
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2 months ago
If 3.18 x 10^23 atoms of iron react with 67.2 L of chlorine gas at STP, what is the maximum
Tems11 [2777]

84.34 grams of iron (III) chloride is the maximum produced since iron is the limiting reagent, and chlorine gas is in excess.

Explanation:

Balanced equation:

2 Fe + 3 Cl2 → 2 FeCl3​

DATA PROVIDED:

iron =  atoms

mass of chlorine = 67.2 liters

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The number of moles of iron will be calculated as

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number of moles = \frac{mass}{molar mass of 1 mole}

Substituting the values into the equation:

n = \frac{67200}{70.96}               (molar mass of chlorine gas = 70.96 g/mol)

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As iron is the limiting reagent therefore

2 moles of Fe lead to 2 moles of FeCl3

0.52 moles of Fe will yield

\frac{2}{2} = \frac{x}{0.52}

0.52 moles of FeCl3 is produced.

To express this in grams:

mass = n x molar mass

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2 months ago
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Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

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Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

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Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

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4) acetic acid reacting with zinc:

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Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

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Answer:

\%\ Composition\ of\ iron=69.92\ \%

Explanation:

The percent composition indicates the mass percentage of a specific element within the compound.

The chemical formula for chromium(III) nitrate is Cr(NO_3)_3.

The molar mass for chromium(III) nitrate is calculated at 238.011 g/mol.

Each mole of chromium(III) nitrate includes 9 moles of oxygen.

The molar mass of oxygen is 16 g/mol.

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\%\ Composition\ of\ iron=69.92\ \%

5 0
2 months ago
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