What is the enthalpy change(Q) when 60.0 g of NaOH is dissolved in one litre of water, given that the temperature of the solution increased by 15.8 °C?

Answers

Answer 1

Answer:

ΔH = -44.029 KJ/mol

Explanation:

We are told that 60.0 g of NaOH is dissolved in one litre of water.

From conversion, 1 litre = 1000 g

From tables, specific heat capacity of water; c = 4.18 J/g/°C

We are given ΔT = 15.8 °C

Formula for the heat absorbed by the solution is given as;

Q = mcΔT

Where;

Q = Heat gained by the water

m = Mass of the water

c = Specific heat of water

ΔT = change in temperature

Thus;

Q = 1000 × 4.18 × 15.8

Q = 66044 J

The solution absorbed 66044 J and therefore it means that the dissolution of the salt gave off 66044 J.

Thus;

Enthalpy is; ΔH = -66044 J

Now, Mass of NaOH is 60 g.

We know that molar Mass of NaOH is 39.997 g/mol. Thus;

Converting to moles, we have;

n = 60/39.997

n = 1.5 mol

Now, enthalpy when 1.5 moles of NaOH are dissolved in water will be;

ΔH = -66044 J/1.5 moles

ΔH = -44029.33 J/mol

ΔH = -44.029 KJ/mol


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True
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[tex]\huge\boxed{\sf True}[/tex]

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[tex]\rule[225]{225}{2}[/tex]

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The type of potential energy related to an object's height

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Since the gravitational potential energy of an object is directly proportional to its height above the zero position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling of the height will result in a tripling of the gravitational potential energy.

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What does the article “Absolute Zero” tell us about what temperature really means?

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Answer:

Absolute zero, technically known as zero kelvins, equals −273.15 degrees Celsius, or -459.67 Fahrenheit, and marks the spot on the thermometer where a system reaches its lowest possible energy, or thermal motion. There's a catch, though: absolute zero is impossible to reach.

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g Hydrogen iodide, HI, decomposes at moderate temperatures according to the equation The amount of I2 in the reaction mixture can be determined from the intensity of the violet color of I2; the more intense the color, the more I2 in the reaction vessel. When 3.80 mol HI was placed in a 5.00-L vessel at a certain temperature, the equilibrium mixture was found to contain 0.443 mol I2. What is the value of Kc for the decomposition of HI at this temperature

Answers

Answer:

Equilibrium constant, Kc = 0.023

Explanation:

Equation for the decomposition of Hydrogen iodide is given below:

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n(HI) = 2 * 3.800 moles = 7.600 moles

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n(I₂) = 0.000 moles

At equilibrium, the equation becomes: 2HI <----> H₂ + I₂

Number of moles of the reactant and the products are given as follows:

n(HI) = 7.600 - (0.886 + 0.886) moles = 5.828 moles

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n(I₂) = 2 * 0.443 = 0.886 moles

Equilibrium constant, Kc = [H₂] [I₂] / [HI]²

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Answer:

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According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

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A.P.E.X

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See explanation

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Answer:

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2.62x10⁻³g NiCl₂ / 250mL * 100 =

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