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Softa [21]
3 years ago
8

A chemist carefully measures the amount of heat needed to raise the temperature of a 278.0g sample of C5H12S from 0.8degree cels

ius to 15.9 degree celsius . The experiment shows that 8.46X10^3J of heat are needed. What can the chemist report for the molar heat capacity of C5H12S ? Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
Alex_Xolod [135]3 years ago
8 0

Answer:

The answer to your question is    C = 2.01 J/g°C

Explanation:

Data

mass = m = 278 g

Temperature 1 = T1 = 0.8°C

Temperature 2 = T2 = 15.9 °C

Heat = Q = 8.46 x 10³ J

Heat capacity = C = ?

Process

1.- Write the formula to calculate the heat

        Q = mC(T2 - T1)

2.- Solve for C

        C = Q / m(T2 - T1)

3.- Substitution

         C = 8.46 x 10³ / 278 (15.9 - 0.8)

4.- Simplification

          C = 8.46 x 10³ / 4197.8

5.- Result

          C = 2.01 J/g°C

         

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The volume of the balloon is approximately 2652 liters.

<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass (M), in kilograms per kilomole, and mass of the gas (m), in kilograms. The volume occupied by the gas (V), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}   (1)

Where:

  • R_{u} - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
  • T - Temperature, in Kelvin
  • P - Pressure, in kilopascals

If we know that m = 0.239\,kg, R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}, T = 273.15\,K, P = 101.325\,kPa and M = 2.02\,\frac{kg}{kmol}, then the volume of the balloon is:

V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}

V = 2.652\,m^{3} (2652\,L)

The volume of the balloon is approximately 2652 liters. \blacksquare

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4 0
2 years ago
in the process of electricity, what flows through the wires?A) protons B)neutronsC)electronsD)positrons
gizmo_the_mogwai [7]

Electrons is your answer

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4 years ago
25cc of 5 % NaOH solution neutralized 30cc of h2sO4 solution. Whatis normality of H2SO4?
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The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.

<h3>What is the molarity of 5% NaOH?</h3>

The molarity of 5% NaOH is 1.32 M

25 cc of NaOH neutralized 30cc of H₂SO₄ solution.

Equation of reaction is given below:

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Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M

  • Normality = Molarity × moles of H⁺ ions per mole of acid

moles of H⁺ ions per mole of H₂SO₄ = 2

Normality of H₂SO₄ = 0.55 x 2 = 1.1 N

In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.

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7 0
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garik1379 [7]

Answer:

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Explanation:

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n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 23° C + 273 = 296 K).

<em>∴ n = PV/RT =</em> (197.4 atm)(20.0 L)/(0.0821 L.atm/mol.K)(296 K) = <em>162.5 mol.</em>

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