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vitfil [10]
3 years ago
11

A155-g sample of a unknown substance was heated from 25.0°C to 40.0°C.In the process, the substance absorbed 5696 J of energy. W

hat is the specific heat of the substance?
Chemistry
1 answer:
Iteru [2.4K]3 years ago
3 0
The answer is: (5696 J) / (155 g) / (40.0 - 25.0)°C = 2.45 J/g·°C 
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5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?
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By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's  Law</h3>

Given Data

  • number of moles of Ne = 5.1169 mol
  • Pressure = 0.9148 atm
  • Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

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V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

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Wat are representative elements
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The molar solubility of C a ( O H ) 2 C a ( O H ) 2 was experimentally determined to be 0.019 M. Based on this value, what is th
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Answer:

Ksp = 2.74 x 10⁻⁵

Explanation:

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According to the ICE table, the expression for the solubility product constant (Kps) is:

Ksp = [Ca²⁺] x ([OH⁻])² = s x (2s)² = 4s³

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