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Leya [2.2K]
2 years ago
6

Find the density of an 8 gram rock if the water in a graduated cylinder rises from 25.0 mL to

Chemistry
1 answer:
Olin [163]2 years ago
6 0

To measure the density of the stone placed in a graduated cylinder let us follow these steps bellow

  1. Measure the volume of water poured into a graduated cylinder
  2. Place the object in the water and remeasure the volume.
  3. The difference between the two volume measurements is the volume of the object.
  4. Divide the mass by the volume to calculate the density of the object.

<em>We know that the formula for density is given as </em>

Given data

Mass = 8gram

Initial Volume of water in cylinder = 25mL

Final Volume of water in cylinder  = 29mL

Hence the volume of the rock  = 29-25 = 4mL

Therefore the density of the rock = 8/4 = 2 g/mL

Learn more:

brainly.com/question/17336041

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The bonds of the product store 27 KJ more energy than the bonds of the reactants how is energy conserved during this reaction
Blababa [14]

If the bonds of the product store 27 KJ more energy than the bonds of the reactants, It means the surroundings absorb 27 kj of energy from the reaction system Hence, Option (D) is the correct answer

<h3>What is the Exothermic reaction ?</h3>

An exothermic process releases heat, causing the temperature of the immediate surroundings to rise.

The bonds of the product store 27 KJ more energy than the bonds of the reactants, It means that energy has been absorbed by the surrounding as the product formed is more stable due to more stronger bond

This can be inferred from more stored energy with in the bonds and Thus, It is a exothermic reaction.Hence, Option (D) is the correct answer

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2 years ago
3. Determine the uses of the following materials:
hichkok12 [17]

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D

Explanation:

D

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How spectral lines are formed
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Balance the following chemical equation.<br><br> CCl4 -&gt; ___ C+ ___ Cl2
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Answer:

Explanation:

CCl4 => C + 2Cl2

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2 years ago
Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

  • P₁ = 1.3 kPa
  • P₂ = 5.3 kPa
  • T₁ = 85.8°C = 358.96 K
  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

  • ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
  • ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
  • ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

3 0
3 years ago
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