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Vlad1618 [11]
2 years ago
10

What could you do to increase the electrical force between two charged particles by a factor of 16

Chemistry
1 answer:
kumpel [21]2 years ago
4 0

Increasing one charged particle by 16 will lead to increasing the electrical force between two charged particles by a factor of 16.

<h3>What is Electrical force?</h3>

This is referred the attractive or repulsive interaction between any two charged bodies present.

F= kq1q2/r²

F1/F2= kq1q2/r² / 16kq1q2/r²

F1/F2 = 1/16

F2 = 16F1

This is therefore how to increase the electrical force between two charged particles by a factor of 16.

Read more about Electrical force here brainly.com/question/1498868

#SPJ1

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  • First choice: 99 J/K

Explanation:

<u>1) First law of thermodynamic (energy balance)</u>

  • Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter

<u>2) Energy change of each substance:</u>

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Heat released or absorbed = mass × Specific heat × change in temperature

  • density of water: you may take 0.997 g/ ml as an average density for the water.

  • mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g

  • Specif heat of water: 1 cal / g°C

  • Heat released by the hot water:

       Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)

  • Heat absorbed by the cold water:

       Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)

  • Heat absorbed by the calorimeter

       Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)

<u>4) Balance</u>

  • Heat₁ = Heat₂ + Heat₃

49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)

  • Solve for Ccal

Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K

Ccal = 23.6 cal/ K

  • Convert to cal / K to Joule / K

  • 1 cal = 4.18 Joule

       23.6 cal / K × 4.18 J / cal = 98.6 J/K

Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.

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