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Gwar [14]
2 years ago
13

In two or more complete sentences, explain how to calculate the amount of energy that is transferred when 150 g of cooper cools

from 63.0 °C to 21.0 °C. Is the energy absorbed or released? The specific heat of copper is 0.387 J/g*°C.
Chemistry
1 answer:
ella [17]2 years ago
8 0

Answer:

-2,438 J energy is released

Heat energy is released

Explanation:

The formula for energy absorbed or released is Q = mT(Cp)

Get temperature in degrees T = 21 - 63  = -42°C

Mass of the cooling cooper m = 150g

Specific heat of copper Cp = 0.387J/g-°C

Put all the above parameters into the formula for energy absorbed or released

Q = mT(Cp)

    = 150 x -42(0.387)

    = -2438J  same as -2.438 KJ

-2438J energy is released

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DanielleElmas [232]
1) Calculate the number of moles of Cu SO4 . 5H20 by dividing the specified mass by the molar mass.

2) The ratio of production given by the equation is 1 mol of Cu SO4 . 5 H2O to 1 mol of Cu SO4=> 1:1, meaning that the number of moles of Cu SO4 produced is the same number of moles of Cu SO4.5H20 heated.

3) Finally mutiply the number of moles of Cu SO4 by its molar mass and there you have the mass of Cu SO4 produced.


5 0
2 years ago
Read 2 more answers
The human eye is able to detect as little as 2.35 × 10–18 j of green light of wavelength 510 nm. Calculate the minimum number of
Mariulka [41]

Answer:

  • <em>The minimum number of photons that can be detected by the human eye is </em><u><em>6.03 × 10 ¹⁶</em></u><em> photons.</em>

Explanation:

The energy of one photon of light is related to the wavelength by the equation:

  • E = h×c/λ

Where, E is the energy of one photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the light.

You are given with <em>λ = 510 nm</em> (nanometers), which you must convert to m (meters), to use SI units ⇒ λ = 510 × 10⁻⁹ m.

The <em>physical constansts </em>needed are:

  • Planck's constant, h = 6.63 × 10⁻³⁴ J.s
  • Speed of light, in vacuum, c = 3.0 × 10⁻⁸ m/s

Now you can substitute in the formula can compute for the value of E:

  • E = 6.63×10⁻³⁴ J.s × 3.0 × 10⁻⁸ m/s / (510×10⁻⁹ m) = 0.039 × 10⁻³³ J

Since, that is the energy of one photon of green light, to calculate the number of photons that can be detected by the human eye, you need to divide the amounf of <em>energy the human eye is able to detect, 2.35 × 10⁻¹⁸ J , </em>by the energy of a photon:

  • number of photons = 2.35×10 ⁻¹⁸J / 0.039 × 10⁻³³ J/ photon

  • number of photons = 60.3 × 10¹⁵ photons = 6.03 × 10¹⁶photons
6 0
2 years ago
Which two scenarios illustrate the relationship between pressure and volume as described by Boyle’s law?
Kruka [31]

Answer:

2) The volume of an underwater bubble increases as it rises and the pressure decreases

hope it was useful for you

stay at home stay safe

5 0
2 years ago
3350 J of heat is required to raise the temperature of a sample of AlF3 from 250C to 800C. What is the mass of the sample?
Aleonysh [2.5K]

Answer:

  1. Look up the specific heat capacity of AlF₃
  2. Calculate ΔT
  3. Calculate the mass of AlF₃

Explanation:

The formula for for the heat (q) absorbed by an object is

 q = mCΔT, where

m = the mass of the sample

 C = the specific heat capacity of the sample. and

ΔT = the change in temperature

1. What you must do

  • Look up the specific heat capacity of AlF₃
  • Calculate ΔT
  • Calculate the mass of AlF₃

2. Sample calculation

For this example, I assume that the specific heat capacity of AlF₃ is 1.16 J·K⁻¹mol⁻¹ .

(a) Calculate ΔT  

\Delta T = T_{\text{f}} - T_{\text{i}} = 800 \, ^{\circ}\text{C} -250 \, ^{\circ}\text{C} = 550 \, ^{\circ}\text{C}

(b) Calculate m

\begin{array}{rcl}\text{3350 J} & = & m \times 1.16 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}\times \text{550 K}\\3350 & = & m \times \text{638 g}^{-1}\\m & = &\dfrac{3350}{\text{638 g}^{-1}}\\\\ & = & \text{ 5.2 g}\\\end{array}\\

6 0
2 years ago
A liquid that occupies a volume of 4.7 liters has a mass of 5.1 kilograms what is the density of the liquid in kg/L
kotykmax [81]
Divide each by 4.7 to get ur kg/L which is 1.085kg/1L
4 0
2 years ago
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