The quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.
<h3>What is the speed of light?</h3>
Speed of light is the rate of speed though the light travels. To find the speed of light in any medium, the following formula is used.
Here, (n) is the index of reaction and (c) is the speed of light in the vacuum. The speed of light in the vacuum is almost equal to the 3.0×10⁸ m/s.
Now the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water has to be find out.
The above formula can be written as,
Here, the product of index of refraction and speed of light is equal to the speed of light in vacuum. This will be true for water as well.
Thus, the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.
Learn more about the speed of light here;
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Answer:
The change in internal energy is - 1.19 kJ
Explanation:
<u>Step 1:</u> Data given
Heat released = 3.5 kJ
Volume calorimeter = 0.200 L
Heat release results in a 7.32 °C
Temperature rise for the next experiment = 2.49 °C
<u>Step 2:</u> Calculate Ccalorimeter
Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C
Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C
<u>Step 3:</u> Calculate energy released
Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ
<u>Step 4:</u> Calculate change in internal energy
ΔU = Q + W W = 0 (no expansion)
Qreac = -Qcal = - 1.19 kJ
ΔU = - 1.19 kJ
The change in internal energy is - 1.19 kJ
Volume of O2 : 168 L
<h3>Further explanation</h3>
Given
Reaction
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
5 moles of KCIO3
Required
volume of O2
Solution
From the equation, mol O2 :
= 3/2 x moles KClO3
= 3/2 x 5 moles
= 7.5 moles
Assumed at STP( 1 mol = 22.4 L) :
= 7.5 x 22.4 L
= 168 L
Trace fossils may consist of impressions made on or in the substrate by an organism: for example, burrows, borings (bioerosion), urolites (erosion caused by evacuation of liquid wastes), footprints and feeding marks, and root cavities.
