The precaution to be taken while measuring the temperature of a liquid in a beaker is applying proper heat balance and taking all the required precautions.
- A beaker with an open top contains a sample of liquid. It exposes this sample to light.
- That liquid absorbs the light energy, turning it into heat energy. As a result, the liquid becomes warmer and evaporation is accelerated. As a result, there is less liquid in the beaker.
- Since it is well known that the surface temperature of a liquid, along with air movement above the liquid surface, is one of the dominant factors affecting evaporation, I want to measure the evaporation rate as a function of surface temperature.
- This can be done by applying a heat balance.
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The mass of aluminium foil is calculated as follows
mass = density x volume
density = 2.70 g/cm^3
volume 54 cm^3
mass of aluminium foil is therefore = 2.70 g/cm^3 x 54 cm^3 =145.8 grams
cm^3 cancel out each other
Grasshopper population will increase since they have less predators
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>

The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
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