Answer: C. The number of collisions of gas particles decreases
Explanation: As temperature decreases so does everything else same with increase.
5.6 Al(OH)3
5.6 Al, 16.8 O, 16.8 H
16.8 mols of oxegyn in 5.6 mols of Al(OH)3
Answer:
15.0 µm
Step-by-step explanation:
Density = mass/volume
D = m/V Multiply each side by V
DV = m Divide each side by D
V = m/D
Data:
m = 1.091 g
D = 7.28 g/cm³
l = 10.0 cm
w = 10.0 cm
Calculation:
<em>(a) Volume of foil
</em>
V = 1.091 g × (1 cm³/7.28 g)
= 0.1499 cm³
(b) <em>Thickness of foil
</em>
The foil is a rectangular solid.
V = lwh Divide each side by lw
h = V/(lw)
= 0.1499/(10 × 10)
= 1.50 × 10⁻³ cm Convert to millimetres
= 0.015 mm Convert to micrometres
= 15.0 µm
The foil is 15.0 µm thick.
Rise, decrease, away from ocean, towards land
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.
Learn more here:brainly.com/question/15962928