Kepler’s third law exhibits the relationships between the distance of a planet from the sun and the period of its revolution. Kepler’s third law is also sometimes referred to as the law of harmonies.
Kepler’s third law compares the orbital period and the radius of an orbit of a planet to the distance of the planet to the sun. It states mathematically that the more distant a planet is from the sun the greater its orbital period will be. The period of revolution of a planet is measured in days, weeks, months or years. For example, Earth’s period of revolution is 365 days.
Answer:
Q = -14322.77 J
Explanation:
Given data:
Mass of water = 55.0 g
Initial temperature = 87.3°C
Final temperature = 25.0 °C
Heat given off = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 25.0 °C - 87.3°C
ΔT = - 62.3 °C
Q = 55.0 g×4.18 J/g.°C × - 62.3 °C
Q = -14322.77 J
Answer:
Mass 1=3M
Mass 2=?
Volume1=75mL
Volume2=250mL
By using molarity formula:
<u>mass1*volume 1=mass2*volume 2</u>
3M*75=mass2*250
mass2=225/250
mass2:0.9M
<h3>the molarity of a solution is 0.9M.</h3>
Answer:
hope the inserted image will help :)
Explanation:
Answer:
a) The wavelength is around 33.8 nm = 3.38*10⁻¹ nm
b) Ultraviolet
Explanation:
a) The energy (E) of a photon is related to its wavelength (λ) by the Planck's equation:
where h = Planck's constant = 6.626*10^-34 Js
c = speed of light = 3*10^8 m/s
E = 3.55*10^6 J/mol
The energy in terms of J/photon is:
Based on eq(1)
\lambda = h\frac{c}{E}=6.626*10^{-34}Js*\frac{3*10^{8}m/s}{5.89*10^{-18}J}=3.38*10^{-8}m
The wavelength is around 33.8 nm = 3.38*10⁻¹ nm
b) In the electromagnetic spectrum the ultraviolet range extends from 390 nm-8.82 nm
The calculated wavelength of 33.8 nm should fall in the UV range.
