Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
<span>ATP,O2 and NADPH are the </span>products<span>. H2O,NADP,ADP and Pi are the reactants. acts as an electron carrier between the cytochrome b6f and </span>photosystem 1 (PS1) complexes in the photosynthetic electron-transfer chain.
Photosystem II<span> (or water-plastoquinone oxidoreductase) is the first protein complex in the light-dependent reactions of oxygenic photosynthesis. It is located in the thylakoid membrane of plants, algae, and cyanobacteria.</span>
Answer:
The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.
Explanation:
To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.
It is given by Bohr's Theory as:
1/λ = Rh (1/n1² - 1/n2²)
where,
λ = wavelength of photon
n1 = initial state = 1 (ground-state of hydrogen)
n2 = final state = ∞ (since, electron goes far away from atom after ionization)
Rh = Rhydberg's Constant = 1.097 x 10^7 /m
Therefore,
1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)
λ = 9.115 x 10^-8 m = 91.15 nm
Now, for energy (E) we know that:
E = hc/λ
where,
h = Plank's Constant = 6.625 x 10^-34 J.s
c = speed of light = 3 x 10^8 m/s
Therefore,
E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)
<u>E = 2.18 x 10^-18 J</u>
E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)
<u>E = 13.6 eV</u>
