The population would decrease, the more predators there are the more food needed for the species .
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℃.
Both are oxidation reactions. Burning is just a lot faster than rusting.
Answer:Explanation:
In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species.
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>
