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Answer:
Chelate, any of a class of coordination or complex compounds consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. An example of a chelate ring occurs in the ethylenediamine-cadmium complex:
The ethylenediamine ligand has two points of attachment to the cadmium ion, thus forming a ring; it is known as a didentate ligand. (Three ethylenediamine ligands can attach to the Cd2+ ion, each one forming a ring as depicted above.) Ligands that can attach to the same metal ion at two or more points are known as polydentate ligands. All polydentate ligands are chelating agents.
Chelates are more stable than nonchelated compounds of comparable composition, and the more extensive the chelation—that is, the larger the number of ring closures to a metal atom—the more stable the compound. This phenomenon is called the chelate effect; it is generally attributed to an increase in the thermodynamic quantity called entropy that accompanies chelation. The stability of a chelate is also related to the number of atoms in the chelate ring. In general, chelates containing five- or six-membered rings are more stable than chelates with four-, seven-, or eight-membered rings.
Explanation:
If it’s multiple answers then it’s A,C,D. If it’s only one answer then your best answer is A. Because you always report accidents to an authority so that they can figure out how to solve the problem.
Answer:
Explanation:
We shall find volume of gas at NTP or at 273 K , 760 mm of Hg .
Pressure of given gas = 1.06 x 760 mm of Hg less vapor pressure of water .
= 805.6 - 23.76 = 781.84 mm of Hg
For it we use gas law formula ,
P₁V₁ / T₁ = P₂V₂ / T₂
781.84 x 136.1 / ( 273 + 25 ) = 760 x V₂ / 273
= 128.26 mL .
= 128.26 x 10⁻³ L .
22.4 L of oxygen will have mass of 32 g
128.26 x 10⁻³ L of oxygen will have mass of 32 x 128.26 x 10⁻³ / 22.4 g
= 183.22 mg .
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℃.
