Answer:... in this set (11). Which type of wetland has large numbers of cattails and grasslike plants? ... Swamp. Mangrove trees would most likely be found in which wetland? ... spring rains. Sometimes called the "River of Grass," ---- is a vast wetland.
Explanation:
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Answer: The statement it is a weak acid is true for the substance.
Explanation:
An acid that dissociates completely when dissolved in water to give hydrogen 
or hydronium 
ions is called a strong acid.
For example, HCl is a strong acid.
An acid that dissociates partially or weakly when dissolved in water to given hydrogen or hydronium ions is called a weak acid.
For example, 
is a weak acid.
A strong base is a base which when dissolved in water then it dissociates completely to give hydroxide ions.
For example, NaOH is a strong acid.
A weak base is a base which when dissolved in water then it dissociates partially or weakly to give hydroxide ions.
For example, 
is a weak base.
Hence, in an aqueous solution where 42% of a substance dissociates to release hydronium ions shows that the dissociation is less than 50%. This means that substance is dissociating weakly so, it is a weak acid.
Thus, we can conclude that the statement it is a weak acid is true for the substance.
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℃.
In earths surface or the bottom of the Ocean
A burning splint will burn more vigorously in pure oxygen than in air because <span>oxygen is a reactant in combustion and concentration of oxygen is higher in pure oxygen than is in air.
Oxygen concentration in air is approximately 20%, the rest of are nitrogen, carbon dioxide and other gases. Oxygen is oxidazing reactant, that means oxygen give electrons in chemical reactions.
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