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Complete Question
A sample of aluminum, which has a specific heat capacity of 0.897 JB loc ! is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The aluminum sample starts off at 85.6 °C and the temperature of the water starts off at 16.0 °C. When the temperature of the water stops changing it's 20.1 °C. The pressure remains constant at 1 atm. Calculate the mass of the aluminum sample.
Answer:
Explanation:
From the question we are told that:
Heat Capacity
Mass of water
Initial Temperature of Aluminium
Initial Temperature of Water
Final Temperature of Water
Generally
Heat loss=Heat Gain
Therefore
In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.
Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.
For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.
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