Answer:
Baking soda has a ph of 9
Explanation:
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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The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are,
= 9.25g/mL,
= 8.634g/mL and
= 9.130g/mL respectively.
The equation of the standard deviation is √{∑(x -
)÷N}
Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.
The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.
The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.
The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.
The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be
. Thus the standard deviation of the density value is 0.265g/mL.
Answer: isotopes
Explanation: I remember this from when I took honors chemistry, they are know as isotopes.
Answer:
You could observe the cell wall in all kingdoms except for the animals. That would be cells of plants, protists (algae), fungi, bacteria and achaea.