The answer is (b). As, vanadium is attached to five fluoride atoms, each flouride containing -1 oxidation state, hence five fluoride contains -5, to neutralize, vanadium should have +5 oxidation state.
You need to look at the electronegativity and decide wheter the difference of both of the numbers are significant enough to form a polar bond
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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Answer:
1 Atm
Explanation:
Dalton's law
The total pressure is 3 Atm so all you have to do is subtract the other partial pressures from 3
Answer:
See Explanation
Explanation:
Given
(a) to (d)
Required
Determine whether the given parameters can calculate the required parameter
To calculate either Density, Mass or Volume, we have



(a) 432 g of table salt occupies 20.0 cm^3 of space
Here, we have:


The above can be used to calculate Density as follows;



(b) 5.00 g of balsa wood, density of balsa wood : 0.16 g/cm^3
Here, we have:


This can be used to solve for Volume as follows:



(c) 32 cm^3 sample of gold density of 19.3 g/cm^3
Here, we have:


This can be used to calculate Mass as follows:



(d) 150 g of iron, density of Iron = 79.0 g/cm^3
Here, we have


This can be used to calculate volume as follows:


<em>Approximated</em>