6.022 x 10 23 titanium atoms
In 47.88 grams of titanium, there is one mole, or 6.022 x 1023 titanium atoms.
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Molar solubility of AgCl will be 0.59 ×
M.
The amount of a chemical that can dissolve in one liter of a solution before reaching saturation is known as its molar solubility. This implies that the quantity of a substance it can disintegrate in a solution even before the solution becomes saturated with that particular substance is determined by its molar solubility.
A compound's molar solubility would be the measure of how many moles of such a compound must dissolve to produce one liter of saturated solution. The molar solubility unit will be mol L-1.
Calculation of molar solubility:
Given data:
M = 0.30 M
= 1.77 × 
The reaction can be written as:
AgCl ⇔ 
s s (s+0.30)
= [
]+ [
]
1.77 ×
= s (0.30)
s = 1.77 ×
/ 0.3
s = 0.59 ×
M
Therefore, molar solubility of AgCl will be 0.59 ×
M.
To know more about molar solubility
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The molecules or atoms that are formed by gain or loss of one or more valence electrons are said to be ions.
When atom loss one or more valence electrons, results in formation of cation whereas when atom gain one or more valence electrons, then formation of anion occurs. Cations carry positive charge and anions carry negative charge.
In general, cations are smaller than the neutral atoms from which they are formed and anions are larger than the neutral atoms.
As cations are smaller than the related neutral atoms because the valence electrons are lost which are farthest away from the nucleus. After that, taking more electrons distant from the cation results in reduction of radius of the ion.
Thus, aluminium cation consist of few electrons which results in fewer occupied energy levels by the electrons further results in reduction of radius i.e. smaller size.
Hence, given statement is true i.e. aluminium atom is larger than the aluminium cation as cation has fewer occupied energy levels.
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of H = 18 %
Percentage of N = 82 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 18 g
Mass of N = 82 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 5.8 moles.
For Hydrogen = 
For Nitrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of H : N = 3 : 1
Hence, the empirical formula for the given compound is 