Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.
Sn(s)+2HF(g)→SnF2(s)+H2(g)
Sn(s)+2HF(g)→
SnF
2
(s)+
H
2
(g)
How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?
Step 1: List the known quantities and plan the problem.
Known
given: 75.0 g Sn
molar mass of Sn = 118.69 g/mol
1 mol Sn = 2 mol HF (mole ratio)
Unknown
mol HF
Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.
g Sn → mol Sn → mol HF
Step 2: Solve.
75.0 g Sn×1 mol Sn118.69 g Sn×2 mol HF1 mol Sn=1.26 mol HF
75.0 g Sn×
1
mol Sn
118.69
g Sn
×
2
mol HF
1
mol Sn
=1.26 mol HF
Step 3: Think about your result.
The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
Yes it is polluting the river with everything the manufaturing plant gives off such as the chemicals released in the air
Answer:
The correct options are;
C. The magnitude of attraction from its nucleus
D. The distance between the electrons and its nucleus
Explanation:
The atomic radius reduces, within a given period, as we move from left to right, the number of protons increases alongside the number of electrons and the while the quantum shell to which the extra electrons are added to is the same. Therefore, the radius of the atom is dependent on the magnitude of the attraction from the nucleus
Similarly, as we progress to the next period, with an extra quantum shell, the atomic radius is seen to increase.
Therefore, the atomic radius is determined by the distance between the electrons and its nucleus.
Format Method - Writing the symbol of the cation and then the anion. Add whatever subscripts in order to balance the charges.
Crisscross Method - The numerical value of the charge of each ion is crossed over and becomes the subscripts for the other ion.
When an oxygen atom is attached to a carbon atom, the carbon atom becomes reduced.
