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.
Answer:
The concentration of the pyridinium cation at equilibrium is 1.00×10⁻³ M
Explanation:
In water we have
C₅H₅NHBr + H₂O ⇒ C₅H₅NH+ + Br−
Pyridinium Bromide (C₅H₅NHBr) Dissociates Completely Into C₅H₅NH+ And Br− as such it is a strong Electrolyte.
Therefore the number of moles of positive ion produced per mole of C₅H₅NHBr is one
pH = - log [H₃O⁺] Therefore 10^-pH = [H₃O⁺] = concentration of C₅H₅NHBr
= 10⁻³ = 0.001M = concentration of C₅H₅NHBr
The concentration of C₅H₅NHBr is = 1.00×10⁻³ M to two places of decimal
Answer:
3.40g/mL
Explanation:
Density is a measure of mass over volume, so to get the density all we have to do is divide the mass by the volume.
21.35g ÷ 6.28 mL = 3.40g/mL
I think the answer would be 1.58 g.
