6.11% w/v of Cu2+ implies that 6.11 g of Cu2+ is present in 100 ml of the solution
therefore, 250 ml of the solution would have: 250 ml * 6.11 g/100 ml = 15.275 g
# moles of Cu2+ = 15.275 g/63.546 g mole-1 = 0.2404 moles
1 mole of CuCl2 contain 1 mole of Cu2+ ion
Hence, 0.2404 moles of Cu2+ would correspond to 0.2404 moles of CuCl2
Molar mass of CuCl2 = 134.452 g/mole
The mass of CuCl2 required = 0.2404 moles * 134.452 g/mole = 32.32 grams
Answer:
Dynamic
Explanation:
When rocks grind and squeeze past each other, dynamic metamorphism can occur.
The final temperature = 36 °C
<h3>Further explanation</h3>
The balanced combustion reaction for C₆H₆
2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ
MW C₆H₆ : 78.11 g/mol
mol C₆H₆ :
Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol
Heat transferred to water :
Q=m.c.ΔT
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
Answer:
0.106 mol (3s.f.)
Explanation:
To find the number of moles, divide the mass of glucose (in grams) by its Mr. Glucose has a chemical formula of C6H12O6. To find the Mr, add all the Ar of all the atoms in C6H12O6.
Ar of C= 12, Ar of H= 1, Mr of O= 16
These Ar values can be found on the periodic table.
Mr of glucose= 6(12)+ 12(1) + 6(16)= 180
Moles of glucose
= mass ÷ mr
= 19.1 ÷ 180
= 0.106 mol (3 s.f.)
