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Furkat [3]
2 years ago
9

Can anyone help me? thank you :))

Chemistry
1 answer:
Alex_Xolod [135]2 years ago
6 0
8. 63360
9. 1000
if I got anything wrong please let me know thank you :)
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Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re
bearhunter [10]

Answer:  28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}   

\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles

The balanced chemical equuation is:

4Al+3O_2\rightarrow 2Al_2O_3  

According to stoichiometry :

4 moles of Al produce == 2 moles of Al_2O_3

Thus 0.556 moles of Al will produce=\frac{2}{4}\times 0.556=0.278moles  of Al_2O_3

Mass of Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0
2 years ago
Using the thermodynamic information in the ALEKS Data tab, calculate the boiling point of titanium tetrachloride . Round your an
ddd [48]

Answer:

The boiling point is 308.27 K (35.27°C)

Explanation:

The chemical reaction for the boiling of titanium tetrachloride is shown below:

TiCl_{4(l)} ⇒ TiCl_{4(g)}

ΔH°_{f} (TiCl_{4(l)}) = -804.2 kJ/mol

ΔH°_{f} (TiCl_{4(g)}) = -763.2 kJ/mol

Therefore,

ΔH°_{f} = ΔH°_{f} (TiCl_{4(g)}) - ΔH°_{f} (TiCl_{4(l)}) = -763.2 - (-804.2) = 41 kJ/mol = 41000 J/mol

Similarly,

s°(TiCl_{4(l)}) = 221.9 J/(mol*K)

s°(TiCl_{4(g)}) = 354.9 J/(mol*K)

Therefore,

s° = s° (TiCl_{4(g)}) - s°(TiCl_{4(l)}) = 354.9 - 221.9 = 133 J/(mol*K)

Thus, T = ΔH°_{f} /s° = [41000 J/mol]/[133 J/(mol*K)] = 308. 27 K or 35.27°C

Therefore, the boiling point of titanium tetrachloride is 308.27 K or 35.27°C.

5 0
3 years ago
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