Can someone help me answer these ?

How many moles of UF6 would have to be decomposed to provide enough fluorine to prepare 8.99 mol of CF4? (Assume sufficient carb

Basile [38]

Answer:

5.99 moles of $UF_6$

Explanation:

In this case, we can start with the decomposition of $UF_6$ , so:

$UF_6~->~U~+~3F_2$ (Reaction 1)

The $F_2$ can react with carbon to produce $CF_4$ :

$CF_4~+~2F_2~->~CF_4$ (Reaction 2)

If we 8.99 mol of $CF_4$ , we can calculate the moles of $F_2$ that we need. In reaction 2 we have a molar ratio of 1:2 (2 moles of $F_2$ will produce 1 mol of $CF_4$ ):

$8.99~mol~CF_4\frac{2~mol~F_2}{1~mol~CF_4}~=~17.98~mol~F_2$

With this value and using the molar ratio in reaction 1 (3 moles of $F_2$ are producing by each mol of $UF_6$ ), so:

$17.98~mol~F_2\frac{1~mol~UF_6}{3~mol~F_2}~=~5.99~mol~UF_6$

So, we will need 5.99 moles of $UF_6$ to produce 8.99 mol of $CF_4$ .

I hope it helps!

7 0

3 years ago

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water at 20°C. Becaus

g100num [7]

Answer:

1. BaS Ksp = 0.169

2. CaSO₃ Ksp = 2.4 x 10⁻⁷

3. As₂O₃ Ksp = 6.7 x 10⁻⁴

3. Sr(IO₃)₂ Ksp = 1.7 x 10⁻⁷

Explanation:

The solubility constant product for a solid AB is given the expression for the equilibrium

AnBm (s) ⇄ A⁺(aq) + B⁻(aq) Ksp = [A⁺(aq)]^n x [B⁻ (aq)]^m

where the subscripts (s) and (aq) denote the solid and dissolved ions, [A⁺(aq)] and [B⁻ (aq)] are the concentrations of the dissolved ions in moles/Liter.

1. BaS

We are given the solubility in grams per 100 mL, since we need the solubility in moles per liter we will first convert the given solubility from g/ 100 mL to g/ L , and then divide this value by the molar mass to get molar solubility. From here it will be straight forward to calculate Ksp from the formula above

s = 6.97 g/ 100 mL = 6.97 g/0.1 L = 69.7 g/L

MW BaS = 169.39 g/mol

s BaS = (69.7 g/169.39 g/mol) / L = 0.41 mol/L = 0.41 M

Ksp = (0.41) x (0.41) = (0.41)² = 0.169

2. CaSO₃

s = 5.9 x 10⁻³ g / 100 mL = 5.9 x 10⁻³ g/ 0.1 L = 5.9 x 10⁻² g/L

MW CaSO₃ = 120.17 g/mol

s CaSO₃ = ( 5.9 x 10⁻² g / 5.9 x 10⁻² / 120.17 g/mol ) /L = 4.9 x 10⁻⁴ M

Ksp = (4.9 x 10⁻⁴) x ( 4.9 x 10⁻⁴) = (4.9 x 10⁻⁴)² = 2.4 x 10⁻⁷

3. As₂O₃

s = 1.80 g/ 100 mL = 1.80 g/ 0.100 L = 18.0 g /L

MW As₂O₃ = 197.84 g/mol

s = (18.0 g/ 197.84 g/mol) / L = 9.1 x 10⁻² mol/L = 9.1 x 10⁻² M

Ksp = (2 x 9.1 x 10⁻² )² x ( 3 x 9.1 x 10⁻² )³ = 6.7 x 10⁻⁴

Note we have to take into account that since the solubility of the compound As₂O₃ is 9.1 x 10⁻² mole/L we will have molar solubility concentrations of 2 x 9.1 x 10⁻² mol As³⁺ and 3 x 9.1 x 10⁻² . We did not need to do this in parts 1 and 2 since the ratio of ions were 1:1

4. Sr(IO₃)₂ 0.152 g/100 mL

s = 0.152 g / 100 mL = 0.152 g/0.1L = 1.52 g/L

MW Sr(IO₃)₂ = 437.43 g/mol

s = (1.52 g/437.43 g/mol)/L = 3.5 x 10⁻³ M

Ksp = (3.5 x 10⁻³ ) x ( 2 x 3.5 x 10⁻³ )² = 1.7 x 10⁻⁷

7 0

3 years ago

Reduction and oxidation must occur together. The electrons from the oxidized species are transferred to the reduced species. Che

Aneli [31]

Explanation:

When a specie tends to lose electrons in a chemical reaction then it means oxidation has taken place. When a specie tends to gain electrons then it means reduction has taken place.

So, for the given reactions the value of $E_{o}$ are as follows.