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spin [16.1K]
2 years ago
3

The volume of fluorine gas required to react with 2.67 g of calcium bromide to form calcium fluoride and bromine at 41.0°c and 4

.31 atm is __________ ml.
Chemistry
1 answer:
Ad libitum [116K]2 years ago
0 0
Balance Equation is as followm

                                       F₂  +  CaBr₂   →   CaF₂  +  Br₂

According to equation 1 mole of F₂ reacts with 1 mole of CaBr₂. So,
As,
                           199.89 g CaBr₂ reacts with   =   1 mole of F₂
Then,
                       2.67 g of CaBr₂ will react with   =   X mole of F₂
Solving for X,
                                        X  =  (2.67 g × 1 mole) ÷ 199.89 g

                                        X  = 0.0133 moles of F₂

Now,
Covert moles of F₂ into Volume, Assuming it as Ideal gas,
So, Acc. to Ideal Gas Equation

                                          P V  = n R T
Solving for V,
                                             V  =  n R T / P

Putting Values,

               V  =  (0.0133 mol × 0.0820 atm.L.mol⁻¹.K⁻¹ × 313 K) ÷ 4.31 atm

               V  =  0.0792 L

Converting into mL,
               
                   =  0.0792 × 1000

                   =  79.2 mL

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Answer:

\triangle G= -6.7 KJ/mol

Explanation:

From the question we are told that:

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X=A⇌B,ΔG= 14.8 kJ/mol

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Answer:

V = 0.0327 L.

Explanation:

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In this case, according to the given information, it turns out possible for us to calculate the liters of C3H6O by the definition of density. We can tell the density of this substance as that of acetone (0.784 g/mL) and therefore calculate the liters as shown below:

V=25.6g*\frac{1mL}{0.784g}*\frac{1L}{1000mL}\\\\V=0.0327L

Regards!

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3 years ago
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