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2 years ago

A compound composed of only c and f contains 17.39 c by mass. what is its empirical formula

1 answer:

4 0

Since the Carbon C is 17.39% by mass hence the Fluorine F is 82.61% by mass. Divide each mass % by the respective molar masses, that is:

C = 17.39 / 12 = 1.45

F = 82.61 / 19 = 4.35

Divide the two by the smaller number, so divide by 1.45

C = 1.45 / 1.45 = 1

F = 4.35 / 1.45 = 3

So the empirical formula is:

CF3

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Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l

Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

$2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

We are given:

$\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]$

$\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol$

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0

2 years ago

Cups ans glasses are taking too long to air dry. This could be caused by

SOVA2 [1]

Answer: Servsafe 9

Explanation:

Remove food from the surface

clean the surface

rinse the surface

sanitize the surface

allow the surface to air dry

3 0

2 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

Answer: The value of $K_c$ for the given reaction is 1.435

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial: 0.20

At eqllm: 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

2 years ago

If 200mL of 0.60 M MgCl2 (aq) is added to 400mL of distilled water, what is the concentration of Mg2+(aq) in the resulting solut

lara [203]

Answer:

A.0.20M

Explanation:

c 1 V 1 = c 2 V 2

Initial Volume, V1 = 200 mL

Final Volume, V2 = 200 + 400 = 600 mL

Initial Concentration, c1 = 0.60 M

Final Concentration, c2= ?

Solving for c2;

c2 = c1v1 / v2

c2 = 0.60 * 200 / 600

c2 = 0.20M

3 0

1 year ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago