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anzhelika [568]

3 years ago

14

If 250. ML of water are poured into the measuring cup, the volume reading is 8.45 oz . This indicates that 250. ML and 8.45 oz a

re equivalent. How many milliliters are in a fluid ounce based on this data?

1 answer:

Tamiku [17]3 years ago

7 0

Answer:

$1oz=29.589ml$

Explanation:

From the question we are told that:

Initial Volume $v_1=250ml$

Final Volume $v_2=8.45oz$

Generally the equation for one ounce is mathematically given by

$1oz=\frac{v_1}{v_2}$

Therefore

$1oz=\frac{250}{8.45}$

$1oz=29.589ml$

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$Rate=k[A]^x[B]^y$

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1: $1.2\times 10^{-2}=k[0.10]^x[0.20]^y$ (1)

From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (2)

Dividing 2 by 1 : $\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}$

$4=2^y,2^2=2^y$ therefore y=2.

b) From trial 2: $4.8\times 10^{-2}=k[0.10]^x[0.40]^y$ (3)

From trial 3: $9.6\times 10^{-2}=k[0.20]^x[0.40]^y$ (4)

Dividing 4 by 3: $\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}$

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Thus rate law is $Rate=k[A]^1[B]^2$

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3 years ago

How many moles (of molecules or formula units) are in each sample?

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The number of moles in each sample will be 0.391 moles, 30.7 moles, 0.456 moles, and 1350 moles

<h3>What is the number of moles?</h3>

The number of moles of a substance is the ratio of the mass of the substance to the molar mass.

In other words; mole = mass/molar mass.

Thus:

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3 years ago

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Answer:

a) $\triangle G^{0} = 7.31 kJ/mol$

b) $K_{-1} = 0.0594 m^{-1} s^{-1}$

Explanation:

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$2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)$

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Pressure change 2P 1P 2P

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Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

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Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

$K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }$

$K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}$

Standard free energy:

$\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol$

b) value of k−1 at 27 °C, i.e. 300K

$K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}$

$K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}$

$K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}$

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3 years ago