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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?
Chemistry
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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Lady_Fox [76]
How are you going to experiment means how will you show your project in a real life situation like a penny being cleaned with different acids.
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3 years ago
Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen
olga55 [171]

Answer: Rate law=k[A]^1[B]^2, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is 3L^2mol^{-2}s^{-1}

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

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}

2=2^x,2=2^1, x=1

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

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1:  1.2\times 10^{-2}=k[0.10]^1[0.20]^2

k=3 L^2mol^{-2}s^{-1}.



6 0
3 years ago
How many moles (of molecules or formula units) are in each sample?
Alexus [3.1K]

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:

  • moles of 18.0 g NO_2 = 18.0/46

                                   = 0.391 moles

  • moles of 1.35 kg CO_2 = 1350/44

                                         = 30.7 moles

  • moles of 46.1 g KNO_3 = 46.1/101.1

                                          = 0.456 moles

  • moles of 191.8 kg Na_2SO_4 = 191800/142

                                                 = 1350 moles

More on the number of moles of substances can be found here: brainly.com/question/1445383

#SPJ1

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2 years ago
Future of geothermal energy
s2008m [1.1K]

Answer:

Explanation:

Among the most recent systems for the exploitation of geothermal energy, the most promising are the third generation ones, also called EGS (Enhanced Geothermal Systems). Their technology allows to dramatically improve energetic efficiency of both geothermal wells and dry rocks.

Present and Future of Geothermal Energy - Cividachttp://www.cividac.com › news › present-and-future-of-g...

7 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

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

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

Explanation:

Equation of reaction:

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

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

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)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

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}

6 0
3 years ago
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