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Daniel [21]
2 years ago
15

A 3.81-gram sample of NaHCO3 was completely decomposed in an experiment. 2NaHCO3 → Na2CO3 + H2CO3 In this experiment, carbon dio

xide and water vapors combine to form H2CO3. After decomposition, the Na2CO3 had a mass of 2.86 grams. a. Determine the mass of the H2CO3 produced. b. Calculate the percentage yield of H2CO3 for the reaction. Show your work or describe the calculation process in detail.
Chemistry
1 answer:
Oksanka [162]2 years ago
5 0

Answer:

The percent yield of H_2CO_3  is, 24.44 %

Explanation:

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A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas
diamong [38]
There are 3 parts in this question:
1) To find the initial Boyle's constant k_{i}
2) To find the final Boyle's constant k_{f}
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = V_{i} = 12.0 L
The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm

The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
The final volume of the cylinder(in litres) = V_{f} = 6.0 L

First you need to know what Boyle's law is:
<span>Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.
</span>
The Mathematical form of Boyle's law is:
P =  \frac{k}{V}

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1. 
The initial volume of the cylinder(in litres) = V_{i} = 12.0 L
The initial pressure(in atmospheric pressure) = P_{i} = 4.0 atm
The Boyle's constant = k_{i} = ?

According to the Boyle's law,

P_{i} = \frac{k_{i}}{V_{i}}

=> k_{i} =  P_{i}V_{i}
Plug-in the values in the above equation, you would get:
k_{i} = 4.0 * 12.0 = 48

Ans-1) k_{i} = 48

2.
The final pressure(in atmospheric pressure) = P_{f} = 8.0 atm
The final volume of the cylinder(in litres) = V_{f} = 6.0 L
The Boyle's constant = k_{f} = ?

According to the Boyle's law,

P_{f} = \frac{k_{f}}{V_{f}}

=> k_{f} =  P_{f}V_{f}
Plug-in the values in the above equation, you would get:
k_{f} = 8.0 * 6.0 = 48

Ans-2) k_{f} = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

k_{i} = k_{f}

Since,
k_{i} = 48
k_{f} = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

-i

4 0
3 years ago
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If a dozen clementine oranges have a mass of 744g, what will be the mass of 15 Clementines?
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930 g for 15 clementines
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A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate miner
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Ferromagnesian silicate minerals (i looked it up)
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Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu
Nitella [24]

Answer:

\Delta \:H_{vap}=40383.88\ J/mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)

Given:

P_1 = 13.95 torr

P_2 = 144.78 torr

T_1 = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K

T_1 = 298.15 K

T_2 = 75°C  = 348.15 K

So,

\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)

\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}

\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=40383.88\ J/mol

4 0
2 years ago
The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution i
bija089 [108]

Answer:

110ml

Explanation:

<em>Using the dilution equation, C1V1 = C2V2</em>

<em>Where C1 is the initial concentration of solution</em>

<em>C2 is final concentration of solution</em>

<em>V1 is intital volume of solution</em>

<em>V2 is final volume of solution.</em>

From the question , C1=6M, C2=0.5M, V1=10ml, V2=?

V2 =\frac{C1V1}{C2}

V2 =\frac{10*6}{0.5}

V2 =120ml

volume of water added = final volume -initial volume

                                    = 120-10

                                   =110ml

3 0
2 years ago
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