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olga55 [171]
3 years ago
10

A 1.20 g sample of water is injected into an evacuated 5.00 l flask at 65°c. part of the water vaporizes and creates a pressure

of 187.5 mmhg. what percentage of the water vaporized?
Chemistry
1 answer:
Mkey [24]3 years ago
3 0
Assume that the water vapor is an ideal gas. So,


PV = nRT

For conversion, 760 mmHg = 101325 Pa and 1,000 L = 1 m³
(187.5 mmHg)(101325 Pa/760 mmHg)(5 L)(1 m³/1,000 L) = n(8.314 m³Pa/molK)(65+273 K)
Solving for n,
n = 0.0445 mole water

Since the molar mass of water is 18 g/mol,
Mass of water vaporized = 0.0445*18 = 0.8 g water vaporized

Hence,
Percentage of water vaporized = 0.8/1.2 * 100 =<em> 66.7%</em>
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At a birthday pool party, the temperature is 28.50°C and the atmospheric pressure is 755.4 mmHg. One of the decoration helium ba
Iteru [2.4K]

Answer:

The volume of the balloon will be 5.11L

Explanation:

An excersise to solve with the Ideal Gases Law

First of all, let's convert the pressure in mmHg to atm

1 atm = 760 mmHg

760 mmHg ___ 1 atm

755.4 mmHg ____ (755.4 / 760) = 0.993 atm

922.3 mmHg ____ ( 922.3 / 760) = 1.214 atm

T° in K = 273 + °C

28.5 °C +273 = 301.5K

26.35°C + 273= 299.35K

P . V = n . R .T

First situation: 0.993atm . 6.25L = n . 0.082 . 301.5K

(0.993atm . 6.25L) / 0.082 . 301.5 = n

0.251 moles = n

Second situation:

1.214 atm . V = 0.251 moles . 0.082 . 301.5K

V = (0.251 moles . 0.082 . 301.5K) / 1.214 atm

V = 5.11L

7 0
3 years ago
Read 2 more answers
Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica
Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = d_{ideal} = \dfrac{P\times M}{RT}

here:

P = pressure of the O₂ gas = 310 bar

= 310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂  gas = 32 g/mol

∴

d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}

d_{ideal} = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol²    &

b = 0.0319  L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}

310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}

310V^3 -44.389V^2+1.382V-0.044=0

After solving;

V = 0.1152 L

∴

d_{Van \ der \ Waal} = \dfrac{32}{0.1152}

d_{Van \ der \ Waal} = 277.77  g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0
3 years ago
Aqueous solutions of Na2CO3 and Ca(NO3)2, 0.10 M each, are combined. A white precipitate is observed in the container after mixi
anzhelika [568]

The question is incomplete. Here is the complete question.

Aqueous solutions of Na_{2}CO_{3} and Ca(NO_{3})_{2}, 0.10 M each, are combined. A white precipitate is observed in the container after mixing. he precipitate is filtered andcarefully rinsed with distilled water to remove other ions. A sample of the precipitate is added to 100 mL of 0.1 M NaCl. A second sample of the precipitate is then added to 100 mL of 0.1 M HCl. What would be observed in each case?

                 Observation upon                                         Observation upon

               addition of precipitate                                  addition of precipitate

                      to NaCl(aq)                                                       to HCl(aq)

(A)     additional precipitates forms                        no visible reaction occurs

(B)     no visible reaction occurs                            gas is produced and some                                                                                        precipitate dissolves

(C)     no visible reaction occurs                             no visible reaction occurs

(D)     additional precipitates forms                       gas is produced and some

                                                                                    precipitate dissolves

Answer: (B) No visible reaction occurs; Gas is produced and some precipitate dissolves

Explanation: When aqueous solutions of Na_{2}CO_{3} and Ca(NO_{3})_{2} are combined, it reacts according to the following balanced equation:

Na_{2}CO_{3}+Ca(NO_{3})_{2} → CaCO_{3}_{(s)}+2NaNO_{3}_{(aq)}

forming calcium carbonate (CaCO_{3}), which, as it is insoluble in water, precipitates as a solid of the color white. This process is <u>Precipitation</u> and this reaction is a <u>Precipitation</u> <u>Reaction</u>.

When calcium carbonate reacts with NaCl it produces:

CaCO_{3}+2NaCl → CaCl_{2}+Na_{2}CO_{3}

Now, calcium chloride is an inorganic compound very soluble in water, so, in this reaction, there are no precipitate and <u>no visible reaction occurs</u>.

When CaCO_{3} reacts with hydrochloridric acid, the balanced reaction is

CaCO_{3}+2HCl → CaCl_{2}+H_{2}CO_{3}

which, also produces calcium chloride and carbonic acid.

Both are soluble in water but, when carbonic acid is in an "aqueous state", carbonic acid, it dissociates, forming carbon dioxide and water. Therefore, <u>gas is produced and some precipitate dissolves</u>.

In conclusion, sentence B is the correct alternative.

5 0
3 years ago
Some important points for hadpicking<br>​
Svet_ta [14]

Answer:

Handpicking right?

It is a very simple and easy method of separation. It doesn't require any kind of equipment to proceed. It takes very less time when performed for small quantity of mixture. It doesn't require any kind of preparation

8 0
2 years ago
A wooden block has a mass of 125 g and a volume of 55 cm3. What is its density?
leonid [27]

Answer:

d = 2.27 cm3

Explanation:

m = 125 g

V = 55 cm3

m = V × d

=> d = m/V = 125/55 = 2.27 g/cm3

4 0
2 years ago
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