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Nadusha1986 [10]
3 years ago
14

A gas sample is collected in a 0.279 L container at 22.7 °C and 0.764 atm. If the sample has a mass of 0.320 g, what is the iden

tity of the gas? Group of answer choices g
Chemistry
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

HCl

Explanation:

<em>Choices:</em>

<em>CO: 28.01g/mol</em>

<em>NO₂: 46g/mol</em>

<em>CH₄: 16.04g/mol</em>

<em>HCl: 36.4g/mol</em>

<em>CO₂: 44.01g/mol</em>

<em />

It is possible to identify a substance finding its molar mass (That is, the ratio between its mass in grams and its moles). It is possible to find the moles of the gas using general ideal gas law:

PV = nRT

<em>Where P is pressure of gas 0.764atm; V its volume, 0.279L; n moles; R gas constant: 0.082atmL/molK and T its absolute temperature, 295.85K (22.7°C + 273.15).</em>

Replacing:

PV = nRT

PV / RT = n

0.764atm*0.279L / 0.082atmL/molKₓ295.85K = n

<em>8.786x10⁻³ = moles of the gas</em>

<em />

As the mass of the gas is 0.320g; its molar mass is:

0.320g / 8.786x10⁻³moles = 36.4 g/mol

Based in the group of answer choices, the identity of the gas is:

<h3>HCl</h3>

<em />

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3<br>Solve and derive the unit for Power​
Mars2501 [29]

Answer:

Unit of power is watt

Explanation:

Power = rate of doing work

or

Power, P = W/t

Where

W is work done and t is time

The unit of work done is Joules and that of time is seconds.

Power, P = J/s

We know that, J/s = watts

Hence, the unit of power is watts.

7 0
3 years ago
A ball is rolled at a velocity of 3 m/s and rolls for 5 seconds. How far does<br> the ball roll? *
MrRa [10]

Answer 15m

Explanation: Distance =  Speed x Time

3 x 5 =15

6 0
2 years ago
A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being
Igoryamba

Answer:

<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>

Explanation:

This question is about solubility.

Regarding solubility, the solutions may be classified as:

  • Unsaturated: the concentration is below the maximum concentration permited at the given temperature.

  • Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.

  • Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.

Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.

  • In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.

  • In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g  of water.

  • Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ

       115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O  ⇒ x =  57.5 g NaNO₃

  • <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>

<em />

3 0
3 years ago
The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
maw [93]
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
a student determined that it requires 106220 j of energy to vaporize 47g of water. is the student is right​
shusha [124]
<h2>Answer:</h2>

He is right that the energy of vaporization of 47 g of water s 106222 j.

<h3>Explanation:</h3>

Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.

In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.

It means for vaporizing 18 g, 40.65 kJ energy is needed.

So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ

Hence the student is right about the energy of vaporization of 47 g of water.

7 0
3 years ago
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