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2 years ago

A flask with a volume of 250.0 mL contains air with a density of 1.164 g/L. What is the mass of the air contained in the flask?

Chemistry

1 answer:

satela [25.4K]2 years ago

6 0

Answer:

convert 250.0 mL in Liters :250. 0 / 1000 = 0,25 LDensity = 1.240 g/LMass

Explanation:

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What is the lightest particle in the following 1)electron 2)proton 3)neutron 4)photon

Anuta_ua [19.1K]

Answer:

1) Electron

Explanation:

It carries a negative charge of 1.602176634 × 10−19 coulomb, which is considered the basic unit of electric charge. The rest mass of the electron is 9.1093837015 × 10−31 kg, which is only 1/1,836the mass of a proton.

5 0

3 years ago

Read 2 more answers

A solution is created by dissolving 11.0 grams of ammonium chloride in enough water to make 235 mL of solution. How many moles o

Amanda [17]

Answer:

The correct answer is 0.206 moles

Explanation:

According to the given scenario, the calculation of the number of moles of ammonium chloride is available in the resulting solution is given below:

Given that

Amount of $NH_4Cl$ is 11.0 grams

And, the volume is 235 mL

Now the molar mass of $NH_4Cl$ is 53.49g/mol

So, the number of moles presented is

= 11.0 ÷ 53.49

= 0.206 moles

hence, the number of moles of ammonium chloride are available in the resulting solution is 0.206 moles

7 0

2 years ago

A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water

S_A_V [24]

Answer:

494.49 g of NaCl.

Explanation:

Data obtained from the question include the following:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mass of NaCl =.?

Next, we shall determine the number of mole of NaCl in the solution.

Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as

Molality = mole of solute /Kg of solvent

With the above formula, we can obtain the number of mole NaCl in the solution as follow:

Molality of NaCl = 3.140 m

Mass of water = 2.692 kg

Mole of NaCl =..?

Molality = mole of solute /Kg of solvent

3.140 = mole of NaCl /2.692

Cross multiply

Mole of NaCl = 3.140 x 2.692

Mole of NaCl = 8.45288 moles

Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:

Mole of NaCl = 8.45288 moles

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =.?

Mole = mass /Molar mass

8.45288 = mass of NaCl /58.5

Cross multiply

Mass of NaCl = 8.45288 × 58.5

Mass of NaCl = 494.49 g.

Therefore, 494.49 g of NaCl are present in the solution.

8 0

3 years ago

Differences and similarities between liquid and gas

timurjin [86]

<span>Various gases and liquids have different densities and combustion points.</span>

5 0

3 years ago

Suppose an ice cube weighing 36.0 g at a temperature of 10°C is placed in 360 g water at a temperature of 20°C. Calculate the te

Scilla [17]

Answer:

10.44 °C

Explanation:

When the thermal equilibrium is reached, both of the substances have the same final temperature (T). The liquid water will lose heat, and the ice cube will absorb this heat. The temperature of the ice will increase until it reaches 0°C, at this temperature, it will change of phase for liquid, absorbing heat, but without a change in the temperature. Then the temperature will increase until the equilibrium.

By the energy conservation, the total amount of heat must be equal to 0:

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

Liquid 1 is the ice after melting, and liquid 2 the liquid that was already at the flask. When there's a change of temperature:

Q = n*c*ΔT, where n is the number of moles, c is the heat capacity and ΔT is the temperature change (final - initial). The temperature variation in °C is equal in K, so the temperature may be used in °C.

The melting heat is:

Q = n*Hfus, Hfus = 6007 J/mol

The molar mass of the water is 18 g/mol, so the number of moles of the water and the ice are:

nwater = nliquid1 = 360/18 = 20 moles

nice = 36/18 = 2 moles

Qice + Qmelting + Qliquid1 + Qliquid2 = 0

2*38*(0 - (-10)) + 2*6007 + 2*75*(T - 0) + 20*75*(T - 20) = 0

760 + 12014 + 150T + 1500T - 30000 = 0

1650T = 17226

T = 10.44 °C

4 0

3 years ago