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xz_007 [3.2K]
2 years ago
12

The volume of a gas at 99.6 kPa and 24°C is 4.23 L. What volume will it occupy at 93.3 kPa ?

Chemistry
1 answer:
yuradex [85]2 years ago
6 0

The volume is increased to 4.52 L on decreasing the pressure to 93.3 kPa.

Explanation:

As per Boyle's law, the volume occupied by gas particles will be inversely proportional to the pressure experienced by those particles at constant temperature.

V=\frac{1}{P}

So in the present problem, the volume of gas at pressure P₁ = 99.6 kPa is given as V₁ = 4.23 L. The temperature is kept constant at 24°C. Then, if the pressure is decreased to 93.3 kPa, then the volume is tend to increase due to Boyle's law.

So let us consider the new pressure be P₂ = 93.3 kPa and the new volume has to be found.

Then using Boyle's law, P_{1} V_{1} = P_{2}  V_{2}

Then, V_{2}=\frac{P_{1} V_{1} }{P_{2} }

So, V_{2}=\frac{99.6*1000*4.23}{93.3*1000}=4.52 L

Thus, the volume is increased to 4.52 L on decreasing the pressure to 93.3 kPa.

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Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, a
geniusboy [140]

Answer:

1) 41.16 % = 0.182 grams

2) The alkali cation is K+ , to form the salt K2SO4

Explanation:

Step 1: Data given

Mass of unknown sulfate salt = 0.323 grams

Volume of water = 50 mL

Molarity of HCl = 6M

Step 2: The balanced equation

SO4^2- + BaCl2 → BaSO4 + 2Cl-

Step 3: Calculate amount of SO4^2- in BaSO4

The precipitate will be BaSO4

The amount of SO4^2- in BaSO4 = (Molar mass of SO4^2-/Molar mass BaSO4)*100 %

The amount of SO4^2- in BaSO4 = (96.06 /233.38) * 100

= 41.16%

So in 0.443g of BaSO4 there will be 0.443 * 41.16 % = <u>0.182 grams</u>

<u />

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2. If it is assumed that the salt is an alkali sulfate determine the identity of the alkali cation.

The unknown sulphate salt has 0.182g of sulphate. This means the alkali cation has a weight of 0.323-0.182 = 0.141g grams

An alkali cation has a chargoe of +1; sulphate has a charge of -2

The formula will be X2SO4 (with X = the unknown alkali metal).

Calculate moles of sulphate

Moles sulphate = 0.182 grams (32.1 + 4*16)

Moles sulphate = 0.00189 moles

The moles of sulphate = 0.182/(32.1+16*4)

The moles of sulphate = 0.00189 moles

X2SO4 → 2X+ + SO4^2-

For 2 moles cation we have 1 mol anion

For 0.00189 moles anion, we have 2*0.00189 = 0.00378 moles cation

Calculate molar mass

Molar mass = mass / moles

Molar mass = 0.141 grams / 0.00378 grams

Molar mass = 37.3 g/mol

The closest alkali metal is potassium. (K2SO4 )

3 0
2 years ago
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
2 years ago
Consider the following information. The lattice energy of CsCl is Δ H lattice = − 657 kJ/mol. The enthalpy of sublimation of Cs
AlladinOne [14]

Answer:

<ERROR>-----------------------------------------------------------------------------<ERROR>

Explanation:

8 0
2 years ago
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That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p
vitfil [10]
<span>The density of an iceberg is less than that of water and that is why it floats. It is the same as ice cubes floating on water too. Water is a very unique substance in that it is one of few compounds where cooling it past freezing point decreases its density (study hydrogen bonds). The possible answers are therefore A or C. If the majority of the ice is below the water then it should be clear from common sense that A is the correct answer as it would mostly float on the top with just a little under the surface if the answer were as low as C. See Archimedes Principle for an explanation of how much of the ice floats and how much is underwater.</span>
3 0
1 year ago
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A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

             = 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0
1 year ago
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