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3 years ago

What is the relationship between a scientific theory and a scientific law

1 answer:

5 0

A scientific theory focuses on explaining a phenomenon, while a law will have a generalised statement about said phenomenon. Often times, we use laws to help us develop theories. =)

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A sample of 5 moles of gas is held at a constant temperature of 500 k inside a container with a volume of 0.25 l. What pressure

pshichka [43]

Answer: 41 atmosphere

Explanation:

Given that:

Volume of water V = 0.5 L

(since 1 liter = 1dm3

0.5L = 0.5dm3)

Temperature T = 500K

Pressure P = ?

Number of moles of gas = 5 moles

Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1

Then, apply ideal gas equation

pV = nRT

p x 0.5dm3 = 5 moles x (0.0082 atm dm3 K-1 mol-1 x 500K)

p x 0.5dm3 = 20.5 atm dm3

p = (20.5 atm dm3 / 0.5dm3)

p = 41 atm

Thus, the gas is under a pressure of 41 atmosphere

5 0

3 years ago

The heat of vaporization of acetic acid is . Calculate the change in entropy when of acetic acid condenses at .

Umnica [9.8K]

The question is incomplete, the complete question is;

The heat of vaporization ΔHv of acetic acid HCH3CO2 is 41.0 /kJmol. Calculate the change in entropy ΔS when 954.g of acetic acid condenses at 118.1°C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits.

Answer:

-1.67 JK-1

Explanation:

Since Heat of vaporization of acetic acid = 41.0 kJ/mol

Therefore:

Heat of condensation of acetic acid = -41.0 kJ/mol

Mass of acetic acid = 954 g

Temperature of condensation = 118.1 °C or 391.1 K

Number of moles of acetic acid = 954 g/60g/mol = 15.9 moles

Heat evolved during condensation = 15.9 moles * -41.0 kJ/mol = -651.9 KJ

Entropy change (ΔS) = Heat evolved/ Temperature = -651.9 KJ/391.1 K

Entropy change (ΔS) = -1.67 JK-1

8 0

3 years ago

What is the number of mole equivalents of H2 produced by the reaction of 1 mole equivalent of Mg (m) with H2SO4 (aq)?

Inga [223]

Answer:

We'll have 1.00 mol H2

Explanation:

Step 1: Data given

Number of moles of Mg = 1.00 moles

Step 2: The balanced equation

Mg(s) + H2SO4(aq) → MgSO4(aq) + H2(g)

Step 3: Calculate moles of H2

For 1 mol Mg we need 1 mol H2SO4 to produce 1 mol MgSO4 and 1 mol H2

We'll have 1.00 mol H2

8 0

3 years ago

skelet666 [1.2K]

Answer:

(ii) 1 dye

(iii) Food coloring F is insoluble in the solvent

(iv) 'E' and 'H'

(b) Food colouring G

Explanation:

Paper chromatography principle is based on the rates of migration of chemicals across a sheet of paper which are different and it consists of a stationary phase such as the water in the paper and a mobile phase such as the solvent resulting in the partitioning of the components of the mixture across the paper

The solution components are positioned to start in one place from where they migrate and separate out on the chromatography paper

(ii) The number of components into which the food colouring 'H' separates into = 1

Therefore, the number of dyes in food colouring 'H' = 1 dye

(iii) Food coloring 'F' does not move because it is insoluble in the solvent, which is the mobile phase

(iv) The food colouring that contains the dye that is likely to be most soluble in the solvent are does for which the dyes travel furthest, which are;

Food coloring 'E' and 'H'

(b) Using a similar question solution found on 'tutor my self' website, we have;

The $R_f$ values are given as follows;

$R_f = \dfrac{Distance \ moved \ by \ dye}{Distance \ moved \ by \ solvent}$

The distance moved by the solvent = 5 units

The distance moved by dyes in food colouring 'E' and 'H' = 4 units

The distance moved by dye in food colouring 'G' = 3.3 units

The distance moved by the second dye in food colouring 'E' = 2.7 units

By inspection, we get;

$R_f$ dye in food colouring 'G' = 3.3/5 = 0.66,

Therefore, the dye with $R_f$ value closest to 0.67 is the dye in food colouring 'G'.

5 0

3 years ago

When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively b

Arte-miy333 [17]

Answer:

B) hydroxide concentration

Explanation:

Hello,

In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.

After that, due to the fact that the pH is related with the pOH as shown below:

pH=14-pOH

We can directly compute the pH.

Best regards.

5 0

3 years ago