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Yuliya22 [10]
3 years ago
14

Two jars are placed on a counter with a McDonald's French Fry inside, one has a lid, the other does not. They are left alone to

see which one decays faster, after 2 days the fry in the closed jar looks fresher. What is the independent variable? What is the dependent variable? What is the control variable?
Chemistry
1 answer:
Agata [3.3K]3 years ago
7 0

Answer: days

Explanation: controlled variable is that they put it in 2 jars for 2 days one with a lid and one with no lid. dependent is how long it was there (2) days. Independent is you left it in one place for the whole 2 days. i think.

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Which is the scientific notation for 8,002.5 seg?
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Indicate the changes (increases, decreases, does not change) in its volume when the pressure undergoes the following changes at
Inga [223]

The question is missing information. Here is the complete question.

A gas at a pressure of 2.0 atm is in a closed container. Indicate the changes (if any) in its volume when the pressure undergoes the following changes at constant temperature and constant amount of gas. Match the words in the left with the column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.

1. The pressure increases to 6.0 atm. The volume ________

2. The pressure drops to 0.40 atm. The volume _________

3. The pressure remains at 2.0 atm. The volume _________

Answer: 1. Decreases

2. Increases

3. Does not change

Explanation: According to the Ideal Gas Law, <u>Pressure</u>, <u>Volume</u> and <u>Temperature</u> of an ideal gas is related, as the following: PV = nRT.

In this case, since temperature (T) and amount of gas (n) are constant, the <em><u>Boyle's</u></em> <em><u>Law</u></em> can be used.

The law states that the volume of a given gas, under the conditios of temperature and amount of it are constant, is inversely proportional to the applied pressure: P₁.V₁ = P₂.V₂

  • For case 1.)

Initial P (P₁) = 2

Initial V (V₁) = V

Final P (P₂) = 6

P₁.V₁ = P₂.V₂

2.V = 6.V₂

V₂ = 1/3V

When the pressure increases to 6 atm, volume <em><u>decreases</u></em> by 1/3.

  • For case 2.)

P₁ = 2

V₁ = V

P₂ = 0.4

2.V = 0.4V₂

V₂ = 5V

When pressure drops to 0.4 atm, volume <em><u>increases</u></em> by 5.

  • For case 3.)

Since there are no change in the pressure, the volume is the same from the beginning, so <em><u>does not change</u></em>.

8 0
3 years ago
State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and expl
Strike441 [17]

Answer:

a) ΔS is negative

b) ΔS is positive

c) ΔS is positive

d) ΔS is negative

Explanation:

Entropy (S) is a thermodynamic parameter which measures the randomness or the disorder in a system. Greater the disorder more positive will be the value of entropy.

The extent of disorder increases as substances transition from the solid to the gaseous state. i.e.

S(solid) < S(liquid) < S(gas)

The entropy change for a given reaction is:

\Delta S = S(products)-S(reactants)----(1)

a) PCl3(l) + Cl2(g) \rightarrow PCl5(s)

Here  the reactants are in the liquid and gas phase which have higher entropy than the product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

b) 2HgO(s) \rightarrow 2Hg(l) + O2(g)

Here the products are in the liquid and gas phase which have higher entropy than the reactant which is in the solid phase i.e. lower entropy.

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

c) H2(g) \rightarrow 2H(g)

Here the products and reactants are in the gas phase. However the number of moles of products is greater than the reactants

Since S(product) > S(reactant), based on equation 1, ΔS will be positive.

d) U(s) + 3F2(g) \rightarrow UF6(s)

Here  one of the reactants is in the gas phase which corresponds to a more positive entropy compared to the  product which is in the solid phase i.e. lower entropy.

Since S(product) < S(reactant), based on equation 1, ΔS will be negative.

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