Which of these is an element?

Why can a liquid change to take the shape of its container but NOT expand to fill the container itself?

tiny-mole [99]

Answer: Option (C) is the correct answer.

Explanation:

Molecules in a liquid have less force of attraction as compared to solids. But liquid molecules have more force of attraction as compared to gases.

Since molecules of a gas are held together by weak Vander waal forces, therefore, they expand to fill the container whereas molecules in a liquid are not expanded in a container like gases because of more force of attraction within molecules of liquids as compared to gases.

Hence, a liquid can take the shape of container in which it is kept.

Thus, we can conclude that out of the given options, a liquid change to take the shape of its container but NOT expand to fill the container itself because the particles of a liquid are held together loosely enough to flow, but not so loose that they expand.

4 0

3 years ago

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During the combustion of 2.00 g of coal, the temperature of 500 g of water inside the calorimeter increased from 25.0°c to 43.7°

pantera1 [17]

Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.

6 0

3 years ago

How many of the following are found in 15.0 kmol of xylene (C8H10)? (a) kg C8H10; (b) mol C8H10; (c) lb-mole C8H10; (d) mol (g-a

77julia77 [94]

Answer:

a) 1592.4 kg C8H10

b) 15*10³ mol C8H10

c) 33.1 lb-mole

d) 1.2 *10^5 mol C

e) 1.5 * 10^5 mol H

f) 1.44 * 10^6 grams C

g) 1.52 * 10^5 grams H

h) 9.03*10^27 molecules C8H10

Explanation:

Step 1: Data given

Number of moles C8H10 = 15.0 kmol =15000 moles

Molar mass of C8H10 = 106.16 g/mol

Step 2: Calculate mass C8H10

Mass C8H10 = moles C8H10 * molar mass C8H10

Mass C8H10 = 15000 * 106.16 g/mol

Mass C8H10 = 1592400 grams = 1592.4 kg

Step 3: Calculate moles C8H10

15.0 kmol = 15*10³ mol C8H10

Step 4: Calculate lb-mol

15.0 kmol = 33.1 lb-mole

Step 5: Calculate moles of C

For 1 mol C8H10 we have 8 moles of C

For 15*10³ mol C8H10 we have 8* 15*10³ =1.2 *10^5 mol C

Step 6: Calculate moles H

For 1 mol C8H10 we have 10 moles of H

For 15*10³ mol C8H10 we have 10* 15*10³ = 1.5 * 10^5 mol H

Step 7: Calculate mass of C

Mass C = moles C * molar mass C

Mass C = 1.2 *10^5 mol C * 12 g/mol

Mass C = 1.44 * 10^6 grams

Step 8: Calculate mass of H

Mass H = moles H * molar mass H

Mass H = 1.5 *10^5 mol H * 1.01 g/mol

Mass C = 1.52 * 10^5 grams

Step 9: Calculate molecules of C8H10

Number of molecules = number of moles * number of Avogadro

Number of molecules = 15*10³ mol C8H10 * 6.022*10^23

Number of molecules = 9.03*10^27 molecules

8 0

3 years ago

list 10 chemical reactions that have benefited your life today. Include the reasons you think each was indeed a chemical reactio

Dennis_Churaev [7]

1) The overall balanced photosynthesis reaction:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.

Plants convert solar energy into the chemical energy of sugars (food).

2) Chemical reaction of cellular respiration:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O.

Organic carbon (glucose) is transformed into inorganic carbon (carbon dioxide) that goes into atmosphere.

3) Gasoline combustion:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O.

Gasoline is a mixture of many different hydrocarbons: alkanes (paraffins), cycloalkanes and alkenes (olefins).

Gasoline is being burn.

4) Balanced chemical reaction of forming rust: 4Fe + 3O₂ → 2Fe₂O₃.

Forming rust usually last for few months and iron changing in another substance with different properties.

5) An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.

In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

Copper is forming from the solution, that is chemical change.

6) Balanced chemical reaction: CaO + H₂O → Ca(OH)₂.

Making limewater (diluted solution of calcium hydroxide).

7) Neutralization is reaction in which an acid and a base react quantitatively with each other. In this reaction, there are no acids and bases, only salts.

Balanced chemical reaction of vinegar and antacid:

2CH₃COOH(aq) + Mg(OH)₂(aq) → (CH₃COO)₂Mg(s) + 2H₂O(l).

Precipation is formed.

8) Balanced chemical reaction of magnesium and hydrochloric acid:

MgO + 2HCl → MgCl₂ + H₂O(g).

Acid dissolves metal oxides.

9) Chemical reaction of silver in air:

4Ag + 2H₂S + O₂ → 2Ag₂S (tarnish) + 2H₂O.

Silver change color in the air.

10) Forming of patina: 2CuO + CO₂ + H₂O → Cu₂CO₃(OH)₂.

Brass (copper) change color to green.

4 0

3 years ago

Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u

Fantom [35]

Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt) (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂ (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

<h2>% = 76.75%</h2><h2>And this is the % of isotope after 1 year</h2>

3 0

3 years ago

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