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3 years ago

Water’s unique properties, high heat capacity, high density, solid phase less dense than liquid phase can be attributed to

Chemistry

1 answer:

Xelga [282]3 years ago

4 0

Answer:

B- The polarity of the molecules and hydrogen bonding between molecules.

Explanation:

Hope this helps:)

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Consider the set of isoelectronic atoms and ions a2–, b–, c, d+, and e2+. which arrangement of relative radii is correct?

faust18 [17]

The term isoelectronic atoms means that if the given atoms are neutral, they would have the same number of electrons, which is relative to their sizes. Ions with positive charges are called cations, losing electrons. On the other hand, ions with negative charges are called anions, gaining electrons. The more electrons that the atoms have, the bigger it is in size. Hence, the arrangement of the radii of the atoms would have to be,

e²⁺, d⁺, c , b⁻, a⁻

The arrangement is from smallest to largest.

7 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

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damaskus [11]

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