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KonstantinChe [14]
3 years ago
14

HELP ASAP!!!!

Chemistry
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

a. ionic, 211.62g/mol

b. molecular, 149g/mol

c. molecular, 342g/mol

Explanation:

Ionic Compound: These are compounds (2 or more elements) where atoms of the element have lost or gained electrons, thus they are ions, thus the name ionic. An easy way to identify ionic compounds is to see if there are any metals. Metals tend to give up their electron to a non-metal. It maybe helpful to familiarize which parts of the periodic table have metals and non-metals.

Molecular Compounds: These are compounds (2 or more elements) that are neutral. An easy way to identify them is that the compound is made up of just non-metals.

Molar Mass is the masses of each individual element in the compound. Refer to the periodic table for the masses and add them up carefully, remember to multiply their mass based on how many molecules there are in the compound, like for a, we can find the mass of NO3, but we need to multiply it by 2 since we have 2 molecules of NO3.

a. Sr(NO3)2

Sr= 87.62

NO3= 14 + (3 x 16)= 62

Sr + 2 (NO3)

87.62 + (2 x 62)= 211.6

b. (NH4)3PO4

NH4= 14 + 4 = 18

PO4= 30.97 + (4 x 16)= 94.9

(NH4 x 3) + PO4

(18 x 3) + 94.9 = 149

c. C12H22O11

(12 x C) + (22 x H) + (11 x O)

(12 x 12) + (22 x 1) + (11 x 16)

144 + 22 + 176= 342

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Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = \frac{Q}{T}

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = \frac{25*1000}{293}

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = \frac{-25*1000}{500}

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = \frac{25*1000}{400}

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = (\frac{25*1000}{400}) + (\frac{-25*1000}{500}

= 12.5 J/K

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