What is the ΔG given the following? ΔH = 20 kJ/mol T = 15°C ΔS = .101 kJ/molK

Why does buck ministerfullerene act as a good lubricant

MakcuM [25]

Answer:

Its molecules are made up of 60 carbon atoms joined together by strong covalent bonds. Molecules of C 60 are spherical. There are weak intermolecular forces between molecules of buckminsterfullerene. These need little energy to overcome, so buckminsterfullerene is slippery and has a low melting point.

Explanation:

5 0

3 years ago

Read 2 more answers

Help me with these questions.

Aloiza [94]

1) C
2) B
3) A/B not sure
4) C I think

5 0

3 years ago

Which activity might help to increase the validity of this experiment?

MArishka [77]

A should be the answer because the more you test an experiment the more data you have to rely on changing the experiment would cause you to have different outcomes making the results different and unreliable so B, C, and D is not going to be the answer Hope this helps

5 0

2 years ago

g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O Oxidation

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ Reduction

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 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

8 0

2 years ago