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finlep [7]
3 years ago
7

Propanol (C3H8O), an alcohol, and methoxyethane (C3H8O), an ether, are what types of isomers?

Chemistry
1 answer:
aleksandrvk [35]3 years ago
7 0
Answer is: <span>A. Functional group isomerism.

</span>Isomers are molecules that have the same molecular formula (in this example C₃H₈O)<span>, but have a different arrangement of the atoms in space. 
</span>Functional group isomerism is when the isomers contain different functional groups, so hey belong to different families of compounds (in this example alcohol and ether).

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What’s the physical properties of a molecular solid, in regards to its physical properties of formation?
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How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?
Karo-lina-s [1.5K]

Answer:

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = <u>58902.9 joules</u>

Explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C  to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

Q = 129 g . 2.10 J/g°C . 23°C → 6230.7 joules

Then, the ice has melted. To be melted and change the state it required:

Q = C lat . m

Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

5 0
3 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 <u>Oxidation</u>

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⁻ <u>Reduction</u>

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
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