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Mars2501 [29]
2 years ago
7

Which functional group is found in an ester?

Chemistry
2 answers:
Klio2033 [76]2 years ago
4 0
<h2>QUESTION:- Which functional group is found in an ester?</h2>

ESTER IS THE COMPOUND IN WHICH OH GROUP IS REPLACED BY THE (O) in the substitution reaction.

so common formula of ester ->

RCOOR'

IN WHICH R AND R' ARE CARBON CHAINS

horrorfan [7]2 years ago
4 0

Answer:

I think it's A let me know if I'm wrong

Explanation:

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How many liters of oxygen are in 8.32 moles of oxygen at STP
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Answer:

186 Liters at STP conditions

Explanation:

1 mole of any gas at STP conditions occupies 22.4 Liters.

Therefore, 8.32 moles O₂(g) = 8.32 moles x 22.4Liters/mole = 186 Liters (3 sig.figs.)

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Give an example of one type of energy conversion (change to another form). Be sure to explain your example.
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A cylinder of argon contains 50 L of Ar at 12.4 atm and 127°C . How many moles of argon are in the cylinder
joja [24]

Answer:

18.9 moles

Explanation:

We have the following data:

V = 50 L

P = 12.4 atm

T= 127°C + 273 = 400 K

R = 0.082 L.atm/K.mol (it is the gas constant)

We use the ideal gas equation to calculate the number of moles n of the gas:

PV = nRT

⇒ n = PV/RT = (12.4 atm x 50 L)/(0.082 L.atm/K.mol x 400 K) = 18.9 mol

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Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)
AURORKA [14]

Explanation:

The net equation will be as follows.

          K(s) + Cl_{2}(g) \rightarrow KCl(s)

So, we are required to find \Delta H_{formation} for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1:  Convert K from solid state to gaseous state

          K(s) \rightarrow K(g),    \Delta H_{1} = 89 kJ

Step 2:  Ionization of gaseous K

           K(g) \rightarrow K^{+}(g) + e^{-},    H_{2} = 418 KJ

Step 3:  Dissociation of Cl_{2} gas into chlorine atom .

            \frac{1}{2} Cl_{2}(g) \rightarrow Cl(g),   \Delta H_{3} = \frac{244}{2} = 122 KJ

Step 4: Iozination of chlorine atom.

              Cl(g) + e^{-} \rightarro Cl^{-}(g),      H_{4} = -349 KJ

Step 5:  Add K^{+} ion and Cl^{-} ion formed above to get KCl .

              K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s),   H_{5} = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

      \Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}

                  = 89 + 418 + 122 - 349 - 717

                  = - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

5 0
3 years ago
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