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valentinak56 [21]
3 years ago
11

The structure shown is represented of which sub

Chemistry
1 answer:
Katarina [22]3 years ago
8 0

Answer:

Option B. pentan-1-ol

Explanation:

We'll begin by writing the name of the compound.

The name of the compound above is pentylpropanoate.

From the name of the compound and the structure, we can suggest the following equation:

CH3CH2COOH + HOCH2CH2CH2CH2CH3 —> CH3CH2COOCH2CH2CH2CH2CH3 + H2O

Thus,

Propanoic acid + pentan-1-ol —> pentylpropanoate + water.

Therefore, the alcohol used in the reaction is pentan-1-ol

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Vol.250 before its to much pressure

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Which element has a gas notation of [Kr]5s2 4d7
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I think it would be Kriptonite
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A shiny, smooth piece of black plastic and an identical piece of white plastic are exposed to visible light and sound waves. Whi
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Answer:

The awnser is D

Explanation:

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2 years ago
A solution of malonic acid, H2C3H2O4 , was standardized by titration with 0.1000 M NaOH solution. If 21.17 mL of the NaOH soluti
MrRa [10]

Answer: The molarity of the malonic acid solution is 0.08335 M

Explanation:

H_2C_3H_2O_4 +2NaOH\rightarrow Na_2C_3H_2O_4+2H_2O

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2C_3H_2O_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=12.70mL\\n_2=1\\M_2=0.1000M\\V_2=21.17mL

Putting values in above equation, we get:

2\times M_1\times 12.70=1\times 0.1000\times 21.17\\\\M_1=0.08335M

Thus the molarity of the malonic acid solution is 0.08335 M

5 0
3 years ago
Consider the reaction given below.
Drupady [299]

Answer:

  • <u>K =  0.167 s⁻¹</u>

Explanation:

<u>1) Rate law, at a given temperature:</u>

  • Since all the data are obtained at the same temperature, the equilibrium constant is the same.

  • Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

        r = K [A]ᵃ [B]ᵇ

<u>2) Use the data from the table</u>

  • Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

         Divide r₂ by r₁:     [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

  • Use the first and second set of data to find the exponent a:

        r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

        r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

        Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

                                  2ᵃ = 2 ⇒ a = 1

         

<u>3) Write the rate law</u>

  • r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

<u>4) Use any set of data to find K</u>

With the first set of data

  • r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K =  0.167 s⁻¹

6 0
3 years ago
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