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IgorLugansk [536]
4 years ago
5

Draw the amide formed when isopropylamine (ch3ch(ch3)nh2) is heated with each carboxylic acid.

Chemistry
1 answer:
Anon25 [30]4 years ago
7 0
Amides are the derivatives of Carboxylic Acid. The are formed when the Hydroxyl group in carboxylic acids is replaced by -NR₂. The simplest way of synthesizing Amides is the condensation of primary or secondary amine with carboxylic acid. This reaction requires heat because it has high activation energy. The Amides formed by reacting the given carboxylic acids with Isopropyl amine are given below,

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Using the model for the formation of Scandium Fluoride (an ionic compound), write the chemical formula.
egoroff_w [7]

Answer:

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4 0
3 years ago
20 grams of ethane (C2H6) in a 15 L container has a pressure of 1.3 bar. What is the temperature, in oC
Zepler [3.9K]

Answer:

The temperature is 42.5 °C

Explanation:

We apply the Law of Ideal Gases to solve this:

P  .  V = n .  R . T

First, we convert the bar into atm, so we make a rule of three.

1.013 bar is 1 atm

1.3 bar is (1.3 . 1) /1.013 = 1.28 atm

1.28atm . 15L = n . 0.082 . T

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20 g / 30 g / mol = 0.666 moles

1.28atm . 15L = 0.666 mol . 0.082 . T

(1.28 atm . 15L) / (0.666 mol . 0.082) = T

315.5 K = T

As this is absolute temperature we must convert to °C

315.5 K - 273= 42.5 °C

3 0
3 years ago
Read 2 more answers
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oksano4ka [1.4K]

Answer:

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8 0
3 years ago
Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the pr
Lyrx [107]

Answer:

The free energy change for the reaction at 37.0°C is -8.741 kJ.

Explanation:

The free energy of the reaction is given by :

\Delta G=\Delta G^o+RT\ln K

where,

\Delta G^o = standard Gibbs free energy

R = Gas constant = 8.314J/K mol

T = temperature in Kelvins

K = equilibrium constant

We have :

\Delta G^o=-16.7 kJ/mol=-16,700 J/mol

1 kJ = 1000 J

T = 37.0 C = 37 +273.15 K = 310.15 K

Ratio of concentrations of the products to the concentrations of the reactants =K = 21.9

\Delta G=-16,700 J/mol+8.314J/K mol\times 310.15 K \ln[21.9]

=-8,741.22 J = -8.741 kJ

The free energy change for the reaction at 37.0°C is -8.741 kJ.

5 0
3 years ago
To one decimal place 1.3511
Jet001 [13]

1.4

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5 0
3 years ago
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