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ale4655 [162]
3 years ago
11

Draw the amide formed when 1-methylethylamine (CH, CH(CH,)NH,) is heated with each carboxylic acid.

Chemistry
1 answer:
Margarita [4]3 years ago
8 0

Answer:

Explanation:

.

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Explain the differences between bronze and brass, and steel and cast iron. Which is best used for cooking wares?
Dennis_Churaev [7]

Answer:

Steel and cast iron

Explanation:

They are all metal but assuming that you are finding the best material for your pan i suggest going for steel or cast iron

8 0
3 years ago
A 0.175 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.27° in a 1-dm sample
-BARSIC- [3]

Answer:

The specific rotation of D is 11.60° mL/g dm

Explanation:

Given that:

The path length (l) =  1 dm

Observed rotation (∝) = + 0.27°

Molarity = 0.175 M

Molar mass = 133.0 g/mol

Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol

Concentration in (g/mL) = 23.275 g/L

Since 1 L = 1000 mL

Concentration in (g/mL) = 0.023275 g/mL

The specific rotation [∝] = ∝/(1×c)

= 0.27°/( 1  dm ×  0.023275 g/mL )

= 11.60° mL/g dm

Thus, the specific rotation of D is 11.60° mL/g dm

3 0
3 years ago
Which of the following metals will react with water to produce a metal hydroxide and hydrogen gas? (2 points)
vazorg [7]
The answer would be the first option - k
8 0
3 years ago
Read 2 more answers
What is the relationship between a mole and Avogadro number
docker41 [41]
A mole contains Avogadro’s number of particles of a substance.
4 0
3 years ago
Read 2 more answers
Suppose the reaction between nitrogen and hydrogen was run according to the amounts presented in Part A, and the temperature and
andrew11 [14]

Explanation:

Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.

This means that, for N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}

 moles of N_{2} + moles of H_{2} = 0.908 mol

Since, 2 moles of N_{2} = 2 \times 0.227 = 0.454 mol

As it is known that the ideal gas equation  is PV = nRT

And, as the temperature and volume were kept constant, so we can write

        \frac{P(_in)}{n_(in)} = \frac{P_(final)}{n_(final)}

          \frac{10.4}{0.908} = \frac{P_(final)}{0.454
}

       P_(final) = 10.4 \times \frac{0.454}{0.908}

                            = 5.2 atm

Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.

7 0
3 years ago
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