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3 years ago

Can somebody please help me

Chemistry

1 answer:

kramer3 years ago

4 0

Answer:

a) carboxylic acid

b) amine

c) ester

d) aldehyde

e) alkene

f) ketone

Explanation:

Most of them are straight forward.

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Which statements describe the elements? Which ones are true and which ones are false?

Alchen [17]

Answer:

Explanation:

Uma declaração verdade é identificada pelo de ela afirmar a verdade diferente das outras opções

8 0

3 years ago

What is bias in an experiment?

11111nata11111 [884]

I’m not sure what the answer is but I hope someone can help you

3 0

2 years ago

Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)

Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0

3 years ago

If there are 80 liters of argon gas in a piston and the piston expands the gas until its volume is 2000 liters with a pressure o

Mice21 [21]

Answer: The correct answer is A. 11.5 atm. The temperature is held constant at 293 K, therefore, we can use Boyle's Law to determine the initial pressure. Boyle's Law states that there is an inverse relationship between pressure and volume of gases. Therefore, as volume increases, the pressure will decrease and vice versa.

Further Explanation:

Boyle's Law can be mathematically expressed as:

$P_{initial}V_{initial}\ = P_{final}V_{final}$

In this problem, we are given the values:

P(initial) = ?

V(initial) = 80 L

P (final) = 0.46 atm

V (final) = 2000 L

Plugging in these values into the equation:

$P_{initial}(80 \ L) = \ (0.46 \ atm)(2000 \ L)\\P_{initial} \ = \frac{(0.46 \ atm) \ (2000 \ L)}{80 \ L}\\P_{initial}\ = 11.5 atm\\$

The initial pressure was 11.5 atm. Since the volume increased or expanded, the space where the gas particles move is bigger, so the frequency of collisions with the wall of the container and with other particles are effectively decreased. This, therefore, decreases the pressure from 11.5 to 0.46 atm.

Learn More

Learn about Charles' Law brainly.com/question/1421697
Learn about Ideal Gas Law brainly.com/question/6534668
Learn about Gay - Lusaac's Law brainly.com/question/1358307

Keywords: gas, Boyle's Law, Ideal Gas Law

7 0

3 years ago

How many atoms are in water is it 24 or 2.<br> Explain

Artemon [7]

There are actually three: One oxygen atom and two hydrogen atoms.

8 0

2 years ago