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KiRa [710]
2 years ago
7

compare the C2-C3 bonds in propane,propene, and propane.Should they be any different with respect to either bond length or bond

strength?If so,how should they vary.​
Chemistry
1 answer:
harkovskaia [24]2 years ago
3 0

Answer: one by one

Explanation:

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In a two-step synthesis where the first reaction has a 55% yield and the second reaction has a 24% yield, the overall percent yi
nekit [7.7K]

Answer:

13,2 %.

Explanation:

Overall yield = 0.55 * 0.24

= 0.132

= 13,2 %.

7 0
2 years ago
Use the internet or your textbook as a reference to compare and contrast the Arrhenius Theory of acids and bases vs. the Brønste
Readme [11.4K]

Answer and Explanation:

1. Arrhenius Theory which describes the concept protonic. The substance that gives H+ ions when diluted in water is called as an acid (e.g. HCl) and the substance that dissociates OH-ions whenever it is diluted in water is called as the base (e.g. NaOH)

on the other hand

Bronsted Lowery Theory describes the concept of a proton donor-acceptor. The proton-donating species is an acid and the proton-accepting species is known as a base.

2. The Chemical name and nature of acid is shown below:-

Nature Chemical Name

a. HCl Acidic Hydrochloric Acid

b. KOH Basic Potassium hydroxide

c. HNO Acidic Nitric Acid

d. Mg(OH)2 Basic Magnesium hydroxide

3 0
2 years ago
Several amino acids are intermediates of the urea cycle, having side ammonia groups that join with free carbon dioxide (CO2) and
mr_godi [17]

Answer:

A. Arginine

Explanation:

The urea cycle is the cycle of the biochemical reactions which produces urea from ammonia.

Steps of the urea cycle:

  • Carbamoyl phosphate, in presence of ornithine transcarbamoylase is converted to citrulline by the denotation of carbamoyl phosphate groupto ornithine and a phosphate group is released.
  • Amino group of the aspartate and carbonyl group of the citrulline are condensed to form argininosuccinate in the presence of enzyme, argininosuccinate synthetase. This condensation reaction is ATP dependent.
  • <u>Argininosuccinate then undergoes cleavage by the argininosuccinase to form intermediate, arginine and fumarate.</u>
  • <u>Arginine is then cleaved by the arginase to form urea as well as ornithine.</u> Ornithine is transported back to mitochondria to begin urea cycle again.
7 0
2 years ago
How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C
NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0
2 years ago
Calculate the entropy change for a process in which 3.00 moles of liquid water at 08c is mixed with 1.00 mole of water at 100.8c
kupik [55]
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:

ΔS = nCln(T₂/T₁)

n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature

We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:

[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K

Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:

ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K

ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K

Now we combine the entropy change of each portion of water to get the total entropy change for the system:

ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K

The entropy change for combining the two temperatures of water is 2.9 J/K.
3 0
2 years ago
Read 2 more answers
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