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Alex_Xolod [135]
3 years ago
11

What two parts of an amino acid are involved in a peptide bond?

Chemistry
1 answer:
Lemur [1.5K]3 years ago
6 0

Answer:

The answer to your question is below.

Explanation:

Amino acids are composed by one amino group, one carboxyl group and one  chain.

The parts of the amino acid that are involved in a peptide group are the amino group (- NH₂) and the carboxyl group (-COOH).

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Initial temperature of metal =
Evgesh-ka [11]

Answer:

the correct answers is 100 22.7 and 24.6

Explanation:

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4 0
4 years ago
Read 2 more answers
A gas occupies 100 mL at 150. kPa. Find its volume at 200. kPa. You must show all your work to receive credit. Be sure to identi
PtichkaEL [24]

The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:

PV = k

where k is a constant.

We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,

P₁V₁ = P₂V₂

Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.

In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa.  Substituting these values in the previous equation and clearing V₂, we have,

P₁V₁ = P₂V₂ → V₂ = \frac{P1V1}{P2}  

→ V₂ = \frac{100 mL x 150 kPa}{200 kPa}

→ V₂ = 75 mL

Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL

6 0
3 years ago
PLZ HELP, GIVING BRAINLIEST!!
andrew-mc [135]

Answer:

Option B. Decreasing the temperature of the solvent

Explanation:

Solubility is mostly enhanced by increasing the temperature of the solvent or solution. This means that am increase in temperature will increase the solubility and decreasing the temperature will decrease the solubility.

5 0
3 years ago
Read 2 more answers
Can someone please help me understand this?
Nataliya [291]

Answer:

French physicist Jacques Charles (1746-1823) studied the effect of temperature on the volume of a gas at constant pressure. Charles's Law states that the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant. The absolute temperature is temperature measured with the Kelvin scale. The Kelvin scale must be used because zero on the Kelvin scale corresponds to a complete stop of molecular motion.

alt

Figure 11.5.1: As a container of confined gas is heated, its molecules increase in kinetic energy and push the movable piston outward, resulting in an increase in volume.

Mathematically, the direct relationship of Charles's Law can be represented by the following equation:

V

T

=k

As with Boyle's Law, k is constant only for a given gas sample. The table below shows temperature and volume data for a set amount of gas at a constant pressure. The third column is the constant for this particular data set and is always equal to the volume divided by the Kelvin temperature.

Explanation:

PLEASS MARK ME AS BRAINLIEST ANSWER

4 0
3 years ago
What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

  • ΔT= final temperature - initial temperature

The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

  • ΔT= 88.9 °C - 58.8 °C = 30.1 °C

Now we know three variables:

  • Q= 4500.0 J
  • c= 0.4494 J/g°C
  • ΔT = 30.1 °C

Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

8 0
2 years ago
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