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3 years ago

Consider the structure of the amino acid alanine. indicate the hybridization about each interior atom.

Chemistry

2 answers:

OLEGan [10]3 years ago

5 0

The structure of Alanine is shown below,
Except the carbon atom of carbonyl group which is Sp² Hybridized, all remaining atoms are Sp³ Hybridized. The hybridization of each element is depicted in a box below,

Brilliant_brown [7]3 years ago

5 0

Answer:

Alanine is composed of three carbon atoms , two oxygen atoms ,one nitrogen atom and seven hydrogen atoms with chemical formula of $C_3H_7NO_2$ , with structure shown in image attached.

$C-1$ atom is $sp^3$ hybridized as it can form four single bonds to give four hybridized orbitals.

$C-2$ atom is $sp^3$ hybridized as it can form four single bonds to give four hybridized orbitals.

$C-3$ atom is $sp^2$ hybridized as it can form two single bonds to give three hybridized orbitals .

$N$ atom is $sp^3$ hybridized as it can form three single bonds to give four hybridized orbitals out of which one hybridized orbitals is occupied by the lone pair of the nitrogen atom.

$O-1$ atom is $sp^2$ hybridized which means there are three hybridized orbitals out of which two of the hybridized orbitals are occupied by the two lone pairs of the oxygen atom.

$O-2$ atom is $sp^3$ hybridized as it form two single bonds to give four hybridized orbitals out of which two of the hybridized orbitals are occupied by the two lone pairs of the oxygen atom .

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When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem

masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

Step 1: Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

Step 2: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

Step 3: Define the formula

Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)

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3 years ago

Which type of molecule is shown below?

zubka84 [21]

Answer:

$\boxed{\mathrm{Alkene}}$

Explanation:

The hydrocarbon shown has a double bond. Hydrocarbons with double bonds are known as alkenes.

Cyclic alkanes have cyclic structure.

Alkanes only have single bonds.

Alkynes have triple bonds.

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3 years ago

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How are gems different from regular minerals? Gems are softer than regular minerals. Regular minerals are more beautiful than ge

just olya [345]

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3 years ago

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Examples of nitrogens unreactivity

REY [17]

WATCH OUT BUDDY:reactivity....Spelling mistake.

ANSWER:
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3 years ago

14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and d

zepelin [54]

The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

$\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M$

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock solution.

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution.

From data (A) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 20mL=1M\times V_2\\\\V_2=100mL$

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 20mL=1M\times V_2\\\\V_2=100mL$

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

$M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 60mL=1M\times V_2\\\\V_2=300mL$

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

$M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?$

Putting values in above equation, we get:

$5M\times 60mL=1M\times V_2\\\\V_2=300mL$

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

5 0

3 years ago