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Evgesh-ka [11]
2 years ago
5

A 1.0 x 102- gram sample is found to be pure alanine, an amino acid found in proteins. How many moles of alanine are in the samp

le
Chemistry
1 answer:
Romashka-Z-Leto [24]2 years ago
4 0

Answer:

1.123x10⁻⁴ moles of alanine

Explanation:

In order to convert grams of alanine into moles, <em>we need to know its molecular weight</em>:

The formula for alanine is C₃H₇NO₂, meaning <u>its molecular weight would be</u>:

  • 12*3 + 7*1 + 14 + 16*2 = 89 g/mol

Then we <u>divide the sample mass by the molecular weight</u>, to do the conversion:

  • 1.0x10⁻² g ÷ 89 g/mol = 1.123x10⁻⁴ moles
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The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ΔH, ΔS, a
Nostrana [21]

The question is incomplete, the complete question is;

The normal boiling point of acetic acid is 118.1°C. If a sample of the acetic acid is at 125.2°C, predict the signs of ∆H, ∆S, and ∆G for the boiling process at this temperature.

A. ∆H > 0, ∆S > 0, ∆G < 0

B. ∆H > 0, ∆S > 0, ∆G > 0

C. ∆H > 0, ∆S < 0, ∆G < 0

D. ∆H < 0, ∆S > 0, ∆G > 0

E. ∆H < 0, ∆S < 0, ∆G > 0

Answer:

A. ∆H > 0, ∆S > 0, ∆G < 0

Explanation:

During boiling, a liquid is converted to vapour. This is a phase change for which heat is absorbed because energy must be taken in to break the intermolecular bonds in the liquid before it can be converted to a gas. Hence ∆H>0

Secondly, a phase change from liquid to gas leads to an increase in entropy hence ∆S>0.

Thirdly, the process is spontaneous. For every spontaneous process ∆G<0

3 0
2 years ago
470,809 mi express in an integer
cupoosta [38]
-470,809 m is in an integer
7 0
2 years ago
Read 2 more answers
Melanie has completed the analysis of her data for the reaction of KMnO4 with malonic acid and data for a reaction of KMnO4 with
WARRIOR [948]

Answer:

See explanation

Explanation:

The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;

k= Ae^-Ea/RT

Where;

k= rate constant

A= pre-exponential factor

Ea=activation energy

R= gas constant

T= temperature

We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.

Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.

5 0
2 years ago
Please please help!
ratelena [41]

 The mass  percent  of potassium chloride  is   1.386%

<u><em>calculation</em></u>

mass  percent = actual mass/ Theoretical mass x 100

Actual mass = 9.35 g

Theoretical mass  is  calculated as below

Step 1 : write the equation for reaction

KCl + H₂O  →   KOH + HCl

Step 2: find the moles of H₂O

moles = mass÷ molar mass

The molar mass of H₂O = (2 x1 ) +(16)  = 18 g/mol

moles is therefore = 162.98 g÷ 18 g/mol =9.054 moles

Step 3: use the mole ratio to determine the moles of KCl

KCl: H₂O  is 1:1 therefore the moles of KCl  is also = 9.054 moles

Step 4:  find the  theoretical mass of KCl

mass = moles x molar mass

from periodic table the  molar mass of KCl = 39 +35.5 =74.5 g/mol

mass = 9.054 moles x 74.5 g/mol =674.5 g


Theoretical mass is therefore = 9.35 g/ 674.5 g x 100 = 1.386%


3 0
2 years ago
Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

Given that:

The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

8 0
2 years ago
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