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3 years ago

Describe the R group found within an amino acid

Chemistry

1 answer:

madreJ [45]3 years ago

5 0

Hello!

An amino acid has five basic parts. An amino group, a carboxyl group, a central carbon atom, a hydrogen, and a R group.

The r group is the side chain specific to each amino acid. The r group varies between each kind of amino acid, while everything else stays the same. Therefore, the r group decides the identity of the amino acid.

Hope this helps! Let me know if you need a more specific explanation.

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Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

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

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

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

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

4 years ago

If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).

11111nata11111 [884]

Answer:

${ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}$

7 0

3 years ago

Calculate the pH of a solution that [H3O4] of 7.22x10-7M

schepotkina [342]

To calculate the pH of a solution that has a [H3O+] of 7.22x10^-7. You would do the following
pH=-log[H3O+]
pH=-log[7.22x10^-7]
pH=?

5 0

3 years ago

At the equivalence point of a titration of the [H+] concentration is equal to:

icang [17]

B. At the equivalence point of a titration of the [H+] concentration is equal to 7.

<h3>What is equivalence point of a titration?</h3>

The equivalence point of a titration is a point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.

At the equivalence point in an acid-base titration, moles of base equals moles of acid and the solution only contains salt and water.

At the equivalence point, equal amounts of H+ and OH- ions combines as shown below;

H⁺ + OH⁻ → H₂O

The pH of resulting solution is 7.0 (neutral).

Thus, the pH at the equivalence point for this titration will always be 7.0.

Learn more about equivalence point here: brainly.com/question/23502649

#SPJ1

5 0

2 years ago

A reaction of 17.0 g of ammonia with 26.6 g of chlorine gas

Elena-2011 [213]

Answer: if doing it on ck-12 the answer is 40.1

Explanation:

4 0

2 years ago