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zhannawk [14.2K]
3 years ago
13

The perspective formula of threonine, an amino acid with two asymmetric centers, is provided below. perspective formula of threo

nine add either an h, oh, or nh2 group to complete the fischer projection for threonine.
Chemistry
1 answer:
Citrus2011 [14]3 years ago
8 0
<span>The threonine add either an h, oh.The provide below standpoint formula of threonine add either an h, oh, or nh2 group to whole the fischer projection for threonine. the threonine aldolase structure there address of the reversible cleavage of several L-3-hydroxy-a-amino acids. like as L-TAs +H3NH. HOH. CH3. L-Threonine (2S, 3R).</span>
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The mass of a jar of sugar is 11. 9 kg. What is the total mass of 13 jars of sugar?.
Masteriza [31]

Answer:

154.7

Explanation:

11.9kg x 13 = 154.7

5 0
2 years ago
Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T
GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0
3 years ago
H2SO4(aq) --&gt; 2H+ + SO4^2–
horsena [70]
I'm guessing
<span>A. The chemical equilibrium will shift to the right.</span>
7 0
3 years ago
Read 2 more answers
When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0
neonofarm [45]

Answer:

The mass of PbSO4 formed 15.163 gram

Explanation:

mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625

mole of Na₂SO₄ = 2 x 0.025 = 0.05

                                      Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃

( Mole/Stoichiometry )    \frac{0.0625}{1}           \frac{0.05}{1}

                                     = 0.0625     = 0.05

From  (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.

Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄

                                            = 0.05 x 303.26 g

                                            = 15.163 g

7 0
3 years ago
High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen. (a) Determine t
Lemur [1.5K]

Answer:

Concentration of Flourine = 24.756%

Explanation:

Given that :

High-density polyethylene may be flourinated by inducing the random substitution of Flourine atoms for hydrogen.

the objective is to determine he concentration of Flourine (in wt%) that must be added if this substitution occurs for 12% of all the original hydrogen atoms.

At standard conditions , the atomic weight of the these compounds are as follows:

Carbon = 12.01 g/mol

Chlorine = 35.45 g/mol

Fluorine = 19.00 g/mol

Hydrogen = 1.008 g/mol

Oxygen = 16.00 g/mol

The chemical formula for polyethylene = (CH₂ - CH₂)ₙ

Therefore, for two carbons, there will be 4 hydrogens;

i.e

(CH₂ - CH₂)₂

( C₂H₄ - C₂H₄ )

Suppose the number of original hydrogen = 4moles

number of moles of Flourine F = 12% of 4

= 0.12 × 4

= 0.48 mol

∴ the number of remaining moles of Hydrogen is:

= 4 - 0.48

= 3.52 moles

number of moles of Carbon = 2 moles

∴ the mass of flourine F = number of moles of F × molar mass of F

= 0.48 × 19

= 9.12

The total mass of the compound now is = (0.48 × 19 ) + (3.52 × 1) + (2× 12)

= 9.12 + 3.52 + 24

= 36.64

Concentration of Flourine = (mass of flourine/total mass) × 100

Concentration of Flourine = (9.12/36.84 ) × 100

Concentration of Flourine = 0.24756 × 100

Concentration of Flourine = 24.756%

3 0
3 years ago
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