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Yuliya22 [10]
3 years ago
9

The combination of ions most likely to produce a precipitate is Group of answer choices Mg2 and C2H3O2-. Fe3 and OH-. Li and PO4

3-. Pb2 and NO3-. NH4 and SO42-.
Chemistry
1 answer:
Paul [167]3 years ago
8 0

Answer:

The combination of ions most likely to produce a precipitate is a group of answer choices:

lead nitrate soluble in water

Mg2+ and C2H3O2-.

Fe3+ and OH-.

Li+ and PO43-.

Pb2+ and NO3-.

NH4+ and SO42-.

Explanation:

Among the given options,

magnesium acetate, lithium phosphate, lead nitrate, ammonium sulfate are soluble in water.

The only one which is insoluble in water is Fe^3+ and OH^- combination.

Fe(OH)_3 is insoluble in water. It forms a precipitate.

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BIOCHEMISTRY HELP! You and a lab partner are performing on several molecular analogs to determine Km. One is a substrate analog,
Alina [70]

Answer:

Michaelis constant is known as km which is the substrate concentration that encourages the compound to work at half maximum velocity represented by Vmax/2. Michaelis constant is inversely related to the substrate and the affinity of the enzyme.  

Induced fit model: The premise of the purported induced fit hypothesis, which expresses that the attachment or association of a substrate or some other atom to an enzyme causes an adjustment to the enzyme in order to fit or restrain its activity.  

In substrate, analog Km or Michaelis constant will be high as the substrate will stay because of analogs inhibit activity.

In the transitional state, analog Km will be in the middle of the substrate and product analogs. Progress state analogs are synthetic mixes with a structure catalyzed reaction that looks like the progressing condition of a substrate atom in a compound enzyme.  

In item simple thus Km is the least.  

0.0013 M = product ananlog,

0.025 M=Transition state, and

0.0045 M = Substrate analog

5 0
3 years ago
Question 7
chubhunter [2.5K]
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5 0
2 years ago
An archaeologist discovered some charcoal at an excavation site. He sent the samples to a lab for C-14 dating, knowing that the
klio [65]
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7 0
3 years ago
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles
Mrrafil [7]

Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃  +  O₂  ⟶  N₂  +  H₂O

The balanced reaction:

4NH₃  +  3O₂  →  2N₂  +  6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

6 0
3 years ago
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