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3 years ago

What is the function of a lyase enzyme?

Chemistry

1 answer:

elena55 [62]3 years ago

5 0

Answer:

c. To facilitate a reaction of one substrate to form two products without the use of water

Explanation:

A lyase is an enzyme that catalyzes - accelerates the chemical reaction - in which a substrate is broken into two molecules. The reaction does not involve hydrolysis or oxidation, so the water molecule is not included in the chemical reaction. Thus, the enzyme facilitates the reaction in which a molecule (substrate) is decomposed into two molecules with the elimination of chemical bonds.

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When the following equation is balanced with the lowest whole number coefficients possible, what is the coefficient in front of

jasenka [17]

2Al + 3ZnCl₂ ----> 3Zn + 2AlCl₃

What is the coefficient in front of AlCl₃? ==>> 2

4 0

3 years ago

How do you explain the relatively high conductivity of tap water compared to a low or zero conductivity for distilled water?

gladu [14]

Tap water contains many dissolved ions that are required to carry an electric current, whilst distilled water contains no or relatively low amounts of dissolved ions. the absence of ions in the distilled water accounts for the low to zero conductivity of the water

hope that helps!

4 0

3 years ago

Read 2 more answers

A gas exerts a pressure of 0.62 atm. Convert this to kPa and mmHg. Be sure to show your work.

IRINA_888 [86]

Answer:

The answer to your question is 0.62 atm = 62.82 kPa = 471.2 mmHg

Explanation:

Data

P = 0.62 atm

P = ? kPa

P = ? mmHg

Process

1.- Look for the conversion factor of atm to kPa and mmHg

1 atm = 101.325 kPa

1 atm = 760 mmHg

2.- Do the conversions

1 atm ----------------- 101.325 kPa

0.62 atm ------------ x

x = (0,62 x 101.325) / 1

x = 62.82 kPa

1 atm ------------------ 760 mmHg

0.62 atm ------------ x

x = (0.62 x 760)/1

x = 471.2 mmHg

4 0

3 years ago

An aluminum block has a density of 2.70 g/mL. If the mass of the block is 24.60 g, find the volume of the substance.

harina [27]

Volume of a substance can be determined by dividing mass of the substance by its density.

That can be mathematical shown as:

Density=Mass/Volume

So, Volume=Mass/Density

Here mass of the substance given as 24.60 g

Whereas density of the substance is 2.70 g/mL

So,

Volume=Mass/Density

=24.6/2.7

=9.1 mL

So volume of the substance is 9.1 mL.

8 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago