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3 years ago

The synthesis of 5-phosphoribosylamine from PRPP is

Chemistry

1 answer:

mel-nik [20]3 years ago

5 0

Answer:

The correct answer is option B: "I, II".

Explanation:

The synthesis of 5-phosphoribosylamine from phosphoribosyl pyrophosphate (PRPP) is inhibited allosterically by ATP and GTP and activated by PRPP via feed forward activation. The enzyme that regulate this process is the enzyme amidophosphoribosyltransferase (AMPRT), which is stimulated by increased PRPP concentrations and is inhibited allosterically by ATP and GTP as well as IMP, AMP, and GMP.

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MaRussiya [10]

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Paladinen [302]

Explanation:

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3 years ago

Read 2 more answers

Im not sure how to do this can someone help with these?

IceJOKER [234]

Answer:

1. 280 g of CO

2. 16.4 g of O₂

3. 42 g of Cl₂

Explanation:

Ans 1

Data Given:

moles of O₂= 5 moles

mass of CO = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2CO + O₂ -----------> 2CO₂

2 mol 1 mol

So if we look at the reaction 2 mole of CO react with 1 mole of O₂ then how many moles of CO will react with 5 moles of O₂

For this apply unity formula

2 mole of CO ≅ 1 mole of O₂

X mole of CO≅ 5 mole of O₂

By Doing cross multiplication

moles of CO = 2 moles x 5 moles / 1 mol

moles of CO = 10 mole

Now calculate mass of 10 moles of CO

Formula used

mass in grams = no. of moles x Molar mass

Molar mass of CO = 12 + 16 = 28 g/mol

Put values in above formula

mass in grams = 10 moles x 28 g/mol

mass in grams = 280 g

So,

280 g of CO will react with 5 moles of O₂

_________________________

Ans 2

Data Given:

mass of C₃H₈ = 22.4 g

moles of O₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

C₃H₈ + 5O₂ -----------> 3CO₂ + 4H₂O

1 mol 5 mol

Convert moles to mass

Molar mass of C₃H₈ = 3(12) + 8(1)

Molar mass of C₃H₈ = 36 + 8 = 44 g/mol

and

molar mass of O₂ = 2(16) = 32 g/mol

So,

C₃H₈ + 5O₂ -----------> 3 CO₂ + 4H₂O

1 mol (44 g/mol) 5 mol (32 g/mol)

44 g 160 g

So if we look at the reaction 44 g of C₃H₈ react with 160 g of O₂, then how many grams of O₂ will react with 22.4 g of ethane

For this apply unity formula

44 g of C₃H₈ ≅ 60 g of O₂

grams of O₂ ≅ 22.4 g of ethane

By Doing cross multiplication

grams of O₂ = 22.4 g x 44 g/ 60 g

grams of O₂ = 16.4 g

16.4 g of O₂ react with 22.4 grams of ethane

______________________

Ans 3

Data Given:

mass of Rubidium Chlorate = 10 g

moles of O₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2 RbClO₃ ------------ 2 RbCl + 3O₂

2 mol 3 mol

Convert moles to mass

Molar mass of RbClO₃ = 85.5 + 35.5 + 3(16)

Molar mass of RbClO₃ = 169

and

molar mass of O₂ = 2(16) = 32 g/mol

So,

2 RbClO₃ ------------> 2 RbCl + 3O₂

2 mol ( 169 g/mol) 3 mol (32 g/mol)

338 g 96 g

So if we look at the reaction 338 g of RbClO₃ gives 96 g of O₂, then how many grams of O₂ will be given by 10 g of RbClO₃

For this apply unity formula

338 g of RbClO₃ ≅ 96 g of O₂

grams of O₂ ≅ 10 g of RbClO₃

By Doing cross multiplication

grams of O₂ = 338 g x 10 g/ 96 g

grams of O₂ = 35.2 g

35.2 g of O₂ will be produce by 10 grams of RbClO₃

______________________

Ans 4

Data Given:

mass of K = 46 g

moles of Cl₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2K + Cl₂ ------------> 2KCl

2 mol 1 mol

Convert moles to mass

Molar mass of K = 39 g/mole

and

molar mass of Cl₂ = 2(35.5) = 71 g/mol

So,

2K + Cl₂ ------------> 2KCl

2 mol ( 39 g/mol) 1 mol (71 g/mol)

78 g 71 g

So if we look at the reaction 78 g of K react wit 71 g of Cl₂, then how many grams of Cl₂ will react with 46 g of K

For this apply unity formula

78 g of K ≅ 71 g of Cl₂

46 g of K ≅ X grams of Cl₂

By Doing cross multiplication

grams of Cl₂ = 71 g x 46 g/ 78 g

grams of Cl₂ = 42 g

42 g of Cl₂ will react with 46 grams of K

4 0

3 years ago

Question 3

zimovet [89]

The answer is 4 hope it helped

8 0

3 years ago

A 35.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following v

marissa [1.9K]

Answer:

a) 2.79

b) 4.75

c) 6.59

d) 8.81

e) 11.03

f) 12.42

Explanation:

Step 1: Data given

Volume of acetic acid = 35.0 mL

Molarity of acetic acid = 0.150 M

Molarity of NaOH = 0.150 M

Step 2: pH after adding 0 mL of NaOH

HA + H2O ⇆ A- + H3O+

Initial concentration

[HA] = 0.150 M

[A-] = 0M

[H3O+] = 0M

Concentration at the equilibrium

[HA] = 0.150 - x M

[A-] = xM

[H3O+] = xM

Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5

0.150 >>> x

We can write it like this

x²/(0.150) = 1.76*10^-5

x² =0.00000264

x = 0.00162

[H3O+] = 0.00162 M

pH = -log [H3O+] = -log (0.00162) = 2.79

Step 3: pH after adding 17.5 mL of NaOH

This is the point of half-neutralization.

pH = pKa = -log (1.76*10^-5- 4.75

Step 4: pH after adding 34.5 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.0345 L

Moles NaOH = 0.005175 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.005175 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.005175 = 0.000075 moles

[OH-] = 0.005175 - 0.005175 = 0 moles

[A-] = 0.005175 moles

pH = pKa + log([A-]/[HA]

pH = 4.75 +log(0.005175/0.000075)

pH = 4.75 + 1.84 = 6.59

Step 4: pH after adding 35.0 mL of NaOH

HA + OH- ⇆ A- + H2O

Moles acetic acid = molarity * volume

Moles acetic acid = 0.150 M * 0.035 L

Moles acetic acid = 0.00525 moles

Moles NaOH = 0.150 M *0.035 L

Moles NaOH = 0.00525 moles

Initial moles

[HA] = 0.00525 moles

[OH-] = 0.00525 moles

[A-] = 0 moles

[H2O] = 0 moles

Moles at the equilibrium

[HA] = 0.00525 - 0.00525 = 0 moles

[OH-] = 0.00525 - 0.00525 = 0 moles

[A-] = 0.00525 moles

We have a solution of 0.00525 mol of A - in 70.0 mL

[A-] = 0.00525 moles / 0.070 L = 0.075 M

A- + H20 ⇆ HA + OH-

Initial moles A- = 0.0750 moles

initial moles HA = 0 moles

Initial moles OH- = 0 moles

Moles A- at the equilibrium = 0.0750 - x

Moles HA at the equilibrium = x

Moles OH- at the equilibrium = x

Kb = Kw/Ka = 10^-14 / 1.76*10^-5

Kb = 5.68 *10^-10

Kb = [HA][OH-] / [A-] = x² / 0.0750 - x = 5.68*10^-10

Since 0.0750 >>> x we can write

5.68*10^-10 = x²/0.0750

x = 6.53 *10^-6

[OH-] = x = 6.53 * 10^-6 M

pOH = - log (6.53*10^-6) = 5.19

pH = 14 - pOH = 14 - 5.19 = 8.81

Step 5: pH after adding 35.5 mL of NaOH

We add an excess moles of NaOH

We have an excess of 0.5 mL NaOH

Moles excess = 0.0005 L *0.150 M = 0.000075 moles

[OH-] = 0.000075 moles / (0.035+0.0355L)

[OH-] = 1.06 * 10^-3 M

pOH = -log( 1.06 *10^-3)

pOH = 2.97

pH = 14 -2.97 = 11.03

Step 6: pH after adding 50.0 mL of NaOH

We have an excess of 15.0 mL NaOH

Moles excess = 0.015 L *0.150 M = 0.00225 moles

[OH-] = 0.00225 moles / (0.035+0.050L)

[OH-] = 0.0265 M

pOH = -log( 0.0265M)

pOH = 1.58

pH = 14 -1.58 = 12.42

4 0

3 years ago