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jekas [21]
3 years ago
7

What kind of inhibitors are transition state analogs usually classified as? uncompetitive inhibitors competitive inhibitors nonc

ompetitive inhibitors
Chemistry
1 answer:
Masteriza [31]3 years ago
3 0
<h2>Competitive inhibitors.</h2>

Explanation:

▪ Transition state analogs can be used as inhibitors in enzyme-catalyzed reactions which are done by blocking the active site of the enzyme.

▪ A transition state analog is somewhat the same as that of the transition state.

▪These are better inhibitors than the substrate analogs in competitive inhibition, the reason is that they bind tighter to the enzyme rather than the substrate.

▪ Thus, they are classified as competitive inhibitors.

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How many grams of dextrose are needed to make 725 mL of a 26.0% (w/v) dextrose solution? Note that mass is not technically the s
Diano4ka-milaya [45]

Answer:

188.5g of dextrose are needed

Explanation:

In Weight per volume percentage - %(w/v) -, the concentration is defined as the mass of solute in grams -In this case, dextrose-, in 100mL of solution.

As you want to prepare 725mL of a 26.0% (w/v) solution. you need:

725mL * (26g / 100mL) = 188.5g of solute =

<h3>188.5g of dextrose are needed</h3>
3 0
3 years ago
Please help me with this.<br><br><br>And show all work as well ASAP!!​
qaws [65]

Answer: The partial pressure of oxygen is 187 torr.

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

p_1=x_1p_{total}  

where, x = mole fraction  

p_{total} = total pressure  

x_{oxygen}=\frac{\text {moles of oxygen}}{\text {total moles}},  

x_{oxygen}=\frac{3.0}{12.33}=0.243,  

p_{oxygen}=0.243\times 770torr=187torr  

Thus the partial pressure of oxygen is 187 torr.

6 0
3 years ago
13. List the following ions in order of increasing radius: Li+, Mg2+, Br"
Sergeeva-Olga [200]

Answer:

The answer would be B, PC13

Explanation:

5 0
2 years ago
Read 2 more answers
For which of the following reactions is S° &gt; 0. Choose all that apply. N2(g) + 3H2(g) 2NH3(g) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2
Lostsunrise [7]

Answer: 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

NH_4I(s)\rightarrow NH_3(g)+HI(g)

2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

Explanation:

Entropy is the measure of randomness or disorder of a system.

A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products,  the randomness is decreasing and the entropy is negative

2. 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products,  the randomness is increasing and the entropy is positive.

3. NH_4I(s)\rightarrow NH_3(g)+HI(g)

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. 2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. 2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

5 0
3 years ago
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is
Airida [17]

Answer:

the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C  

Explanation:

Given the data in the question;

First we calculate the Volume of the steel cylinder;

V = πr²h

radius r = Diameter / 2 = 27 cm / 2 = 13.5 cm

height h = 32.4 cm

so we substitute

V = π × ( 13.5 cm )² × 32.4 cm

V  = π × 182.25 cm × 32.4 cm

V = 18550.79 cm³  

V = 18.551 L

given that; maximum safe pressure P = 3.10 MPa = 30.5946 atm

vessel contains 0.218kg or 218 gram of carbon monoxide gas

molar mass of carbon monoxide gas is 28.010 g/mol

so

moles of carbon monoxide gas n = 218 gram /  28.010 g/mol = 7.7829 mol

we know that;

PV = nRT

solve for T

T = PV / nR

we know that gas constant R = 0.0820574 L•atm•mol⁻¹ K⁻¹

so we substitute

T = ( 30.5946 × 18.551 ) / ( 7.7829 × 0.082 )

T = 567.5604 / 0.6381978

T = 889.317387 K

T = ( 889.317387 - 273.15 ) °C

T = 616.167 ≈ 616 °C  { 3 significant digits }

Therefore, the maximum safe operating temperature the engineer should recommend for this reaction is 616 °C  

6 0
3 years ago
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