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3 years ago

Some enzymes can transfer a hydrogen atom without lowering the activation energy barrier. how is this possible?

1 answer:

4 0

It is true that enzymes speed up the reaction by lowering the activation energy. However, the main concept here is the compatibility of an enzyme to its substrate. This is achieved through lock-and-key theory or the induced fit theory. Either way, the enzymes manipulate the substrates near to the amino acids that can transfer protons (hydrogen ions). This is how it speeds up the reaction.

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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l

Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law: $PV=nRT$ where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units: $R=0.0821\frac{L\cdot atm}{mol\cdot K}$

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means $T=273.15K$ and $P=1atm$

Lastly, we must calculate the number of moles of $O_2(g)$ there are. Given 0.80g of $O_2(g)$ , we will need to convert with the molar mass of $O_2(g)$ . Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: $32g\text{ }O_2=1mol\text{ }O_2$

Thus, $\frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n$

Isolating V in the Ideal Gas Law:

$PV=nRT$

$V=\frac{nRT}{P}$

...substituting the known values, and simplifying...

$V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}$

$V=0.56L \text{ } O_2$

So, 0.80g of $O_2(g)$ would occupy 0.56L at STP.

5 0

2 years ago

Read 2 more answers

What is formed by the hydrolysis of Starch?

Ymorist [56]

Answer:

soluble starch, maltose and various dextrins.

Explanation:

3 0

3 years ago

A megamole of titanium contains __________ moles of titanium.

Gala2k [10]

Answer is: 1 megamole is equal to 1000000 (one million) mole or 10⁶ moles.
The SI base unit for amount of substance is the mole. The SI prefix "mega" represents a factor of 10⁶, or in exponential notation, 1E6.
The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 12 grams of carbon C-12.

8 0

4 years ago

Read 2 more answers

Identify the most common reaction conditions for the dehydration of 2,3-dimethyl-2-pentanol.Select all that apply.A. The presenc

defon

Answer:

D. The presence of a base catalyst.

E. The presence of an acid catalyst. and

F. The presence of heat.

Explanation:

8 0

4 years ago

When 50.0 mL of 1.27 M of HCl(aq) is combined with 50.0 mL of 1.32 M of NaOH(aq) in a coffee-cup calorimeter, the temperature of

sergij07 [2.7K]

Answer:

-55.9kJ/mol is the change in enthalpy of the reaction

Explanation:

In the reaction:

HCl(aq) + NaOH(aq) → H₂O(l) + NaCl

Some heat is released per mole of reaction.

To know how many moles reacts we need to find limiting reactant:

Moles HCl = 0.050L ₓ (1.27mol / L) = 0.0635 moles HCl

Moles NaOH = 0.050L ₓ (1.32mol / L) = 0.066 moles NaOH

As there are more moles of NaOH than moles of HCl, HCl is limiting reactant and moles of reaction are moles of limiting reactant, 0.0635 moles

Using the coffee-cup calorimeter equation we can find how many heat was released thus:

Q = C×m×ΔT

Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)

Replacing:

Q = 4.18J/g°C×100g×8.49°C

Q = 3548.8J of heat are released in the reaction

Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:

ΔH = Heat / mol of reaction

ΔH = -3548.8J / 0.0635 moles of reaction

Negative because is released heat.

ΔH = -55887J / mol

ΔH =

<h3>-55.9kJ/mol is the change in enthalpy of the reaction</h3>

3 0

3 years ago