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alex41 [277]
3 years ago
7

What product or ratio of properties remains constant (PxT, TxV, P/V, V/T, etx)?When you have decided, label the "constant?" colu

mn with proper term and fill in the values.
Chemistry
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

In this phenomenon we talk about ideal gases, that is why in these equations the constant is the number of moles and the constant R, which has a value of 0.082

Explanation:

The complete equation would have to be P x ​​V = n x R x T

where n is the number of moles, and if it is not clarified it is because they remain constant, as the question was worded.

On the other hand, the symbol R refers to the ideal gas constant, which declares that a gas behaves like an ideal gas during the reaction, and its value will always be the same, which is why it is called a constant. The value of R = 0.082.

The ideal gas model assumes that the volume of the molecule is zero and the particles do not interact with each other. Most real gases approach this constant within two significant figures, under pressure and temperature conditions sufficiently far from the liquefaction or sublimation point. The real gas equations of state are, in many cases, corrections to the previous one.

The universal constant of ideal gases is not a fundamental constant (therefore, choosing the temperature scale appropriately and using the number of particles, we can have R = 1, although this system of units is not very practical)

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The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

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3 years ago
Duncan takes a break from studying and goes to the gym to swim laps If swimming burns amount of time ? 85 * 10 ^ 5 cal per hour
Volgvan

When 6.85×10⁵ cal is converted to kilojoules, the result obtained is 2866.04 KJ

<h3>Data obtained from the question </h3>
  • Energy (cal) = 6.85×10⁵ cal
  • Energy (KJ) =?

<h3>Conversion scale </h3>

1 cal = 0.004184 KJ

<h3>How to convert 6.85×10⁵ cal to kilojoules</h3>

1 cal = 0.004184 KJ

Therefore,

6.85×10⁵ cal = 6.85×10⁵ × 0.004184

6.85×10⁵ cal = 2866.04 KJ

Thus, 6.85×10⁵ cal is equivalent to 2866.04 KJ

Learn more about conversion:

brainly.com/question/2139943

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