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ludmilkaskok [199]
3 years ago
9

A piston in a heat engine does 500 joules of work, and 1,400 joules of heat are added to the system. Determine the change in int

ernal energy and explain how this example demonstrates the conservation of energy.
Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
8 0
The change is that the piston gets hotter? (Theres more heat in it)
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Other than bacteria, which factor leads to nitrogen fixation?
alina1380 [7]
Lightning rain can lead to nitrogen fixation other than bacteria
4 0
3 years ago
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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

8 0
3 years ago
What is the reason for heat transfer from one substance to another?
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Hey man its difference in temperature
3 0
3 years ago
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If 0.200 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, what is the mass in grams o
morpeh [17]

Answer:

31.2 g of Ag₂SO₄

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AgNO₃(aq) + H₂SO₄ (aq) → Ag₂SO₄ (s) + 2HNO₃ (aq)

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄ to produce 1 mole of Ag₂SO₄ and 2 moles of HNO₃.

Next, we shall determine the limiting reactant.

This can obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted with 1 mole of H₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react with = (0.2 x 1)/2 = 0.1 mole of H₂SO₄.

From the calculations made above, only 0.1 mole out of 0.155 mole of H₂SO₄ given is needed to react completely with 0.2 mole of AgNO₃. Therefore, AgNO₃ is the limiting reactant.

Next,, we shall determine the number of mole of Ag₂SO₄ produced from the reaction.

In this case we shall use the limiting reactant because it will give the maximum yield of Ag₂SO₄ as all of it is consumed in the reaction.

The limiting reactant is AgNO₃ and the number of mole of Ag₂SO₄ produced can be obtained as follow:

From the balanced equation above,

2 moles of AgNO₃ reacted to produce 1 mole of Ag₂SO₄.

Therefore, 0.2 moles of AgNO₃ will react to produce = (0.2 x 1)/2 = 0.1 mole of Ag₂SO₄.

Therefore, 0.1 mole of Ag₂SO₄ is produced from the reaction.

Finally, we shall convert 0.1 mole of Ag₂SO₄ to grams.

This can be obtained as follow:

Molar mass of Ag₂SO₄ = (2x108) + 32 + (16x4) = 312 g/mol

Mole of Ag₂SO₄ = 0.1

Mass of Ag₂SO₄ =?

Mole = mass /Molar mass

0.1 = Mass of Ag₂SO₄ /312

Cross multiply

Mass of Ag₂SO₄ = 0.1 x 312

Mass of Ag₂SO₄ = 31.2 g

Therefore, 31.2 g of Ag₂SO₄ were obtained from the reaction.

5 0
3 years ago
How to convert 5.2 km to mm
ZanzabumX [31]
5.2 km to mm.
1km = 1000m
= 5.2 * 1000m.
=5200 m.                  1m = 1000mm

= 5200 * 1000mm
= 5200000 mm
4 0
3 years ago
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