Theory is often used as a synonym for idea in everyday life.

How does the law of conservation of mass relate to the number of atoms of each element that are present before a reaction vs. th

TiliK225 [7]

The law of conservation of mass or principle of mass conservation states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time, as system's mass cannot change, so quantity cannot be added nor removed. Hence, the quantity of mass is conserved over time.

The law implies that mass can neither be created nor destroyed, although it may be rearranged in space, or the entities associated with it may be changed in form. For example, in chemical reactions, the mass of the chemical components before the reaction is equal to the mass of the components after the reaction. Thus, during any chemical reaction and low-energy thermodynamic processes in an isolated system, the total mass of the reactants, or starting materials, must be equal to the mass of the products.

According to the Law of Conservation, all atoms of the reactant(s) must equal the atoms of the product(s).
As a result, we need to balance chemical equations. We do this by adding in coefficients to the reactants and/or products. The compound(s) itself/themselves DOES NOT CHANGE.

6 0

2 years ago

At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubil

8090 [49]

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

$\frac{S_{1}}{P_{1}}=\frac{S_{2}}{P_{2}}$

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

$\frac{0.725}{3.08}=\frac{S_{2}}{8}$

On rearranging,

$S_{2}=\frac{0.725\times 8}{3.08}=1.88$

Therefore, solubility will be 1.88 mg of $N_{2}$ gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.

3 0

2 years ago

Why are the oxidation and reduction half-reactions separated in an electrochemical cell?

Aleksandr [31]

Answer:

<h2>It makes the current viable enough to pass through an exterior wire.</h2>

Explanation:

Electrochemical cells primarily comprise of two half-cells. These half-cells assist in isolating the oxidation and reduction half-reactions. These two reactions are linked by a wire which allows the current to move from one edge to the other. The oxidation at the anode and the reduction take place at the cathode and the addition of a salt bridge helps in completing the circuit and permits the current to flow and leads to the generation of electricity.

7 0

2 years ago

At the end of mitosis the cells produced are identical to each other is it True or False

My name is Ann [436]

True:

The result is two genetically identical daughter nuclei.

6 0

2 years ago

Which of the following are true statements about equilibrium systems? For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(

Grace [21]

Answer:

The first, third and fourth statements are correct.

Explanation:

1) For the following reaction at equilibrium: CaCO3(s) ⇌ CaO(s) + CO2(g) adding more CaCO3 will shift the equilibrium to the right.

⇒ Le Chatellier says As the CaCO3 concentration is increased, the system will attempt to undo that concentration change by shifting the balance to the right. This statement is true.

2) For the following reaction at equilibrium: CaCO3(s)⇌ CaO(s) + CO2(g) increasing the total pressure by adding Ar(g) will shift the equilibrium to the right.

⇒ Le chatellier says that if we increase the pressure, the equilibrium will shift to the side with the least number of particles.

Since the molar densities of CaO and CaCO3 are constant, they don't appear in the equilibrium expression. This is why only changes to the pressure (concentration) of CO2 affect the position of the equilibrium.

If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of CO2, CaO or CaCO3. The added gas won't affect the partial pressure of CO2. This statement is false.

3)For the following reaction at equilibrium: 2 H2(g) + O2(g) ⇌ 2 H2O(g) the equilibrium will shift to the left if the volume is doubled.

⇒ Le Chatellier says if we increase the pressure, the equilibrium will shift to the side with the most particles.

In this case we have 2 moles of H2 and 1 mole of O2 on the left side and 2 mole of H2O on the right side. This means on the left side are more particles. So the equilibrium will shift to the left, so this statement is true.

4) For the following reaction at equilibrium: H2(g) + F2(g) ⇌ 2HF(g) removing H2 will increase the amount of F2 present once equilibrium is reestablished. Increasing the temperature of an endothermic reaction shifts the equilibrium position to the right.

⇒ Le chatellier says if H2 will be removed (this means the left side will get less particles) so the equilibrium will shift to the left, to increase the amount of F2.

⇒Le chatelier says if we increase the temperature of an exotherm reaction , there will be less energy released. The equilibrium will shift to the side of the reactants (the left side).

If we increase the temperature of an endotherm reaction, the equilibrium will shift to the side of the products (the right side). This statement is true.

4 0

2 years ago