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Neko [114]
3 years ago
13

Which observation does not indicate that a chemical reaction has occurred?

Chemistry
1 answer:
castortr0y [4]3 years ago
6 0

Answer:

change in the total mass of substances

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What is the molarity of a solution that contains 122g of MgSO4 n 3.5L of solution?​
Semmy [17]

Answer:

0.29mol/L or 0.29moldm⁻³

Explanation:

Given parameters:

Mass of MgSO₄ = 122g

Volume of solution = 3.5L

Molarity is simply the concentration of substances in a solution.

Molarity = number of moles/ Volume

>>>>To calculate the Molarity of MgSO₄ we find the number of moles using the mass of MgSO₄ given.

Number of moles = mass/ molar mass

Molar mass of MgSO₄:

Atomic masses: Mg = 24g

S = 32g

O = 16g

Molar mass of MgSO₄ = [24 + 32 + (16x4)]g/mol

= (24 + 32 + 64)g/mol

= 120g/mol

Number of moles = 122/120 = 1.02mol

>>>> From the given number of moles we can evaluate the Molarity using this equation:

Molarity = number of moles/ Volume

Molarity of MgSO₄ = 1.02mol/3.5L

= 0.29mol/L

IL = 1dm³

The Molarity of MgSO₄ = 0.29moldm⁻³

8 0
3 years ago
Explain why we experience seasons. *
REY [17]

Answer:

Seasons occur because Earth is tilted on its axis relative to the orbital plane, the invisible, flat disc where most objects in the solar system orbit the sun. ... In June, when the Northern Hemisphere is tilted toward the sun, the sun's rays hit it for a greater part of the day than in winter.

4 0
3 years ago
Read 2 more answers
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
3 years ago
The symbol for the element in period 2, group 13
denis-greek [22]
Element: Boron
Symbol: B
So answer is B
8 0
2 years ago
Someone please help!!!
sveticcg [70]

Answer: The last electron will be filled in first orbital of 3p sub-shell.

Explanation: Filling of electrons in orbitals is done by using Hund's Rule.

Hund's rule states that the electron will be singly occupied in the orbital of the sub-shell before any orbital is doubly occupied.

For filling up of the electrons in Sulfur atom having 16 electrons. First 10 electrons will completely fill according to Aufbau's Rule in 1s, 2s and 2p sub-shells and last 6 electrons are the valence electrons which will be filled in the order of 3s and then 3p.

3s sub-shell will be fully filled and the orbitals of 3p sub-shell will be first singly occupied and then pairing will take place. Hence, the last electron will be filled in the first orbital of 3p-sub-shell.

5 0
3 years ago
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