When the crystals of potassium permanganate are preserved in water, the purple-coloured crystals of potassium permanganate break further into smaller particles that populate the distance between the molecules of water imparting a purple colour to the water. This is an example of diffusion.

<h3>What are the two conclusions given out in the method of diffusion?</h3>

Diffusion is the process of movement of solvent from higher concentration to lower concentration through a semipermeable membrane. So, we can form the decision that it cannot occur through a thick membrane from which small molecules cannot pass through.

<h3>What is difference between osmosis and diffusion?</h3>

Osmosis is the direction of solvent particles from a solution that is diluted to a more concentrated one. In contrast, diffusion is the movement of particles from a higher concentration region to a part of lower concentration.

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6 0

2 years ago

why can an increase in temperature lead to more effective collisions between reactant particles and an increase in the rate of a

Tasya [4]

The reason why is because particles are moving faster and possess greater kinetic energy due to the increase in temperature, this causes greater frequency or number of collisions to take place and also allows for a higher likelihood that upon reactant molecules colliding they will undergo the reaction process as they will have achieved the required energy needed for the reaction to start. Thus overall having a greater change in quantity conversion of reactants to products within a shorter period of time, thus increasing the reaction rate.

5 0

3 years ago

) B5H9(l) is a colorless liquid that will explode when exposed to oxygen. How much heat is released when 0.211 mol of B5H9 react

Tom [10]

<u>Answer:</u> The amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

<u>Explanation:</u>

The chemical equation for the reaction of $B_5H_9$ with oxygen gas follows:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.4kJ/mol\\\Delta H_f_{(B_2O_3(s))}=-1272kJ/mol\\\Delta H_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ$

To calculate the amount of heat released for the given amount of $B_5H_9(l)$ , we use unitary method, we get:

When 2 moles of $B_5H_9(l)$ reacts, the amount of heat released is 5259 kJ

So, when 0.211 moles of $B_5H_9(l)$ will react, the amount of heat released will be = $\frac{5259}{2}\times 0.211=554.8kJ$

Hence, the amount of heat released when 0.211 moles of $B_5H_9(l)$ reacts is 554.8 kJ

7 0

3 years ago