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nata0808 [166]
3 years ago
7

Reaction rates are affected by how often the particles collide true false

Chemistry
2 answers:
elena55 [62]3 years ago
7 0
True..........................
SIZIF [17.4K]3 years ago
4 0
This is true.

The reaction rates are affected by how often the particles collide.
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Mixing salt with water is an example of a _________change. *<br> physical change<br> chemical change
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Answer:

physical

Explanation:

no chemical reaction is happening

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3 years ago
Select the correct answer from each drop-down menu. At chemical equilibrium, the amount of because .
Elena-2011 [213]

Answer:

The answer that completes the question are in BOLD:

At chemical equilibrium, the amount of PRODUCT AND REACTANT REMAIN CONSTANT because the RATES OF THE FORWARD AND REVERSE REACTIONS ARE EQUAL.

Explanation:

In a reversible chemical reaction, an equilibrium is said to be achieved when the rates of the forward reaction is equal to that of the reverse reaction. A reversible reaction is one in which products are formed from reactants simultaneously with the formation of reactants from products.

The combination of two or more substances called REACTANTS gives rise to another substance called PRODUCT, which can in turn give rise to Reactants again. With time, the rate at which the reactants give rise the products, which is called the FORWARD REACTION will be equal to the rate at which the products give rise to the reactants, which is called REVERSE REACTION. At this point, the chemical reaction is said to be in a STATE OF EQUILIBRIUM.

When the rate at which both reaction occurs becomes equal i.e. at an equilibrium state, the concentration of both the reactants and the products becomes constant i.e. no longer changes. Hence, the amount of the reactants forming the products is the same as the amount of products forming the reactants.

N.B: At chemical equilibrium, the amount of the reactants and products does not necessarily equals zero (0). It simply means that there is no net change in the concentration/amount of both reactants and products.

3 0
3 years ago
In this prelab exercise, you will calculate how to prepare solutions that you will make in lab. Below is an example of how to pr
Anastasy [175]

Explanation:

Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution (L)}}

Moles of compound =

\frac{\text{mass of compound}}{\text{Molar mass of compound}}

We have ;

Volume of solution = 600 mL = 0.600 L ( 1 mL = 0.001 L)

Moles of NaOH = n

Molarity of the solution = 3 M

3M=\frac{n}{0.600 L}

n = 3 M × 0.600 L = 1.800 mol

Mass of 1.800 mole sof NaOH :

1.800 mol × 40 g/mol = 72.0 g

Preparation:

Weight 72.0 grams of sodium hydroxide and add it to the 500 mL of volumetric flask along with some water. Dissolve the all the solute by adding small proportion of water. After the solution becomes clear make the water upto the mark of 500 ml.

Transfer the solution to a bigger beaker  and 100 mL of water more to it.

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