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3 years ago

5

What is the difference between a mixture, a solution, and a pure substance?

2 answers:

konstantin123 [22]3 years ago

5 0

A mixture is a material made up of two or more different Chemical substance/substances which are not chemically combined, a solution is a homogenous mixture of two or more substances in relative amounts that can be varied continuously up to what is called the limit of solubility, and a pure substance is a substance that has a fixed chemical composition

vichka [17]3 years ago

4 0

A mixture is a compound of two or more substances a pure substance is only made of one kind of atom and a solution is composed of a liquid substance made of two or more types of atoms

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Which of these is the final result of secondadry succession?

Grace [21]

Answer:

a

Explanation:

ADAPTATIOnn but if thhere would be an option o all the above it would be that

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3 years ago

Read 2 more answers

Which letter represents the activated complex?<br> 1. A<br> 2. B<br> 3. F<br> 4. G

lana66690 [7]

Answer:

2. B.

Explanation:

The letter B represents the activated complex.

The activated complex is the intermediate that is formed between the states of reactants "F" and products "G".

Letter A represents the activation energy of the reaction, that is the difference in potential energy between the reactants "F" and activated complex "B".

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3 years ago

A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati

prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

$k=\frac{2.303}{48.0}\log\frac{53500}{42520}$

$k=9.98\times 10^{-2}hr^{-1}$

Now we have to calculate the half-life.

$k=\frac{0.693}{t_{1/2}}$

$9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=6.94hr$

Therefore, the half life of 28-Mg in hours is, 6.94

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3 years ago

What is △n for the following equation in relating Kc to Kp?

Nimfa-mama [501]

Answer:

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Explanation:

The relation between Kp and Kc is given below:

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Where,

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Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}$

<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1 </u>

<u></u>

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3 years ago

Why the size of Na+ is smaller then Na <br><br>

Rufina [12.5K]

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3 years ago