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dezoksy [38]
3 years ago
8

Which example indicates that a chemical change has occurred? *

Chemistry
1 answer:
andrezito [222]3 years ago
7 0

Answer:

A. When two aqueous solutions are mixed, a precipitate is formed.

Explanation:

The precipitate (a solid substance that falls from the liquid) is the result of a chemical reaction taking place between the liquids.

The other three answer choices are indicative of physical changes (temperature change, phase change, color change).

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Make a poem about compounds​
OleMash [197]

Answer:

<h3>A Werewolf</h3>

___________

Poem with Compound Words

------------------------------------------

whenever there's a full moon,

I cannot overlook,

some alterations in my ways,

and changes in my look.

my werewolf hair grows everywhere,

my werewolf teeth get long,

my eyesight gets much keener and

I'm muscular and strong.

I get to roam around outside,

the moonlight makes me howl,

these otherworldly sound effects

mean I am on the prawl.

I see the moon is round and full,

I'm moonstruck by the sight,

I've made some telltale changes-so, you'd best stay in tonight.

Explanation:

I HOPE IT HELPS YOUR ENGLISH SUBJECT ;) ★

4 0
3 years ago
Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose mol
Sonja [21]

Answer: The given statement is true.

Explanation:

When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.

Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.

Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.

8 0
3 years ago
What is the half-life of 20 g of a radioactive sample if 5 g remain after 8 minutes?
Anni [7]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 20 g

m (final mass after time T) = 5 g

x (number of periods elapsed) = ?

P (Half-life) = ? (in minutes)

T (Elapsed time for sample reduction) = 8 minutes

Let's find the number of periods elapsed (x), let us see:

m =  \dfrac{m_o}{2^x}

5 =  \dfrac{20}{2^x}

2^x = \dfrac{20}{5}

2^x = 4

2^x = 2^2

\boxed{x = 2}


Now, let's find the half-life (P) of the radioactive sample, let's see:

T = x*P

8 = 2*P

2\:P = 8

P = \dfrac{8}{2}

\boxed{\boxed{P = 4\:minutes}}\Longleftarrow(Half-Life)\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

5 0
3 years ago
The concentration of dye in Solution A is 26.609 M. You have 13 mL of water at your disposal to make the dilutions. The solution
exis [7]

Answer:

6,613 M

Explanation:

Dilution is the process of reducing the concentration of a solute in solution, mixing initial solution with more solvent.

The concentration of Solution B is:

23,881 M ×  = 9,552 M

Because the initial eight parts are diluted to 12+8 parts.

Thus, concentration of solution C is:

9,552 M ×  = 6,613 M

I hope it helps!

8 0
3 years ago
The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles
sergejj [24]

<u>Answer:</u> The expression for equilibrium constant is K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

For the general chemical equation:

aA+bB\rightleftharpoons cC+dD

The expression for K_c is given as:

K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}

For the given chemical reaction:

2HgO(s)+H_2O(l)+2Cl_2(g)\rightleftharpoons 2HOCl(aq.)+HgO.HgCl_2(s)

The expression for K_{eq} is given as:

K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}

The concentration of solid is taken to be 0.

So, the expression for K_{eq} is given as:

K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}

3 0
3 years ago
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