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Snezhnost [94]
3 years ago
13

Characteristic of a chemical compound

Chemistry
1 answer:
Artist 52 [7]3 years ago
4 0

Answer:

Hello Adam Here!!!

Explanation:

A chemical compound has the following characteristics: (i) A chemical compound is obtained by the chemical combination of two or more elements in a definite proportion by mass. (ii) Compounds are homogeneous, i.e. their properties are the same throughout.

Happy to Help! =)

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Idk how to do this at all!!!
sergey [27]
Hello there!

I am not really sure which part you are talking about but let me answer some of them.

First, you need to read the information they gave you about glucose etc..

13- What is produced in photosynthesis?
"Photosynthesis is essential for all life on earth, because it provides food and oxygen."

Thus, the answer is: Photosynthesis produced food and oxygen.

14- What is the glucose used for?
"The glucose produced is used by the plant for energy and growth."

Thus, the answer is: The glucose used for energy and growth of the plants.

15- What is the oxygen used for?
"The oxygen produced is released into the air for us to breath."

Thus, the answer is: The oxygen used for us to breath.


I hope this helps!

Let me know if you have questions about the answer.

 
3 0
3 years ago
Give 1 example of a chemical reaction that is involved in coral<br> bleaching.
const2013 [10]
2HCO3 - + Ca2+ CaCO3 + CO2 + H2O Bicarbonate (HCO3-) combines with calcium ions in the water to make calcium carbonate (CaCO3, limestone). This process can occur both within organisms such as corals or as a simple chemical reaction in the water itself.
3 0
1 year ago
Metallic crystals are excellent conductors of electricity due to the existence of positive and negative ions.
My name is Ann [436]

the answer is FALSE


4 0
3 years ago
Read 2 more answers
A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but additio
zhannawk [14.2K]

Answer:

We can have: Calcium, strontium, or barium

Explanation:

In this case, we have to remember the solubility rules for sulfate SO_4~^-^2 and the chloride Cl^-:

<u>Sulfate</u>

All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.(Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2), which are NOT soluble.

<u>Chloride</u>

All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). (Pb^+^2~,Ag^+~,Hg_2~^+^2), which are NOT soluble.

If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I),  lead" and our possibilities are:

<u>"Calcium, strontium, or barium".</u>

I hope it helps!

7 0
3 years ago
Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
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