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Lelu [443]
3 years ago
11

When considering different hypotheses, usually the ____ one which can account for the ____ is the correct one. |

Chemistry
2 answers:
san4es73 [151]3 years ago
7 0

Answer:

Simplest, Information.

Explanation:

muminat3 years ago
4 0
I think the answer is theory
You might be interested in
B. How do you think the physical properties (strength, flexibility, and viscosity) of the polymer would change if more borate ha
Talja [164]

It became thicker and its viscosity decreased and cannot flow as easily as before.

You ignite a chemical reaction by adding the borax solution to the glue mixture.

In a chemical reaction, the molecules of glue and borax combine to form a flexible, springy new substance. With rubber's vulcanization serving as a model, chemical cross-linking has been extensively employed to change the physical properties of polymeric materials.

Chemical links between polymer chains provide a substance with a more solid structure and perhaps a better-defined shape. It thickened and lost viscosity, making it more difficult to flow than it once could.

Learn more about the chemical reaction  here brainly.com/question/16714866

#SPJ4.

3 0
1 year ago
You need to produce a buffer solution that has a pH of 5.31. You already have a solution that contains 10. mmol (millimoles) of
Karolina [17]

Answer:

37 mmol of acetate need to add to this solution.

Explanation:

Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-

pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]

Here pH is 5.31, pK_{a} (acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.

Plug in all the values in the above equation:

5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]

or, mmol of CH_{3}COO^{-} = 37

So 37 mmol of acetate need to add to this solution.

3 0
3 years ago
Examine the following equations.
Zigmanuir [339]

Answer:

A and B

Explanation:

This is because there was emission of gamma (Y) radiations in both the reactions.

8 0
3 years ago
12. Which compound can act as both a BrønstedLowry
oksian1 [2.3K]
1) The compound which can act as a <span>Bronsted-Lowry acid and a Bronsted-Lowry base is definitely water - H2O. Remember that water is </span><span>amphoteric which means it can </span>either accept protons or donate them, so it is the most proper option among other represented. Here are examples of both base and acid with water : <span>HCl+H20=H30+Cl ; </span><span>NH3+H2O=NH4+OH 

2) The </span><span>acids in this equilibrium reaction CN– + H2O HCN + OH. Acid species always donate </span> H+ to the species with which they react. In the second option you can see how H2O donates an H+ to CN-. If the reaction gets reversed we will obtain<span> HCN that donates an H+ to OH that shows that it is an acid.
</span><span>
3) </span>The products of self-ionization of water are OH⁻ and H₃O⁺. Self-ionization is an ionization reaction during which  H2O deprotonates its hydrogen atoms to become a hydroxide ion -- OH−. After this process OH-  protonates another water molecule forming H3O+<span>. 

4) The type of </span> solution which is one with a pH of 8 is acidic. Here is a little table that can be a prompt for you if you ever come across such tasks - ph : 7 is neutral<span>. </span>pH<span>  </span>lower than 7<span> are acidic, and </span>pH<span>  higher than </span>7<span> basic ones. 

5) </span><span>The acid dissociation constant for an acid dissolved in water is equal to the equilibrium constant. I consider this option correct because we can obtan </span><span>Kw only when dealing with Kb, and we can conclude that the hydrolysis constant of the conujugate base. 

6) </span>A 0.12 M solution of an acid that ionizes only slightly in solution would be termed dilute and weak. You can determine it depending on its concentration. Such value as 0.12M usually defined as a dilute solution of a weak acid due to the fact that acid represents its <span>partial ionization which is a direct characteristic of a weak acid.
 
7) To solve this task we should appeal to H</span>enry's law that says<span> the solubility of a gaz is proportional to its partial pressure. And according to this we can understand that </span>202kPa is the half of 404kPa which means that the needed solubility must be divided by <span>2 7.5/2=3.75 g/L and that's all.

8) I think that the most important points which best show </span><span>how the addition of a solute affects the boiling point, the freezing
point, and the vapor pressure of the solvent are : BOILING: a</span>dditional attractive forces can only exist between solute and solvent and in order to boil they must be overcome for the solution;we should add KE to overcome the forces. FREEZING : to freeze we have to withdrawn KE as the solute particles are surrounded by solvent molecules. VAPOR : WHen <span>solvent shells are being formed  the solute particles reduces the number of solvent particles that have sufficient KE to vaporize.</span>
<span>
9) </span>[H+][OH-]= Kw = 1.0 * 10^-14
[H+]= Kw/ [OH-]= 1.0x 10^-14 / 1 x 10^-11 =1 x 10^-3 mol/L &#10;&#10;pH = - log [H+]= - log 1 x 10^-3 = 3
Since we got Ph of 3 in a result we can define solution as an acidic one, as I mentioned before.

10) Since the formula of the given acid is HA it undergoes like that :<span> HA<=> H+ + A- .
</span><span>ka = [H][A] / [HA].
Now we have only </span>[H+] and to go further you need to write  <span>electroneutrality equation for the reaction :
</span>[H+] = [OH-] + [A-]  (since [H]>>>[OH]), then 
<span>[H+] = [A-] 
</span>Then mass balance equation : 
Ct = 0.5M = [A-] + [HA] 
<span>[HA] = 0.5 - [A-] = 0.5 - [H+] 
</span>Finally here is what we have done and get : 
ka = [H]^2 / (0.5 - [H+]) &#10;
ka = 0.0001*0.0001/(0.5-0.0001) = 2.00x10^-8

11) The main points that are common for acids : they form Hydrogen ions when dissloved in water, - Ex. Vinegar and Lemon, Ph >7, they have <span>Increased hydrogen ions (H+). The facts about bases : they r</span>educe the concentration of hydgoren ions in a solution which is opposite to asids,<span>- Ex. Antiacid,and Ammonia ,</span>Ph valuse above 7, they form hydrogen (OH-).
- The common points of both acids and bases : <span>Hydrogen ions ,</span>
 both not neutral and water based.
5 0
3 years ago
Lead(II) oxide from an ore can be reduced to elemental lead by heating in a furnace with carbon. Calculate the expected yield of
Lelu [443]

Answer:

53kg is the expected yield of lead

Explanation:

Firstly, in order to solve this question, we need to write the equation of reaction correctly. This is as follows:

PbO(s) + C(s) ---> Pb(l) + CO(g)

We proceed from here. We should get the limiting reactant but this can only be obtained by getting the number of moles of each reactant present.

The formula to use across all boards is that the number of moles is the mass of each of the reactant divided by the molar mass of each of the reactant.

For PBO, mass is 57kg = 57000g

Molar mass of PBO = 223.20g/mol

The number of moles is thus 57,000/223.2 = 255.37 moles

For carbon, mass is also 57kg = 57000g

Molar mass is 12g/mol

Number of moles of carbon = 57000/12 = 4750 moles

From the number of moles, we can see that the number of moles of Carbon is greater than that of PbO. This means that PbO is the limiting reagent.

Hence we use it to calculate percentage yield.

The number of moles of lead formed is the same of number of moles of lead oxide = 255.37 since we have mole ratio of 1 to 1

The molar mass of lead is 207.20g/mol

The mass of lead formed is = moles of lead formed * molar mass of lead = 207.20 * 255.37 = 52,912g which is approximately 53kg

Hence the expected yield is 53kg

6 0
3 years ago
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