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AlladinOne [14]
3 years ago
11

Please help me dont just use me for points

Chemistry
1 answer:
JulijaS [17]3 years ago
4 0

Answer:

It's B

Explanation:

A is going to evaporate more quickly causing it to push through the tube to beaker B.

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Na2o + h2so4 ———&gt; na2so4 + h2o<br><br><br> Can anyone balance this please!!??
marysya [2.9K]

Answer:

The equation is already balanced. There's an equal number of materials on each side of the equation.

4 0
3 years ago
5.0 gram of anhydrous calcium carbonate is reacted with an excess of dilute nitric acid. if the reaction is carried out at 25 de
Aneli [31]

Answer:

I don’t know

Explanation:

6 0
3 years ago
If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha
NARA [144]
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
3 0
3 years ago
Read 2 more answers
A saturated solution of baso4 has a concentration of 0.5mol/l. a 55ml sample is taken by you. what is the mass of baso4 in the s
SIZIF [17.4K]

Answer:

6.4 g BaSO₄

Explanation:

You have been given the molarity and the volume of the solution. To find the mass of the solution, you need to (1) find the moles BaSO₄ (via the molarity ratio) and then (2) convert moles BaSO₄ to grams BaSO₄ (via the molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given values.

Molarity (mol/L) = moles / volume (L)

(Step 1)

55 mL / 1,000 = 0.055 L

Molarity = moles / volume                             <----- Molarity ratio

0.5 (mol/L) = moles / 0.055 L                        <----- Insert values

0.0275 = moles                                             <----- Multiply both sides by 0.055

(Step 2)

Molar Mass (BaSO₄): 137.33 g/mol + 32.065 g/mol + 4(15.998 g/mol)

Molar Mass (BaSO₄): 233.387 g/mol

0.0275 moles BaSO₄          233.387 g
---------------------------------  x  -------------------  =  6.4 g BaSO₄
                                                1 mole

6 0
2 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
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