All categories

2 years ago

I AM GIVING BRAINLIEST PLEASEE HELPPPP I NEED HELPPPPPP PLEAEEEEEE

Chemistry

2 answers:

Helga [31]2 years ago

4 0

Answer:

Interestingly enough, I'm not getting

0.0341% w/v

either. Here's why.

Start by calculating the percent composition of chlorine,

, in calcium chloride, This will help you calculate the mass of chloride anions,

−

, present in your sample.

To do that, use the molar mass of calcium chloride, the molar mass of elemental chlorine, and the fact that

mole of calcium chloride contains

moles of chlorine atoms.

35.453

g mol

−

110.98

g mol

−

⋅

100

63.89% Cl

This means that for every

100 g

of calcium chloride, you get

63.89 g

of chlorine.

As you know, the mass of an ion is approximately equal to the mass of the neutral atom, so you can say that for every

100 g

of calcium chloride, you get

63.89 g

of chloride anions,

−

This implies that your sample contains

0.543

g CaCl

⋅

63.89 g Cl

−

100

g CaCl

0.3469 g Cl

−

Now, in order to find the mass by volume percent concentration of chloride anions in the resulting solution, you must determine the mass of chloride anions present in

100 mL

of this solution.

Since you know that

500 mL

of solution contain

0.3469 g

of chloride anions, you can say that

100 mL

of solution will contain

100

mL solution

⋅

0.3469 g Cl

−

500

mL solution

0.06938 g Cl

−

Therefore, you can say that the mass by volume percent concentration of chloride anions will be

% m/v = 0.069% Cl

−

−−−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the volume of the solution.

Explanation:

True [87]2 years ago

3 0

Answer:

0.479 M or mol/L

Explanation:

So Molarity is moles/litres of solution...often written as M=mol/L

So here we are given grams of BaCl2 which we have to convert to moles. To convert to moles of BaCl2 we have to divide 63.2 g BaCl2 by molar mass of BaCl2 which is 208.23 g/mol so you get 63.2/208.23 = 0.3035 moles of BaCl2

Second step is converting the 634mL to litres by simply dividing by 1000 because we know 1 litre has 1000ml so 634/1000 = 0.634L

Now we just plug these guys in our molarity formula M=mol/L

M= 0.3035/0.634 = 0.479 M or mol/L

You might be interested in

12. Which compound can act as both a BrønstedLowry

oksian1 [2.3K]

1) The compound which can act as a Bronsted-Lowry acid and a Bronsted-Lowry base is definitely water - H2O. Remember that water is amphoteric which means it can 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 : HCl+H20=H30+Cl ; NH3+H2O=NH4+OH

2) The acids in this equilibrium reaction CN– + H2O HCN + OH. Acid species always donate 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 HCN that donates an H+ to OH that shows that it is an acid.

3) 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+.

4) The type of 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. pH lower than 7 are acidic, and pH higher than 7 basic ones.

5) 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 Kw only when dealing with Kb, and we can conclude that the hydrolysis constant of the conujugate base.

6) 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 partial ionization which is a direct characteristic of a weak acid.

7) To solve this task we should appeal to Henry's law that says the solubility of a gaz is proportional to its partial pressure. And according to this we can understand that 202kPa is the half of 404kPa which means that the needed solubility must be divided by 2 7.5/2=3.75 g/L and that's all.

8) I think that the most important points which best show how the addition of a solute affects the boiling point, the freezing
point, and the vapor pressure of the solvent are : BOILING: additional 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 solvent shells are being formed the solute particles reduces the number of solvent particles that have sufficient KE to vaporize.

9) $[H+][OH-]= Kw = 1.0 * 10^-14$
$[H+]= Kw/ [OH-]= 1.0x 10^-14 / 1 x 10^-11 =1 x 10^-3 mol/L 

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 : HA<=> H+ + A- .
ka = [H][A] / [HA].
Now we have only [H+] and to go further you need to write electroneutrality equation for the reaction :
[H+] = [OH-] + [A-] (since [H]>>>[OH]), then
[H+] = [A-]
Then mass balance equation :
Ct = 0.5M = [A-] + [HA]
[HA] = 0.5 - [A-] = 0.5 - [H+]
Finally here is what we have done and get :
$ka = [H]^2 / (0.5 - [H+]) 
$
$ka = 0.0001*0.0001/(0.5-0.0001) = 2.00x10^-8$

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 Increased hydrogen ions (H+). The facts about bases : they reduce the concentration of hydgoren ions in a solution which is opposite to asids,- Ex. Antiacid,and Ammonia ,Ph valuse above 7, they form hydrogen (OH-).
- The common points of both acids and bases : Hydrogen ions ,
both not neutral and water based.

5 0

2 years ago

Given the balanced equation:

Svet_ta [14]

Is there any answer choices before i get started on working out the problem

8 0

2 years ago

Plzz help i will give brillants and stars and thx.

dalvyx [7]

Hdhdjejdhebdehxhshwhdb d nonsense

3 0

3 years ago

Read 2 more answers

What can you say about the strength of the intermolecular forces in neon and argon based on the critical points of Ne and Ar? (c

algol13

Explanation:

Since, it is given that critical temperature of Argon is 150.9 K and critical pressure of Argon is 48.0 atm.

It is known that gas phase of neon occurs at 50 K. As the boiling point of Ar is more than the boiling point of neon which means that there is strong intermolecular force of attraction between argon molecules as compared to neon molecules.

This is also because argon is larger in size. As a result, induced dipole-induced dipole forces leads to more strength in Ar as compared to Ne.

8 0

2 years ago

An impurity sometimes found in Ca₃(PO₄)₂ is Fe₂O₃, which is removed during the production of phosphorus as ferrophosphorus (Fe₂P

Citrus2011 [14]

This impurity is troubling from an economic standpoint because it lead to decrease in the yield of phosphorus

Ferrophosphorus is a byproduct of phosphorus production in submerged-arc furnaces , by their reduction with carbon. It is formed from the iron oxide impurities.

Iron impurities present in the calcium phosphate will be precipitated out as the iron phosphate which eventually will lead to the decrease in the yield of phosphorous during the production of phosphorous.

Thus we can conclude that Fe₂P causes decrease in yield

Learn more about production of phosphorus at brainly.com/question/13337198

#SPJ4

6 0

1 year ago