Unlike the earth, the sun ____

write a balanced net ionic equation for the following reaction: BaCl2(aq) + H2SO4 (aq) -- BaSO4(s) + HCl (aq)

Fudgin [204]

The balanced net equation for

BaCl2 (aq) + H2SO4(aq) → BaSO4(s) + HCl (aq) is

Ba^2+(aq) +SO4^2- → BaSO4 (s)

Explanation

Ionic equation is a chemical equation in which electrolytes in aqueous solution are written as dissociated ions.

ionic equation is written using the below steps

Step 1: write a balanced molecular equation

BaCl2 (aq) +H2SO4 (aq)→ BaSO4(s) +2HCl (aq)

Step 2: Break all soluble electrolytes in to ions

= Ba^2+ (aq) + 2Cl^-(aq) + 2H^+(aq) + SO4^2-(aq)→ BaSO4(s) + 2H^+(aq) +2Cl^- (aq)

step 3: cancel the spectator ions in both side of equation ( ions which do not take place in the reaction)

 = 2Cl^- and 2H^+ ions

Step 4: write the final net equation

 Ba^2+(aq) + SO4^2-(aq)→ BaSO4(s)

3 0

3 years ago

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The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D

Westkost [7]

Answer:

a) $K_{2} S$ and $NH_{4} Cl$ :

There are no insoluble precipitate forms.

b) $Ca Cl_{2}$ and $(NH_{4} )_{2} Co_{3}$ :

There are the insoluble precipitates of $CaCo_{3}$ forms.

c) $Li_{2}S$ and $MnBr_{2}$ :

There are the insoluble precipitates of $MnS$ forms.

d) $Ba(No_{3} )_{2}$ and $Ag_{2} So_{4}$ :

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e) $Rb_{2}Co_{3}$ and $NaCl$ :

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- $K_{2} S$ ⇒ soluble, $NH_{4} Cl$ ⇒ soluble.

KCl ⇒ soluble, $(NH_{4})_{2} S$ ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- $Ca Cl_{2}$ ⇒ soluble, $(NH_{4} )_{2} Co_{3}$ ⇒ soluble.

$CaCo_{3}$ ⇒ insoluble, $NH_{4} Cl$ ⇒ soluble.

There are the insoluble precipitates of $CaCo_{3}$ forms.

c)

Solubility rule suggests:- $Li_{2}S$ ⇒ soluble, $MnBr_{2}$ ⇒ soluble.

$LiBr$ ⇒ soluble, $MnS$ ⇒ insoluble.

There are the insoluble precipitates of $MnS$ forms.

d)

Solubility rule suggests:- $Ba(No_{3} )_{2}$ ⇒ soluble, $Ag_{2} So_{4}$ ⇒insoluble.

As $Ag_{2} So_{4}$ is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- $Rb_{2}Co_{3}$ ⇒ soluble, $NaCl$ ⇒ soluble.

$RbCl$ ⇒ soluble, $Na_{2} Co_{3}$ ⇒ soluble.

There are no insoluble precipitates forms.

6 0

3 years ago

A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A

anygoal [31]

this is a dilution equation where 50.0 mL of 1.50 M H₂SO₄ is taken and added to 200 mL of water.

c1v1 = c2v2

where c1 is concentration and v1 is volume of the concentrated solution

and c2 is concentration and v2 is volume of the diluted solution to be prepared

50.0 mL of 1.50 M H₂SO₄ is added to 200 mL of water so the final solution volume is - 200 + 50.0 = 250 mL

substituting these values in the formula

1.50 M x 50.0 mL = C x 250 mL

C = 0.300 M

concentration of the final solution is A) 0.300 M

4 0

3 years ago

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You are given two aqueous solutions with different ionic solutes (Solution A and Solution B). What if you are told that Solution

klio [65]

Answer:

Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.

It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,

% by mass = mass of solute/mass of solution * 100

Now the formula for molality is,

Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams

Now molality of solution A is,

m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)

m = 2.07

Now the molality of solution B is,

m = 15/58.5 * 1000/85

m = 3.02

Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).

5 0

3 years ago

Pls assist if you can. Reasons why hydrogen was chosen as a standard reference to other elements

MrMuchimi

Answer:

See below

Step-by-step explanation:

Hydrogen either reacts with or is formed by reactions with many other elements, so chemists could use it directly to determine their relative masses.
Hydrogen has the smallest atomic mass, so it was convenient to give H a relative atomic mass of 1 and assign those of other elements as multiples of this number.

The O = 16 scale became the standard in 1903 and carbon-12 was chosen in 1961.

4 0

3 years ago

Unlike the earth, the sun _____.