What is the function of the rest of the huge molecule?

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

2 years ago

Limiting line of bakmer series result when electron jumps from orbit: A) 3 to 2 B) infinity to 2 C) 4 to 5 D) infinity to 1

krok68 [10]

Bunkhouse. M jgzistdjfug

5 0

2 years ago

Read 2 more answers

How can you force a solid compound to the bottom of a melting point capillary? a. tap the closed endb. tap the open endc. use a

snow_lady [41]

Answer: option A. tap the closed end

Explanation:

3 0

2 years ago

Relative to protons, electrons have:

saw5 [17]

Answer:

(B.) a significantly lower mass that protons

Explanation:

The mass (in g) of an electron is 9.11 × $10^{-28}$ g, which is significantly lower than the mass of protons and neutrons which are approximately the same. They are extremely small (equal to 0 amu), contributing nothing to the overall mass of the atom.

The nucleus has a mass of 1 amu and is positively charged.

6 0

3 years ago

A student performed a chemical reaction in which 35 grams of hydrogen and 65 grams of oxygen reacted to form water. What is the

sergejj [24]

Answer:

The mass of the product is 73,06 g

Explanation:

The reaction to form water is:

2H2 + O2 ---> 2H2O

35 g 65 g

Molar mass H2 = 2g/m ---> I have 17,5 moles (35g / 2 g/m)

Molar mass O2 = 32g/m ---> I have 2,03 moles (65g / 32g/m)

2 moles of H2 __reacts with __ 1 mol O2

17,5 moles of H2 __ reacts with ___ (17,5 m . 1 m) / 2 m = 8,75 m

I have just 2,03 m of O2 and I need 8,75 m so O2 is my limiting reagent

1 mol of O2 ___ reacts with 2 moles H2

2,03 moles of O2 __ reacts with (2,03 m . 2 m) / 1 m = 4,06 m

I have 17,5 moles of H2 and I need 4,06 m. H2 is my excess reagent

"All operations are done with the limiting reagent"

1 mol O2 are required____ to form 2 H2O

2,03 mol O2 are required __ to form (2,03m . 2m) / 1m = 4,06 m

Molar mass of water: 18 g/m

Mass of water ----> Molar mass . Moles water = 18 g/m . 4,06 m = 73,08g

8 0

2 years ago