What changes within the star A lead to the red giant B?

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

What is the atomic number of carbon

rosijanka [135]

Carbon (from Latin: carbo "coal") is a chemical element with the symbol C and atomic number 6. It is nonmetallic and tetravalent—making four electrons available to form covalent chemical bonds. It belongs to group 14 of the periodic table.

8 0

2 years ago

Calculate the mass of CaCl2•2H2O required to make 100.0 mL of a 0.100 M solution. Each of the calculations below will take you t

Zolol [24]

Answer:

The mass is 1.4701 grams and the moles is 0.01.

Explanation:

Based on the given question, the volume of the solution is 100 ml or 0.1 L and the molarity of the solution is 0.100 M. The moles of the solute (in the given case calcium chloride dihydride (CaCl2. H2O) can be determined by using the formula,

Molarity = moles of solute/volume of solution in liters

Now putting the values we get,

0.100 = moles of solute/0.1000

Moles of solute = 0.100 * 0.1000

= 0.01 moles

The mass of CaCl2.2H2O can be determined by using the formula,

Moles = mass/molar mass

The molar mass of CaCl2.2H2O is 147.01 gram per mole. Now putting the values we get,

0.01 = mass / 147.01

Mass = 147.01 * 0.01

= 1.4701 grams.

4 0

2 years ago

Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a dist

strojnjashka [21]

Answer:

Force of attraction = 35.96 $\times 10^{27}$ N

Explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm = $10^{-9}$ m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

$F = \dfrac{k \times q_1 q_2}{ \r^2}$ ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k = $8.99 \times 10^9 \ Nm^{2}C^{-2}$

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

$F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }$

Force of attraction = 35.96 $\times 10^{27}$ N

6 0

2 years ago

Chalcopyrite is an ore with the composition cufes2. what is the percentage of iron in a 39.6 g sample of this ore? answer in uni

iren2701 [21]

<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass. For this, we look at the atomic masses of the elements present in the compound. Cu has an atomic mass of 63.546 amu Fe has 55.845 amu and S has 36.065 amu Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu So we have not established the mass of the compound in amus 63.546 + 55.845 + 72.13 = 191.521 That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845 So to get the percentage, or fraction of iron, we take 55.845 / 191.521 Which comes out to 29.15% by mass Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>

5 0

3 years ago