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navik [9.2K]
2 years ago
14

Calcium ions, barium ions, and magnesium ions all have two positive charges. What could explain the differences in the way each

reacted with a sodium hydroxide solution?
Chemistry
2 answers:
STALIN [3.7K]2 years ago
8 0

Answer: check explanation.

Explanation:

The three elements/metals, that is Calcium, Barium and Magnesium all belongs to group 2A on the periodic table. Other elements/metals of group 2A in the periodic table are; Beryllium, Strontium and Radium.

As one go down the group, the atomic radius increases from Magnesium to Barium(this is because of the increase in number of shells of electrons). And, as one go down the group the first ionization energy decreases.Because of this decease in ionization energy it makes it easier for the valence electrons to be removed and thus, REACTIVITY INCREASES DOWN THE GROUP.

Ca^2+ + 2OH^- --------> Ca(OH)2.

Mg^2+ + 2OH^- ---------> Mg(OH)2.

PS: The Na^+ is a spectator ion.

BARSIC [14]2 years ago
5 0

C3.

Tes

Example 8

Name each ionic compound.

CaCl2

AlF3

Co2O3

Solution

Using the names of the ions, this ionic compound is named calcium chloride. It is not calcium(II) chloride because calcium forms only one cation when it forms an ion, and it has a characteristic charge of 2+.

The name of this ionic compound is aluminum fluoride.

We know that cobalt can have more than one possible charge; we just need to determine what it is. Oxide always has a 2− charge, so with three oxide ions, we have a total negative charge of 6−. This means that the two cobalt ions have to contribute 6+, which for two cobalt ions means that each one is 3+. Therefore, the proper name for this ionic compound is cobalt(III) oxide.

Test Yourself

Name each ionic compound.

Sc2O3

AgCl

Answers

scandium oxide

silver chloride

How do you know whether a formula—and by extension, a name—is for a molecular compound or for an ionic compound? Molecular compounds form between nonmetals and nonmetals, while ionic compounds form between metals and nonmetals. The periodic table (Figure 3.2 “A Simple Periodic Table”) can be used to determine which elements are metals and nonmetals.

There also exists a group of ions that contain more than one atom. These are called polyatomic ions. Table 3.7 “Common Polyatomic Ions” lists the formulas, charges, and names of some common polyatomic ions. Only one of them, the ammonium ion, is a cation; the rest are anions. Most of them also contain oxygen atoms, so sometimes they are referred to as oxyanions. Some of them, such as nitrate and nitrite, and sulfate and sulfite, have very similar formulas and names, so care must be taken to get the formulas and names correct. Note that the -ite polyatomic ion has one less oxygen atom in its formula than the -ate ion but with the same ionic charge.

Table 3.7 Common Polyatomic Ions

Name Formula and Charge  Name Formula and Charge

ammonium NH4+  hydroxide OH−

acetate C2H3O2−, or CH3COO− nitrate NO3−

bicarbonate (hydrogen carbonate) HCO3− nitrite NO2−

bisulfate (hydrogen sulfate) HSO4− peroxide O22−

carbonate CO32− perchlorate ClO4−

chlorate ClO3− phosphate PO43−

chromate CrO42− sulfate SO42−

cyanide CN− sulfite SO32−

dichromate Cr2O72− triiodide I3−

The naming of ionic compounds that contain polyatomic ions follows the same rules as the naming for other ionic compounds: simply combine the name of the cation and the name of the anion. Do not use numerical prefixes in the name if there is more than one polyatomic ion; the only exception to this is if the name of the ion itself contains a numerical prefix, such as dichromate or triiodide.

Writing the formulas of ionic compounds has one important difference. If more than one polyatomic ion is needed to balance the overall charge in the formula, enclose the formula of the polyatomic ion in parentheses and write the proper numerical subscript to the right and outside the parentheses. Thus, the formula between calcium ions, Ca2+, and nitrate ions, NO3−, is properly written Ca(NO3)2, not CaNO32 or CaN2O6. Use parentheses where required. The name of this ionic compound is simply calcium nitrate. Write the proper formula and give the proper name for each ionic compound formed between the two listed ions. cause the ammonium ion has a 1+ charge and the sulfide ion has a 2− charge, we need two ammonium ions to balance the charge on a single sulfide ion. Enclosing the formula for the ammonium ion in parentheses, we have (NH4)2S. The compound’s name is ammonium sulfide.

Because the ions have the same magnitude of charge, we need only one of each to balance the charges. The formula is AlPO4, and the name of the compound is aluminum phosphate.

Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6. To get 6+, we need three iron(II) ions, and to get 6−, we need two phosphate ions. The proper formula is Fe3(PO4)2, and the compound’s name is iron(II) phosphate.

Test Yourself

Write the proper formula and give the proper name for e

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When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
2 years ago
A sample of sulfur hexafluoride gas occupies a volume of 9.10 L at 198ÁC. Assuming that the pressure remains constant, what temp
alexandr1967 [171]
  As we know that
<span>V1/T1 = V2/T2 
V1 = 9.10 L 
T1 = 471 K 
V2 = 2.50 L 
T2 = 2.5 x 471 / 9.10 = 129.3 K 
T2 = 129.3 - 273 =
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hope it helps</span>
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3 years ago
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Answer:

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Explanation:

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It is non-renewable - once it’s gone it’s gone and cannot be easily replaced.

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