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Nataliya [291]
3 years ago
6

What is the chemical formula for a compound between Li and Br?

Chemistry
1 answer:
natima [27]3 years ago
5 0

LiBr.

<h3>Explanation</h3>

Note that the group number in this answer refers to the new IUPAC group number, which ranges from 1 to 18. Counts from the left. Start with the first two column (group 1 and 2), go on to the transition elements (Sc, Ti, etc. in group 3 through 12), and continue with the nonmetals (group 13 through 18).  

Li is a group 1 metal. As a metal, it tends to form positive ions ("cations"). Metals in group 1 and 2 are <em>main group</em> metals. The charge on main group metal ions tends to be the same as the group number of the metal. Li is in group 1. The charge on an Li ion will be +1. Formula of the Li ion will be \text{Li}^{+}.

Br is a group 17 nonmetal. As a nonmetal, it tends to form negative ions ("anions"). The charge on nonmetal ions excepting for H tends to equal the group number of the nonmetal minus 18. Br is in group 17. The charge on a Br ion will be 17 - 18 = -1. Formula of the Br ion will be \text{Br}^{-}

All the ions in an ionic compound carry charge. However, some of the ions like \text{Li}^+ are positive. Others ions like \text{Br}^{-} are negative. Charge on the two types of ions balance each other. As a result, the compound is <em>overall</em> neutral.

1 × (+1) + 1 × (-1) = 0. The positive charge on one \text{Li}^{+} ion balances the negative charge on one \text{Br}^{-} ion. The two ions would pair up at a 1:1 ratio.

The empirical formula for an ionic compound shows all the ions in the compound. Positive ions are written in front of negative ions. \text{Li}^{+} is positive and \text{Br}^{-} is negative. The formula shall also show the simplest ratio between the ions. For the compound between Li and Br, a 1:1 ratio will be the simplest. The "1" subscript in an empirical formula can be omitted. Hence the formula: LiBr.

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A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is in
tamaranim1 [39]

Answer:

The molarity of the strong base is 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

Explanation:

<u>Step 1:</u> Data given

Molarity of H2SO4 = 0.250 M

The initial buret reading is 5.00 mL

The final reading is 30.00 mL

<u />

<u>Step 2:</u> Calculate volume of H2SO4 used

30.00 mL - 5.00 mL = 25.00 mL

<u>Step 3:</u> Calculate moles of H2SO4

0.250 M = 0.250 mol/L

Since there are 2 H+ ions per H2SO4

0.250 mol/L  * 2 = 0.500 mol/L

The number of moles H2SO4 = 0.500 mol/L * 0.025 L

Number of moles H2SO4 = 0.0125 mol

<u>Step 4</u>: Calculate moles of OH-

For 1 mol H2SO4, we need 1 mol of OH-

For 0.0125 mol of H2SO4, we have 0.0125 mol of OH-

<u>Step 5</u>: Calculate the molarity of the strong base

Molarity = moles / volume

Molarity OH- = 0.0125 mol / 0.02 L

Molarity OH - = 0.625 M

Which procedural error will result in a strong base molarity that is too high?

⇒ Using a buret with a tip filled with air rather than the H2SO4 solution

   

5 0
2 years ago
Determine the concentration of H+ in each solution at 25∘C
puteri [66]

Answer:

1: [H+] = 0.01 M

2: [H+] = 0.0001 M

3: [H+] = 0.0001 M

Explanation:

Step 1: data given

pH = -log[H+]

pH = pOH = 14

Step 2:

1. A solution with pH = 2.0

pH = 2

-log[H+] = 2.0

[H+] = 10^-2

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2. A solution with pH = 4.0

pH = 4

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[H+] = 10^-4

[H+] = 0.0001 M

3. A solution with pOH = 10.0

pH = = 14 - 10 = 4

pH = 4

-log[H+] = 4.0

[H+] = 10^-4

[H+] = 0.0001 M

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3 years ago
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