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katrin [286]
3 years ago
9

Use the periodic table to determine the charge of the ions below. Select the correct answer from the drop-down menus. A lithium

ion has a charge of A calcium ion has a charge of An aluminum ion has a charge of A fluorine ion has a charge of An oxygen ion has a charge of
Chemistry
2 answers:
dezoksy [38]3 years ago
9 0
Lithium ion has a charge of 1+
Calcium ion has a charge of 2+
Aluminum ion has a charge of 3+
Fluoride has a charge of 1-
Oxide has a charge of 2-
Evgesh-ka [11]3 years ago
7 0

Explanation:

When an atom gains an electron then it acquires a negative charge whereas when an atom loses an electron then it acquires a positive charge.

Atomic number of lithium is 3 and it loses one electron in order to attain stability. Thus, lithium ion has +1 charge.

Atomic number of calcium is 20 with electronic distribution as 2, 8, 2 and it loses two electrons in order to attain stability. Thus, calcium ion has +2 charge.

Atomic number of aluminium is 13 and it loses 3 electrons in order to attain stability. Thus, aluminium ion has +3 charge.

Atomic number of fluorine is 9 and it needs one electron in order to attain stability. Thus, fluorine ion has a charge of -1.

Atomic number of oxygen is 8 with electronic distribution as 2, 6 and in order to attain stability oxygen atom gains two electrons. Thus, oxygen ion has -2 charge.

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If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?
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Answer:

V_2=18 \ L \ H_2

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
  • Charles' Law: \frac{V_1}{T_1} =\frac{V_2}{T_2}

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

  1. Substitute:                    \frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}
  2. Cross-multiply:             (5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)
  3. Multiply:                        4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K
  4. Isolate <em>x</em>:                       18.0403 \ L \ H_2 = x
  5. Rewrite:                         x=18.0403 \ L \ H_2

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em />18.0403 \ L \ H_2 \approx 18 \ L \ H_2<em />

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