Question

1. when an element loses electrons during a chemical reaction, it has been... a. oxidized b. turned into an electro - loser. c. reduced
2. if barium ions and nitride ions formed a compound, the formula would be... a. ban b. ba₂n₃ c. ba₂(n₂)₃ d. ba₃n₂
3. the balanced chemical equation for the synthesis of nitrogen and iodine is... a. n + 2i₂ -------> ni₄ b. n₂ + 3i₂ -------> 2ni₃ c. n + 3i -------> ni₃ d. n₂ + 4i₂ -------> 2ni₂
4. acids, when compared with water and bases, have a greater abundance of which ion? a. oh⁻ b. h₃o⁺ c. both a & b d. neither a or b
5. give the ph range: 1.5 pts. acid ______ water ____ base ______
6. balance the three following chemical equations. (3 pts.) ch₄+ o₂ -------> co₂ + h₂o cuso₄+ al -------> al₂(so₄)₃ + cu kclo₃ -------> kcl + o₂

Explanation:

Step1: Recall oxidation - reduction concept

Oxidation is the loss of electrons in a chemical reaction. So when an element loses electrons, it has been oxidized.

Step2: Determine ion - charge for compound formula

Barium has a +2 charge ($Ba^{2 + }$) and nitride has a - 3 charge ($N^{3-}$). To form a neutral compound, we need 3 barium ions and 2 nitride ions, so the formula is $Ba_{3}N_{2}$.

Step3: Balance nitrogen - iodine synthesis reaction

Nitrogen exists as $N_{2}$ and iodine as $I_{2}$. The balanced reaction is $N_{2}+3I_{2}
ightarrow2NI_{3}$.

Step4: Identify ion in acids

Acids donate protons ($H^{+}$) in water. In water, $H^{+}$ combines with $H_{2}O$ to form $H_{3}O^{+}$. So acids have a greater abundance of $H_{3}O^{+}$ ions compared to water and bases.

Step5: Recall pH ranges

Acids have a pH range of 0 - 7, water has a pH of 7, and bases have a pH range of 7 - 14.

Step6: Balance $CH_{4}+O_{2}

ightarrow CO_{2}+H_{2}O$
Balance carbon first: 1 carbon on both sides. Then balance hydrogen: 4 hydrogens on the left, so we need 2 $H_{2}O$ on the right. Then balance oxygen: 2 + 2 = 4 oxygens on the right, so we need 2 $O_{2}$ on the left. The balanced equation is $CH_{4}+2O_{2}
ightarrow CO_{2}+2H_{2}O$.

Step7: Balance $CuSO_{4}+Al

ightarrow Al_{2}(SO_{4})_{3}+Cu$
Balance sulfate groups first. We need 3 $CuSO_{4}$ on the left. Then balance aluminum and copper. The balanced equation is $3CuSO_{4}+2Al
ightarrow Al_{2}(SO_{4})_{3}+3Cu$.

Step8: Balance $KClO_{3}

ightarrow KCl + O_{2}$
Balance oxygen: we have 3 oxygens on the left and 2 on the right. The least - common multiple of 3 and 2 is 6. So we need 2 $KClO_{3}$ on the left and 3 $O_{2}$ on the right. Then balance potassium and chlorine. The balanced equation is $2KClO_{3}
ightarrow2KCl + 3O_{2}$.

Answer:

1. a. oxidized
2. d. $Ba_{3}N_{2}$
3. b. $N_{2}+3I_{2}

ightarrow2NI_{3}$

1. b. $H_{3}O^{+}$
2. acid: 0 - 7; water: 7; base: 7 - 14
3. $CH_{4}+2O_{2}

ightarrow CO_{2}+2H_{2}O$
$3CuSO_{4}+2Al
ightarrow Al_{2}(SO_{4})_{3}+3Cu$
$2KClO_{3}
ightarrow2KCl + 3O_{2}$