6. jill (a not - so - smart scientist) went ahead and put magnesium metal into contact with bromine liquid.crazy enough, a vigorous chemical reaction occurred, producing magnesium bromide powder.
a) given the above statement, write the balanced chemical equation for this reaction below (be sureto include the states of both the reactants and the product). (2 marks)
b) explain using electron configurations and reactivity why bromine liquid is so reactive in thepresence of a metal.(2 marks)
7. balance the following reactions: (2 marks)
a) ko₂(s) + co₂(g) → k₂co₃(s) + o₂(g)
b) c₈h₁₈(l) + o₂(g) → co₂(g) + h₂o(g)

Question

Accepted Answer

### 6. a)
# Explanation:
## Step1: Write the unbalanced equation
Magnesium ($Mg$, solid) reacts with bromine ($Br_2$, liquid) to form magnesium bromide ($MgBr_2$, solid). The unbalanced equation is $Mg(s)+Br_2(l)\to MgBr_2(s)$.
## Step2: Check the balance of atoms
For $Mg$: 1 atom on the left and 1 atom on the right. For $Br$: 2 atoms on the left ($Br_2$) and 2 atoms on the right ($MgBr_2$). The equation is already balanced.

# Answer:
$Mg(s)+Br_2(l)\to MgBr_2(s)$

### 6. b)
# Brief Explanations:
- Electron configuration of $Br$ (atomic number $35$): $[Ar]3d^{10}4s^{2}4p^{5}$. It has 7 valence electrons.
 - Halogens (like $Br$) are reactive as they need 1 more electron to complete their octet (attain a stable noble - gas - like configuration). Metals tend to lose electrons. So, $Br$ readily accepts electrons from metals (oxidizes the metal) to form ionic compounds (like $MgBr_2$).

# Answer:
Bromine ($Br$) has an electron configuration $[Ar]3d^{10}4s^{2}4p^{5}$ with 7 valence electrons. Halogens (including $Br$) are highly reactive with metals because they need one more electron to complete their octet (achieve a stable electron configuration similar to noble gases). Metals, on the other hand, tend to lose electrons. In the reaction with a metal (e.g., $Mg$), $Br$ accepts electrons (oxidizes the metal) to form ionic compounds (e.g., $MgBr_2$).

### 7. a)
# Explanation:
## Step1: Balance $K$ atoms
Let the coefficient of $KO_2$ be $x$ and of $K_2CO_3$ be $y$. For $K$: $x = 2y$. Let $y = 1$, then $x = 2$. The equation becomes $2KO_2(s)+CO_2(g)\to K_2CO_3(s)+O_2(g)$
## Step2: Balance $O$ atoms
Left - hand side: $2\times2 + 2=6$ $O$ atoms (from $KO_2$ and $CO_2$). Right - hand side: $3 + 2z$ ($z$ is the coefficient of $O_2$). From $K_2CO_3$ (3 $O$ atoms) and $O_2$ ($2z$ $O$ atoms).
$6=3 + 2z$, so $z=\frac{3}{2}$. Multiply all coefficients by 2 to get rid of the fraction.
The balanced equation is $4KO_2(s)+2CO_2(g)\to 2K_2CO_3(s)+3O_2(g)$

# Answer:
$4KO_2(s)+2CO_2(g)\to 2K_2CO_3(s)+3O_2(g)$

### 7. b)
# Explanation:
## Step1: Balance $C$ atoms
Let the coefficient of $C_8H_{18}$ be $1$. Then the coefficient of $CO_2$ is $8$ (since there are 8 $C$ atoms in $C_8H_{18}$). The equation becomes $C_8H_{18}(l)+O_2(g)\to 8CO_2(g)+H_2O(g)$
## Step2: Balance $H$ atoms
There are 18 $H$ atoms in $C_8H_{18}$, so the coefficient of $H_2O$ is $9$. The equation is $C_8H_{18}(l)+O_2(g)\to 8CO_2(g)+9H_2O(g)$
## Step3: Balance $O$ atoms
Right - hand side: $8\times2+9\times1 = 25$ $O$ atoms. So the coefficient of $O_2$ is $\frac{25}{2}$. Multiply all coefficients by 2 to get rid of the fraction.
The balanced equation is $2C_8H_{18}(l)+25O_2(g)\to 16CO_2(g)+18H_2O(g)$

# Answer:
$2C_8H_{18}(l)+25O_2(g)\to 16CO_2(g)+18H_2O(g)$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

6. a)

Step1: Write the unbalanced equation

Step2: Check the balance of atoms

Step1: Balance \(K\) atoms

Step2: Balance \(O\) atoms

Answer:

6. b)