Looking at the periodic table, determine the number of valence electrons present in a following neutral atoms.

Question

Column A
1.
Oxygen (O):
2.
Aluminium (Al):
3.
Neon (Ne):
4.
Silicion (Si):
5.
Magnesium (Mg):
6.
Iodine (I):
7.
Lithium (Li):
8.
Nitrogen (N):

Column B
a.7 valence electrons
b.4 valence electrons
c.1 valence electrons
d.5 valence electrons
e.3 valence electrons
f.8 valence electrons
g.6 valence electrons
h.2 valence electrons

Answer 1

Here are the number of valence electrons in each of the specified neutral atoms, matched with the corresponding option from Column B:

1. Oxygen (O): 6 valence electrons (g)
2. Aluminium (Al): 3 valence electrons (e)
3. Neon (Ne): 8 valence electrons (f)
4. Silicon (Si): 4 valence electrons (b)
5. Magnesium (Mg): 2 valence electrons (h)
6. Iodine (I): 7 valence electrons (a)
7. Lithium (Li): 1 valence electron (c)
8. Nitrogen (N): 5 valence electrons (d)

To summarize the matches:

• Oxygen (O): g. 6 valence electrons
• Aluminium (Al): e. 3 valence electrons
• Neon (Ne): f. 8 valence electrons
• Silicon (Si): b. 4 valence electrons
• Magnesium (Mg): h. 2 valence electrons
• Iodine (I): a. 7 valence electrons
• Lithium (Li): c. 1 valence electron
• Nitrogen (N): d. 5 valence electrons

Answer 2

Looking at the periodic table, determine the charge each atom will make when it bonds with other atoms.

Column A
1.
Calcium (Ca):
2.
Francium (Fr):
3.
Bromine (Br):
4.
Sulfur (S):
5.
Xenon (Xe):
6.
Carbon (C):
7.
Phosphorous (P):
8.
Boron (B):

Column B
a.+3
b.-1
c.-3
d.+4/-4
e.-2
f.+2
g.0
h.+1

Answer 3

Here are the charges that each specified atom is likely to adopt when it bonds with other atoms, matched with the corresponding option from Column B:

1. Calcium (Ca): +2 charge (f)
2. Francium (Fr): +1 charge (h)
3. Bromine (Br): -1 charge (b)
4. Sulfur (S): -2 charge (e)
5. Xenon (Xe): 0 charge (g) (inert gas, usually does not bond)
6. Carbon (C): +4 or -4 charge (d) (can form various compounds)
7. Phosphorus (P): -3 charge (c) (commonly forms -3 charge when gaining electrons)
8. Boron (B): +3 charge (a)

To summarize the matches:

• Calcium (Ca): f. +2
• Francium (Fr): h. +1
• Bromine (Br): b. -1
• Sulfur (S): e. -2
• Xenon (Xe): g. 0
• Carbon (C): d. +4/-4
• Phosphorus (P): c. -3
• Boron (B): a. +3

Answer 4

Match the following terms to the correct definition

Column A
1.
chemical change:
2.
physical change:
3.
chemical reaction:
4.
Law of Conservation of Mass:
5.
chemical equation:

Column B
a.a representation of a chemical reaction using chemical symbols and numbers
b.a process that produces a chemical change, in which substances are changed into new substances
c.a change that takes place when atoms of one or more substances are rearranged into new combinations, forming one or more new substances
d.States that mass is neither created nor destroyed by chemical reactions or physical transformations
e.a change in which the chemical makeup of the substance is not changed

Answer 5

Here are the matches of terms from Column A to the correct definitions in Column B:

1. chemical change: c. a change that takes place when atoms of one or more substances are rearranged into new combinations, forming one or more new substances
2. physical change: e. a change in which the chemical makeup of the substance is not changed
3. chemical reaction: b. a process that produces a chemical change, in which substances are changed into new substances
4. Law of Conservation of Mass: d. States that mass is neither created nor destroyed by chemical reactions or physical transformations
5. chemical equation: a. a representation of a chemical reaction using chemical symbols and numbers

To summarize:

• Chemical change: c
• Physical change: e
• Chemical reaction: b
• Law of Conservation of Mass: d
• Chemical equation: a

Answer 6

Are the following examples of chemical or physical changes?

Column A
1.
Digesting lunch:
2.
Cutting your hair:
3.
Crumbling up your snack wrapper:
4.
Grilling chicken:

Column B
a.Chemical Changes
b.Physical Changes

Answer 7

Here are the classifications of the changes listed in Column A as either chemical or physical changes:

1. Digesting lunch: a. Chemical Changes
2. Cutting your hair: b. Physical Changes
3. Crumbling up your snack wrapper: b. Physical Changes
4. Grilling chicken: a. Chemical Changes

To summarize:

• Digesting lunch: a. Chemical Changes
• Cutting your hair: b. Physical Changes
• Crumbling up your snack wrapper: b. Physical Changes
• Grilling chicken: a. Chemical Changes

Answer 8

Balance the following the chemical equation:

Mg + O_2 → MgO

Blank 1:
Blank 2:
Blank 3:

Answer 9

To balance the chemical equation \( \text{Mg} + \text{O}_2 \rightarrow \text{MgO} \), we need to ensure that the number of each type of atom on the reactant side is equal to the number on the product side.

The unbalanced equation is:

\[ \text{Mg} + \text{O}_2 \rightarrow \text{MgO} \]

Steps to Balance:

1. Count the atoms:

   • On the left (reactants), we have:
     • Magnesium (Mg): 1
     • Oxygen (O): 2 (since it is \( \text{O}_2 \))
   • On the right (products), we have:
     • Magnesium (Mg): 1
     • Oxygen (O): 1

2. Balancing Oxygen:

   • To balance the oxygen, we need 2 \( \text{O} \) atoms on the product side.
   • This means we need 2 \( \text{MgO} \) to achieve 2 oxygen atoms:

   \[ \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

3. Balance Magnesium:

   • Now we have 2 Mg on the product side, so we must adjust the left side accordingly:

   \[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

Final Balanced Equation:

\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \]

Fill in the Blanks:

• Blank 1: 2 (for Mg)
• Blank 2: 1 (for O_2, as it's not changed)
• Blank 3: 2 (for MgO)

Thus, the balanced equation is:

\[ \boxed{2} \text{Mg} + \boxed{1} \text{O}_2 \rightarrow \boxed{2} \text{MgO} \]

Answer 10

Question 6
Balance the following the chemical equation:

Al + O_2 → Al_2O_3

Blank 1:
Blank 2:
Blank 3:

Answer 11

To balance the chemical equation \( \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \), we need to ensure that the number of atoms for each element is equal on both sides of the equation.

Steps to Balance:

1. Count the atoms in the unbalanced equation:

   • On the left (reactants), we have:
     • Aluminum (Al): 1
     • Oxygen (O): 2 (from \( \text{O}_2 \))
   • On the right (products), we have:
     • Aluminum (Al): 2 (from \( \text{Al}_2\text{O}_3 \))
     • Oxygen (O): 3 (from \( \text{Al}_2\text{O}_3 \))

2. Balancing Aluminum:

   • We need 2 Al atoms on the reactant side to match the 2 Al atoms in the product side, so we put a coefficient of 2 in front of Al: \[ 2 \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \]

3. Balancing Oxygen:

   • Now we have 3 oxygen atoms on the right side (from \( \text{Al}_2\text{O}_3 \)), but we have 2 on the left side (from \( \text{O}_2 \)). To balance the oxygen atoms, we can put a coefficient of \( \frac{3}{2} \) in front of \( \text{O}_2 \): \[ 2 \text{Al} + \frac{3}{2} \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \] However, to avoid using a fraction, we can multiply the entire equation by 2:

   \[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \]

Final Balanced Equation:

\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \]

Fill in the Blanks:

• Blank 1: 4 (for Al)
• Blank 2: 3 (for O_2)
• Blank 3: 2 (for Al_2O_3)

Thus, the balanced equation is: \[ \boxed{4} \text{Al} + \boxed{3} \text{O}_2 \rightarrow \boxed{2} \text{Al}_2\text{O}_3 \]