make the following mole conversions - show all work in space below and place the answer in blank. (try to round with the correct number of significant figures!)
33. 0.150 moles of na₂co₃ into grams
34. 12.40 grams of potassium bromide into moles
35. 85.0 grams of co₂ into molecules
36. how many oxygen atoms are in 3.5 moles of sulfur dioxide?
37. how many grams of nitrogen are in 1.20 × 10²⁵ molecules of ammonia?
predict products in the blank, if necessary, and balance the following equations. if given a wo equation, you will need to write the formulas before starting. indicate the type of reaction in blank to the far right: decomposition, synthesis, double displacement, single displacement, or combustion
38. __ al (s) + __ o₂ (g) → __ al₂o₃ (s)
rxn type
39. __ c₃h₈ (g) + __ o₂ (g) →
40. __ caco₃ (s) → __ cao (s) + __ co₂ (g)
41. __ mg (s) + __ hcl (aq) →
42. __ na₂so₄ (aq) + __ bacl₂ (aq) →

Question

make the following mole conversions - show all work in space below and place the answer in blank. (try to round with the correct number of significant figures!)
33. 0.150 moles of na₂co₃ into grams
34. 12.40 grams of potassium bromide into moles
35. 85.0 grams of co₂ into molecules
36. how many oxygen atoms are in 3.5 moles of sulfur dioxide?
37. how many grams of nitrogen are in 1.20 × 10²⁵ molecules of ammonia?
predict products in the blank, if necessary, and balance the following equations. if given a wo equation, you will need to write the formulas before starting. indicate the type of reaction in blank to the far right: decomposition, synthesis, double displacement, single displacement, or combustion
38. __ al (s) + __ o₂ (g) → __ al₂o₃ (s)
																								rxn type
39. __ c₃h₈ (g) + __ o₂ (g) →
40. __ caco₃ (s) → __ cao (s) + __ co₂ (g)
41. __ mg (s) + __ hcl (aq) →
42. __ na₂so₄ (aq) + __ bacl₂ (aq) →

Accepted Answer

### Question 33:
# Explanation:
## Step1: Calculate molar mass of $\ce{Na2CO3}$
Molar mass of $\ce{Na}$ = 22.99 g/mol, $\ce{C}$ = 12.01 g/mol, $\ce{O}$ = 16.00 g/mol.
Molar mass = $2	imes22.99 + 12.01 + 3	imes16.00$ = $45.98 + 12.01 + 48.00$ = 105.99 g/mol.
## Step2: Convert moles to grams
Mass = moles × molar mass = $0.150\ 	ext{mol} 	imes 105.99\ 	ext{g/mol}$ ≈ 15.9 g (3 significant figures).
# Answer: 15.9 g

### Question 34:
# Explanation:
## Step1: Calculate molar mass of $\ce{KBr}$
Molar mass of $\ce{K}$ = 39.10 g/mol, $\ce{Br}$ = 79.90 g/mol.
Molar mass = $39.10 + 79.90$ = 119.00 g/mol.
## Step2: Convert grams to moles
Moles = $\frac{	ext{mass}}{	ext{molar mass}}$ = $\frac{12.40\ 	ext{g}}{119.00\ 	ext{g/mol}}$ ≈ 0.1042 mol (4 significant figures).
# Answer: 0.1042 mol

### Question 35:
# Explanation:
## Step1: Calculate moles of $\ce{CO2}$
Molar mass of $\ce{CO2}$ = $12.01 + 2	imes16.00$ = 44.01 g/mol.
Moles = $\frac{85.0\ 	ext{g}}{44.01\ 	ext{g/mol}}$ ≈ 1.931 mol.
## Step2: Convert moles to molecules (use Avogadro’s number $6.022	imes10^{23}$)
Molecules = moles × $6.022	imes10^{23}$ = $1.931\ 	ext{mol} 	imes 6.022	imes10^{23}\ 	ext{molecules/mol}$ ≈ $1.163	imes10^{24}$ molecules (3 significant figures).
# Answer: $1.16	imes10^{24}$ molecules (or $1.163	imes10^{24}$)

### Question 36:
# Explanation:
## Step1: Moles of $\ce{O}$ in $\ce{SO2}$
1 mole $\ce{SO2}$ has 2 moles $\ce{O}$. So, 3.5 mol $\ce{SO2}$ has $3.5\ 	ext{mol} 	imes 2$ = 7.0 mol $\ce{O}$.
## Step2: Convert moles of $\ce{O}$ to atoms (Avogadro’s number)
Atoms = $7.0\ 	ext{mol} 	imes 6.022	imes10^{23}\ 	ext{atoms/mol}$ ≈ $4.215	imes10^{24}$ atoms (2 significant figures, so $4.2	imes10^{24}$ or $4.22	imes10^{24}$).
# Answer: $4.2	imes10^{24}$ atoms (or $4.22	imes10^{24}$)

### Question 37:
# Explanation:
## Step1: Moles of $\ce{NH3}$
Moles = $\frac{	ext{molecules}}{N_A}$ = $\frac{1.20	imes10^{25}}{6.022	imes10^{23}\ 	ext{molecules/mol}}$ ≈ 19.93 mol.
## Step2: Moles of $\ce{N}$ in $\ce{NH3}$
1 mole $\ce{NH3}$ has 1 mole $\ce{N}$. So, moles of $\ce{N}$ = 19.93 mol.
## Step3: Mass of $\ce{N}$ (molar mass of $\ce{N}$ = 14.01 g/mol)
Mass = $19.93\ 	ext{mol} 	imes 14.01\ 	ext{g/mol}$ ≈ 279 g (3 significant figures).
# Answer: 279 g

### Question 38:
# Explanation:
## Step1: Balance $\ce{Al}$ and $\ce{O}$
- $\ce{Al}$: Left = 1, Right = 2 → Multiply $\ce{Al}$ by 2: $2\ce{Al}$.
- $\ce{O}$: Left = 2, Right = 3 → Find LCM of 2 and 3 (6). Multiply $\ce{O2}$ by 3, $\ce{Al2O3}$ by 2: $3\ce{O2}$, $2\ce{Al2O3}$.
- Now $\ce{Al}$: Left = 2, Right = 4 → Multiply $\ce{Al}$ by 4: $4\ce{Al}$.
Balanced equation: $4\ce{Al} + 3\ce{O2}
ightarrow 2\ce{Al2O3}$.
## Step2: Reaction type (Synthesis: combining elements to form a compound).
# Answer: 4, 3, 2; Rxn Type: Synthesis

### Question 39:
# Explanation:
## Step1: Predict products (combustion of hydrocarbon: $\ce{CO2 + H2O}$)
$\ce{C3H8 + O2 -> CO2 + H2O}$.
## Step2: Balance $\ce{C}$, $\ce{H}$, then $\ce{O}$
- $\ce{C}$: Left = 3, Right = 1 → Multiply $\ce{CO2}$ by 3: $3\ce{CO2}$.
- $\ce{H}$: Left = 8, Right = 2 → Multiply $\ce{H2O}$ by 4: $4\ce{H2O}$.
- $\ce{O}$: Right = $3	imes2 + 4	imes1$ = 10 → Multiply $\ce{O2}$ by 5: $5\ce{O2}$.
Balanced equation: $\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$.
## Step3: Reaction type (Combustion: hydrocarbon + $\ce{O2}$ → $\ce{CO2 + H2O}$).
# Answer: 1, 5, $3\ce{CO2} + 4\ce{H2O}$; Rxn Type: Combustion

### Question 40:
# Explanation:
## Step1: Balance (already balanced for $\ce{Ca}$, $\ce{C}$, $\ce{O}$)
$\ce{CaCO3 -> CaO + CO2}$.
## Step2: Reaction type (Decomposition: one compound breaks into two or more).
# Answer: 1, 1, 1; Rxn Type: Decomposition

### Question 41:
# Explanation:
## Step1: Predict products (single displacement: $\ce{Mg}$ displaces $\ce{H}$ in $\ce{HCl}$ → $\ce{MgCl2 + H2}$)
$\ce{Mg + HCl -> MgCl2 + H2}$.
## Step2: Balance $\ce{Cl}$ and $\ce{H}$
- $\ce{Cl}$: Left = 1, Right = 2 → Multiply $\ce{HCl}$ by 2: $2\ce{HCl}$.
- $\ce{H}$: Left = 2, Right = 2 (balanced).
Balanced equation: $\ce{Mg + 2HCl -> MgCl2 + H2}$.
## Step3: Reaction type (Single Displacement: metal displaces $\ce{H}$ from acid).
# Answer: 1, 2, $\ce{MgCl2 + H2}$; Rxn Type: Single Displacement

### Question 42:
# Explanation:
## Step1: Predict products (double displacement: $\ce{BaSO4}$ precipitate + $\ce{NaCl}$)
$\ce{Na2SO4 + BaCl2 -> BaSO4 + NaCl}$.
## Step2: Balance $\ce{Na}$, $\ce{Cl}$
- $\ce{Na}$: Left = 2, Right = 1 → Multiply $\ce{NaCl}$ by 2: $2\ce{NaCl}$.
- $\ce{Cl}$: Left = 2, Right = 2 (balanced).
Balanced equation: $\ce{Na2SO4 + BaCl2 -> BaSO4 + 2NaCl}$.
## Step3: Reaction type (Double Displacement: exchange of ions, precipitate forms).
# Answer: 1, 1, $\ce{BaSO4 + 2NaCl}$; Rxn Type: Double Displacement

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 33:

Step1: Calculate molar mass of $\ce{Na2CO3}$

Step2: Convert moles to grams

Step1: Calculate molar mass of $\ce{KBr}$

Step2: Convert grams to moles

Step1: Calculate moles of $\ce{CO2}$

Step2: Convert moles to molecules (use Avogadro’s number $6.022\times10^{23}$)

Answer:

Question 34: