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koban [17]
3 years ago
5

Aluminum oxide solid reacts with gaseous carbon monoxide to produce aluminum metal and carbon dioxide gas. Write the balanced eq

uation for this reaction.
Chemistry
2 answers:
Igoryamba3 years ago
6 0

Answer:

Al₂O₃(s) +  3CO(g) →  2Al (s)  + 3CO₂ (g)

Explanation:

1 mol of aluminum oxide reacts with 3 moles of carbon monoxide in order to produce 2 moles of solid aluminum and 3 moles of carbon dioxide.

Lorico [155]3 years ago
6 0

Answer:

Al2O3(s) +3CO(g) → 2Al(s) + 3CO2(g)

Explanation:

Step 1: data given

Aluminium oxide = Al2O3

carbon monoxide = CO

aluminium metal = Al

carbon dioxide = CO2

Step 2: The unbalanced equation

Al2O3(s) + CO(g) → Al(s) + CO2(g)

Step 3: Balancing the equation

Al2O3(s) + CO(g) → Al(s) + CO2(g)

On the left side we have 2x Al (in Al2O3), on the right side we have 1x Al. To balance the amount of Al, we have to multiply Al on the right side by 2

Al2O3(s) + CO(g) → 2Al(s) + CO2(g)

On the left side we have 4x O, on the right side we have 2x O. To balance the amount of O on both sides, we have to multiply CO on the left side by 3 and CO2 on the right side by 3. Now the equation is balanced.

Al2O3(s) +3CO(g) → 2Al(s) + 3CO2(g)

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6 0
3 years ago
Slow cooling of hot magma leads to the formation of blank crystals<br> whats the blank
Mashcka [7]
Slow cooling of hot magma leads to the formation of large crystals
8 0
3 years ago
How much energy is needed to vaporize 75.0 g of diethyl ether (c4h10o) at its boiling point (34.6°c), given that δhvap of diethy
malfutka [58]

Answer: 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

Explanation:

First we have to calculate the moles of diethyl ether

\text{Moles of diethyl ether}=\frac{\text{Mass of diethyl ether}}{\text{Molar mass of diethyl ether}}=\frac{75.0g}{74g/mole}=1.01moles

As, 1 mole of diethyl ether require heat = 26.5 kJ

So, 1.01  moles of diethyl ether require heat = \frac{26.5}{1}\times 1.01=26.8kJ

Thus 26.8 kJ of energy is needed to vaporize 75.0 g of diethyl ether

5 0
4 years ago
If all the water in 430.0 mL of a 0.45 M NaCI solution evaporates what is the mass of NaCI will remain
skad [1K]

Answer:

11.31 g.

Explanation:

Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.

M = (no. of moles of solute)/(V of the solution (L)).

<em>∴ M = (mass/molar mass)of NaCl/(V of the solution (L)).</em>

<em></em>

<em>∴ mass of NaCl remained after evaporation of water = (M)(V of the solution (L))(molar mass)</em> = (0.45 M)(0.43 L)(58.44 g/mol) = <em>11.31 g.</em>

6 0
3 years ago
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Colligative properties calculations are used for this type of problem. Calculations are as follows:<span>


ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (0.50 mol/kg)
ΔT(freezing point)  = 0.93 °C
Tf - T = 0.93 °C
<span>T = -0.93 °C</span></span>

4 0
3 years ago
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