Calculate the volume occupied by 32.0 g of O2 gas, the pressure of the O2 gas is 78.5 kPa at 25°C.

What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?

lisabon 2012 [21]

Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

$2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3$ ;

$4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}$ .

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

$3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}$ .

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

$2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}$

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

4 0

3 years ago

If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×

Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0

3 years ago

Identify two acids and two bases that you use or come into contact with in an average week. Identify uses for each substance.

skelet666 [1.2K]

Explanation:

Two acids we come into contact with in an average week

Vinegar is an 10% solution of acetic acid $(CH_3COOH)$ in water. Used in salad dressing and while cooking food. It has a sour taste.
Citric acid present in fruits and vegetables like : lemons, orange, tomatoes etc. It is a weak organic acid with sour taste.

Two bases we come into contact with in an average week.

Baking soda ( $NaHCO_3$ ) is used in baking food like: cakes, cookies, breads. Baking soda is one of the ingredient while baking breads and cakes.
Caustic soda (NaOH) is used for preparation of detergents, papers , soaps etc. We use soaps and detergents for washing.

4 0

3 years ago

Read 2 more answers

The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c

netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is $4.58\times 10^{-3}M$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of oxygen gas

$C_2\text{ and }p_2$ are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used: 1 atm = 760 torr

$C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm$

Putting values in above equation, we get:

$\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M$

Hence, the solubility of oxygen gas at 628 torr is $4.58\times 10^{-3}M$

4 0

2 years ago

What is the mass of 2.90 ×1022 molecules of NaOH (Molar mass = 40.0 g/mol)?

ivann1987 [24]

40.0 g ( 1 mole ) --------------- 6.02x10²³ molecules
? ? --------------------------- 2.90x10²² molecules

mass = 2.90x10²² * 40.0 / 6.02x10²³

mass = 1.16x10²⁴ / 6.02x10²³

mass = 1.9269 g

hope this helps!

3 0

3 years ago