What is the molar mass of 65.18 g of gas exerting a pressure of 2.25 atm on the walls of a 4.81 L container at 350 k?

Answer:

Explanation:

1) Alkali metals and halogens both need to achieve a stable outer electron shell, requiring alkali metals to lose one electron and halogens to gain one.

2) They share an identical count of outer shell electrons.

3) Typically, they have elevated melting points.

4) They exhibit low reactivity or none at all.

5) They belong to group 7.

Answer:

- The coefficients in front of H⁺ and Fe³⁺ are 8 and 5 respectively.

- A total of 5 moles of electrons are exchanged.

Explanation:

This reaction is represented as:

Fe²⁺(aq) + MnO₄⁻(aq) → Fe³⁺(aq) + Mn²⁺(aq)

Analyzing the oxidation states:

Fe²⁺ transitions to Fe³⁺

This indicates an increase in oxidation state → OXIDATION

Meanwhile, Mn in MnO₄⁻ starts with +7 and transforms into Mn²⁺

This suggests a decrease in oxidation state → REDUCTION

Let's formulate the half reactions:

Fe²⁺ → Fe³⁺  +  1e⁻    (it loses 1 mole of electrons)

MnO₄⁻ + 5e⁻ →  Mn²⁺  (it gains 5 moles of electrons)

Next, we will balance the oxygen atoms. In an acidic environment, water is added to balance the oxygens on the opposite side. Since there are 4 oxygens on the reactant side, we add 4 H₂O to the product side.

MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Now, to balance the hydrogen atoms, we have 8 hydrogens in the products, necessitating the inclusion of 8H⁺ in the reactants, yielding the complete half-reaction:

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

Notably, there's 1e⁻ in the oxidation and 5e⁻ in the reduction. To cancel electrons, we must multiply the oxidation half-reaction by 5.

(Fe²⁺ → Fe³⁺  +  1e⁻) x 5

5Fe²⁺ → 5Fe³⁺  +  5e⁻  

8H⁺  + MnO₄⁻ + 5e⁻ →  Mn²⁺  + 4H₂O

By adding both half reactions, we have:

5Fe²⁺  + 8H⁺  + MnO₄⁻ + 5e⁻ →  5Fe³⁺  +  5e⁻   + Mn²⁺  + 4H₂O

The electrons cancel out, resulting in the balanced equation:

5Fe²⁺  + 8H⁺  + MnO₄⁻  →  5Fe³⁺  + Mn²⁺  + 4H₂O

Density is calculated as Mass divided by Volume.

Volume can also be expressed as Mass divided by Density.

Volume becomes 2 divided by 128.

This results in Volume equal to 1/64.

Therefore, each side measures 4 cm.

Answer:

The flask yielding the greatest amount of product is the second flask.

The water molecules produced are

Explanation:

From the prompt we learn that

The chemical reaction is

2CH3OH(l) + 3O2(g)⟶4H2O(g)+2CO2(g)

The first flask contains 3 molecules of C H 3 O H and 3 molecules of O 2

The second flask has 1 molecule of C H 3 O H and 6 molecules of O 2.

The third flask consists of 4 molecules of C H 3 O H and 2 molecules of O 2.

Analyzing the three flasks, it's evident that the flask producing the most considerable total product is the second one.

According to the balanced equation, 2 moles of C H 3 O H react with 3 moles of oxygen, making O 2 the limiting reagent.

Generally, one mole of any substance is of that substance.

Here, represents Avogadro's number valued at .

As seen in the balanced equation, 3 moles of O 2 (3 * molecules of O 2) yield 4 moles of H2O (4 *

Thus, 6 molecules of O 2 would generate X molecules of H 2 O.

So

=>

1.09 moles of tin (Sn)

Explanation:

The reaction we are examining involves tin chloride (SnCl₂) reacting with sodium (Na) to yield tin (Sn) and sodium chloride (NaCl):

SnCl₄ + 4 Na → Sn + 4 NaCl

Considering the chemical equation, this results in:

If 1 mole of SnCl₂ reacts with 4 moles of Na,

then 2.34 moles of SnCl₂ will react with X moles of Na.

X = (2.34 × 4) / 1 = 9.36 moles of Na

So, 2.34 moles of SnCl₂ can react with 9.36 moles of Na, but we have only 4.36 moles of Na on hand. Therefore, Na is the limiting reactant. Based on this, we can conclude:

If 4 moles of Na yield 1 mole of Sn,

then 4.36 moles of Na will produce Y moles of Sn.

Y = (4.36 × 1) / 4 = 1.09 moles of Sn

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limiting reactant