What is the molecular formula of caffeine, given that it has an empirical formula of C_4H_5N_2O and a molar mass of 194 g/mol?

Answer:

1. Sand Casting: Sand casting is a popular method used to create pewter objects. Here's how it's achieved:

- First, a pattern or model of the desired object is created using materials like wood, plastic, or metal.

- The pattern is then pressed into a special type of sand called green sand, leaving an impression.

- Next, the pattern is removed, leaving a cavity in the sand.

- Molten pewter is poured into the cavity and left to cool and solidify.

- Once cooled, the sand is carefully removed, revealing the pewter object.

2. Spin Casting: Spin casting, also known as centrifugal casting, is another way to mold pewter. Here's how it works:

- A rubber mold of the object is created, usually using a master pattern made of wax or another material.

- The rubber mold is placed in a casting machine that rotates at high speed.

- Molten pewter is poured into the mold while it spins.

- The centrifugal force evenly distributes the pewter, filling the mold's cavities.

- As the pewter cools, it solidifies and takes the shape of the mold.

- Once cooled, the mold is opened, and the pewter object is removed.

3. Die Casting: Die casting is a method used for producing high-quality pewter objects. Here's how it's done:

- A metal mold, called a die, is created to match the desired shape of the object.

- Molten pewter is injected into the die under high pressure.

- The pressure ensures that the pewter fills the mold completely and captures all the intricate details.

- After a short cooling period, the die is opened, and the pewter object is ejected.

Explanation:

These are just a few ways to form or mold pewter. Each method has its advantages and is suitable for different types of objects. It's important to choose the appropriate method based on the complexity of the design and the desired quality of the final product.

Answer:

• The third chice: 0.0750 M Na₂SO₄

Explanation:

Assume 100% ionization:

1) 0.100 M solution K₂SO₄

• K₂SO4 (aq) → 2K⁺ (aq) + SO₄²⁻ (aq)

• Mole ratios: 1 mol K₂SO4 : 2 mol K⁺  + 1 mol SO₄²⁻ (aq) : 3 mol ions. This is 1 : 3

• At constant volume, the mole ratios are equal to the concentration ratios (M).

• 1  M K₂SO₄: 3 M ions = 0.100 M K₂SO₄ / x ⇒ x = 0.300 M ions

This means, that you have to find which of the choices is a solution that contains the same 0.300 M ion concentration.

2) 0.0800 M Na₂CO₃

• Na₂CO₃ (aq) → 2 Na⁺ + CO₃⁻

• 1 M Na₂CO₃ / 3 M ions = 0.0800M / x ⇒ x = 0.0267 M ions

This is not equal to 0.300 M, so this solution would not contain the same total concentration as a 0.100 M solution of K₂SO₄, and is not the right answer.

3)  0.100 M NaCl

• NaCl → Na⁺ + Cl⁻

• 1 M NaCl / 2 M ions = 0.100 M NaCl / x ⇒ x = 0.200 M ions

This is not equal to 0.300 M ion, so not a correct option.

4) 0.0750 M Na₃PO₄

• Na₃PO₄ → 3Na⁺ + PO₄³⁻

• 1 M Na₃PO₄ / 4 M ions = 0.0750 M Na₃PO₄ / x ⇒ x = 0.300 M ions

Hence, this ion concentration is equal to the ion concentration of a 0.100 M solution of K₂SO₄, and is the correct choice.

5)  0.0500 M NaOH

• NaOH → Na⁺ + OH⁻

• 1 M NaOH / 2 mol ions = 0.0500 M NaOH / x ⇒ x = 0.100 M ions

Not equal to 0.300 M, so wrong choice.

Changes Occurs when a Barium Atom - Option 3  

When a barium atom loses two electrons it becomes a positive ion and its radius decreases. Barium (Ba) has atomic number 56 so it has 2 electrons in first shell of an atom to become stable according to duplet rule. Then other 52 electrons revolve in the shells according to octet rules.

Another 2 electrons are in the outermost shell. To become stable electrons lose to form barium ions (Ba+2). Hence, by losing 2 electrons the outermost shell will be diminished so its radius decreases and by losing electrons it becomes positive ions.

Answer:

The correct answer is option 3, It becomes a positive ion and its radius decreases

Explanation:

As per the Octet rule, Barium has 2 electrons in its outermost shell. When it loses the two electron it gains two positive charge i.e Ba2+. As the barium loses the two electron from its outermost shell, the outermost shell becomes vacant and thus is no more considered as a part of atomic geometry of the barium atom and since the outermost shell is considered negligible the radius of barium atom reduces automatically.