How do i Calculate the grams of silver chloride produced from 18.5 ml of 4.5M magnesium chloride with 99.2 ml of 1.75 M silver nitrate in a step by step?

4.21 g of AgCl

3.06 g of BaCl₂ will be needed to complete the reaction

Explanation:

The first step is to determine the reaction.

Reactants: BaCl₂ and AgNO₃

The products will be the silver chloride (AgCl) and the Ba(NO₃)₂

The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)

We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-

2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.

We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.

Now, we consider the BaCl₂.

2 moles of nitrate can react to 1 mol of barium chloride

Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:

0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl

Explanation:

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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indium, and thallium have three electrons i think