Calculate the number of atoms in 2.90 gram* of Silver How many oxygen atoms in 5

Question

Calculate the number of atoms in 2.90 gram* of Silver How many oxygen atoms in 55.0 grams of water? Answer all of the questions below based on the following reaction: Solutions of silver nitrate and magnesium bromide react to from a pale yellow precipitate and magnesium nitrate. a) Write the balanced equation below (be sure to include the phases of each of the reactants and products). b) If there are 35.8 grams of silver nitrate, how many moles is this equivalent to? c) If 1.5 moles of magnesium bromide arc available with an excess amount of silver nitrate, how many moles of silver bromide will form? d) How many grams of magnesium bromide arc used, when 11.4 grams of magnesium nitrate are formed? e) How many formula units of magnesium bromide are needed to react with 1.50 grams of silver nitrate?

Explanation / Answer

Question 9.

Apply

Molar mass of Silver = 107.8682 g/mol

so

mol = mass/MW = 2.90/107.8682 = 0.02688 mol of Ag

note that

1 mol = 6.022*10^23 atoms

so

0.02688 mol = 0.02688*6.022*10^23 = 1.618713*10^22 atoms of Silver

Question 10.

find oxygen atoms in 55 g of Water

first change to mol of water

mol = mass/MW = 55/18 = 3.0555 moles of water

1 mol of water = 1 mol of oxygen

3.0555 mol of water = 3.0555 mol of oxygen

1 mol = 6.022*10^23 atoms

3.0555 mol --> 3.0555*6.022*10^23 atoms = 1.840022*10^24 atoms of Oxygen

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