Suppose of iron(II) bromide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) bromide is dissolved in it. Round your answer to significant digits.

The question is incomplete, here is the complete question:

Suppose 3.00 g of iron(II) bromide is dissolved in 350 mL of a 0.042 M aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) bromide is dissolved in it. Round your answer to 2 significant digits.

The final molarity of bromide ions in the solution is 0.042 M

Explanation:

To calculate the number of moles, we use the equation:

Given mass of iron (II) bromide = 3 g

Molar mass of iron (II) bromide = 215.65 g/mol

Putting values in above equation, we get:

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

Molarity of silver nitrate solution = 0.042 M

Volume of solution = 350 mL = 0.350 L     (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

The chemical equation for the reaction of iron (II) bromide and silver nitrate follows:

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of iron (II) bromide

So, 0.0147 moles of silver nitrate will react with = of iron (II) bromide

As, given amount of iron (II) bromide is more than the required amount. So, it is considered as an excess reagent.

Thus, silver nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver bromide

So, 0.0147 moles of silver nitrate will produce = of silver bromide

1 mole of silver bromide produces 1 mole of silver ions and 1 mole of bromide ions

Now, calculating the molarity of bromide ions by using equation 1:

Moles of bromide ions = 0.0147 moles

Volume of solution = 0.350 L

Putting values in equation 1, we get:

Hence, the final molarity of bromide ions in the solution is 0.042 M

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg