If 100.0 ml of 0.453 m na2so4 are added to 100.0 ml of 0.907 m pb(no3)2, how many grams of pbso4 can be produced?
na2so4(aq)+pb(no3)2(aq)⟶2nano3(aq)+ pbso4(s)

The mass of produced will be 13.73 grams.

Explanation:

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

Molarity of the solution = 0.453 mol/L

Volume of solution = 100 mL = 0.1 L     (Conversion factor:  1L = 1000 mL)

Putting values in equation 1, we get:

Molarity of the solution = 0.907 mol/L

Volume of solution = 100 mL = 0.1 L

Putting values in equation 1, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of lead nitrate.

So, 0.0453 moles of sodium sulfate will react with = of lead nitrate.

As, the given amount of lead nitrate is more than the required amount. Thus, it is considered as an excess reagent.

Hence, sodium sulfate is considered as an limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of lead sulfate.

So, 0.0453 moles of sodium sulfate will react with = of lead sulfate.

To calculate the mass of lead sulfate, we use the equation:

Moles of lead sulfate = 0.0453 mol

Molar mass of lead sulfate = 303.26 g/mol

Putting values in above equation, we get:

Hence, the mass of produced will be 13.73 grams.

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).

answer: they produce more slow than unicellular organisms.