Which particle has no charge

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

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

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

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

3 years ago

Lead will float in water.

Fed [463]

Actually, no. While their mass may be the same (1kg), the volume of lead is a lot smaller than that of feathers. As there is the same mass stuffed in a smaller space, it must be denser. The density of water is 1 g/cm3, so if the density of the lead is more than 1g/cm3, it has to sink

7 0

3 years ago

Read 2 more answers

The half-life of cesium-137 is 30 years. Suppose we have a 200-mg sample. (a) Find the mass that remains after t years.

scZoUnD [109]

Given:

Half life(t^ 1/2) :30 years

A0( initial mass of the substance): 200 mg.

Now we know that

A= A0/ [2 ^ (t/√t)]

Where A is the mass that remains after t years.

A0 is the initial mass

t is the time

t^1/2 is the half life

Substituting the given values in the above equation we get

A= [200/ 2^(t/30) ] mg

Thus the mass remaining after t years is [200/ 2^(t/30) ] mg

5 0

3 years ago

The car wit fastest acceleration is the tempest. it has an acceleration of 6.1 m/s. suppose the car accelerates from rest to a f

ziro4ka [17]

Given:

Initial velocity (v1) = 0 m/s

Final velocity (v2) = 30 m/s

Acceleration (a) = 6.1 m/s2

To determine:

The time (t) taken to reach the final speed

Explanation:

Use the relation:

Acceleration (a) = [final velocity(v2) - initial velocity (v1)]/time (t)

t = (v2-v1)/a = 30-0/6.1 = 4.92 s

Ans: Time taken is around 4.9 s

3 0

3 years ago

How many liters are in 5.42 mols of O2 gas? (2 decimal places)

Debora [2.8K]

Explanation:

1 mol = 22.4 l

5.42 mol = 22.4 × 5.42 = 121.408

in two decimal place it is 121.41

7 0

3 years ago