Ask question

Ask question

All categories

2 years ago

5

29. What is the noble gas configuration for tellurium?

1 answer:

Whitepunk [10]2 years ago

5 0

Answer:

[Kr] 4d10 5s2 5p4 is the electronic configuration for tellurium.

Electrons per shell: 2,8,18,18,6

Atomic number: 52

Symbol: Te

You might be interested in

Is Potassium Acetate an acidic or basic salt?

Softa [21]

Cause potassium comes from strong base and acetate comes from weak acid, when they disociate potassium is stronger base or weaker conjugated acid than acetate is cause acetate is weaker acid or stronger conjugated base... i hope you can get it from this what i wrote

8 0

2 years ago

Which one of the following does not represent 1.000 mol of the indicated

Katen [24]

Answer:

B. 26.00 g Fe

Atomic mass of Fe (Iron) is 55.845g

3 0

3 years ago

Read 2 more answers

If you had to pick one organelle that was the most important, which one would you pick? Why would you pick that one?

Advocard [28]

The mitochondrion because it produces ATP energy so the cell can survive.

8 0

3 years ago

Number 14? Please :)

ASHA 777 [7]

Evaporated is the answer

4 0

2 years ago

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]

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

<u>Explanation:</u>

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)

<u>For lead (II) nitrate:</u>

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:

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

<u>For NaI:</u>

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

2 years ago