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2 years ago

By the way...... what do those symbols for the elements have in common (for number 7-11): Pb, Au, Cu, Hg, Na ? *

Chemistry

2 answers:

Keith_Richards [23]2 years ago

8 0

Answer:

those have symbols for their Latin or Greek name

Explanation:

hope it helps

galina1969 [7]2 years ago

8 0

Answer:

Na number of 11

Explanation:

Pb no. 82

Au no. 79

Cu no. 29

Hg no. 80

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Explanation:

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Rank the members of each set of compounds in order of decreasing ionic character of their bonds. Use partial charges to indicate

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Answer:

Ionic character

A. PF₃ > PBr₃ > PCl₃

B. BF₃ > CF₄ > NF₃

C. TeF₄ > BrF₃ > SeF₄

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The most electronegative element is fluorine, followed chlorine, phosphorous nitrogen etc.

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Ionic compounds formed between elements with high electronegativity difference.

% ionic character is directly proportional to electronegativity difference.
According to Pauling Scale E.n for F(4.0), O(3.5), N(3.0), C(2.5), B(2.0), P(2.19), Se(2.55) , Te (2.1), Cl(3.16) and Br(2.96)
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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]

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

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