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Bingel [31]
3 years ago
11

A valid lewis structure of __________ cannot be drawn without violating the octet rule. a valid lewis structure of __________ ca

nnot be drawn without violating the octet rule. sbf3 if3 so42− pf3 nf3
Chemistry
1 answer:
Oxana [17]3 years ago
5 0
A valid Lewis structure of IF3 cannot be drawn without violating the octet rule.

Answer: IF3 (Iodine Trifluoride)

This is because, I (Iodine) and F (Fluorine) both have odd number of valence electrons (7) which also means that there are too many valence electrons in the valence shell.
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Counting atoms worksheet answer key i need help!!
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6 0
3 years ago
Complete the conceptap below . About Chemical Bonds ​
Irina18 [472]

Answer:

1. ionic bonds

2. metallic bonds

3. share

4. metal

5. non-metal

6. metals

7. NaCl ( sodium chloride )

8. CO2 ( carbon dioxide )

9. Cu ( copper )

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

5 0
2 years ago
Please help i am striggling with science so bad
Dennis_Churaev [7]

well, yes it will continue to swing but not forever. it will just a long time but eventually stop. the reason is because of the air resistance which will continue to damp the motion until the bob stops

3 0
3 years ago
Which of the following statements does NOT describe the anomeric carbon? A. This carbon is attached to two oxygens. B. This carb
MaRussiya [10]

Answer:

The answer is E. All of the statements describe the anomeric carbon.

Explanation:

When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.

As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).

It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).

The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)

It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.

8 0
3 years ago
When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.
Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2

And the resulting percent yield:

Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%

Regards!

4 0
3 years ago
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