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katrin2010 [14]
3 years ago
9

How many total carbon atoms are in the structure 2 methyl, 3, 4 diethyl decane?

Chemistry
1 answer:
makvit [3.9K]3 years ago
6 0

Answer:B

15 carbon atoms

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The numerator is the blank number in a fraction, while the denominator is the blank number in a fraction
tigry1 [53]
The top number of a portion is called its numerator whereas the bottom number is its denominator. So a Fraction is the quantity of shaded parts separated by the quantity of a balance of as demonstrated as follows: number of shaded parts is the numerator over the whole part which is the denominator.
8 0
3 years ago
How many grams of HF are needed to react with 3.0 moles of Sn?
Flauer [41]

Answer:

120g

Explanation:

Step 1:

We'll begin by writing the balanced equation for the reaction.

Sn + 2HF —> SnF2 + H2

Step 2:

Determination of the number of mole HF needed to react with 3 moles of Sn.

From the balanced equation above,

1 mole of Sn and reacted with 2 moles of HF.

Therefore, 3 moles Sn will react with = 3 x 2 = 6 moles of HF.

Step 3:

Conversion of 6 moles of HF to grams.

Number of mole HF = 6 moles

Molar Mass of HF = 1 + 19 = 20g/mol

Mass of HF =..?

Mass = number of mole x molar Mass

Mass of HF = 6 x 20

Mass of HF = 120g

Therefore, 120g of HF is needed to react with 3 moles of Sn.

3 0
3 years ago
30 POINTS! ----- How many moles of oxygen gas are needed to completely react with 145 grams of aluminum? Report your answer with
gladu [14]
First a balanced reaction equation must be established:
4Al _{(s)}    +   3 O_{2}  _{(g)}     →    2 Al_{2} O_{3}

Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
                                                                                    =  4.83 mols

Now the mole ratio of Al : O₂ based on the equation is  4 : 3  
                                                                [4Al  + 3 O₂ → 2 Al₂O₃]

∴ if moles of Al = 4.83 moles
  then moles of O₂ = (4.83 mol ÷ 4) × 3
                              =  3.63 mol   (to  2 sig. fig.) 

Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum. 


     
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Draw the orbital shape of s
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