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Komok [63]
3 years ago
15

Which of the following is not a base? A: NH3 B. NaOH C. H3C6H5O7 D. KOH

Chemistry
2 answers:
olga2289 [7]3 years ago
7 0
I think that C is the answer.....
irina1246 [14]3 years ago
7 0

Answer: Option (C) is the correct answer.

Explanation:

A base is a substance or specie that contains a lone pair of electron or a hydroxyl group, that is, OH group.

For example, NH_{3}, NaOH, and KOH are all bases.

But H_{3}C_{6}H_{5}O_{7} is not a base because there is no lone pair of electron and no hydroxyl group.

Thus, we can conclude that out of the given options H_{3}C_{6}H_{5}O_{7} is not a base.

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What is the mole to mole relationship between copper and zinc? Cu2+(aq)+->Cu(s)+Zn2+(aq)
Aliun [14]

The reaction is missing the Zn(s) in the reactants. The stoichiometry of the copper/zinc is 1 mole to 1 mole

5 0
3 years ago
Three carbon atoms are linked by single covalent bonds such that they form the shape of a V. All of the unshared electrons form
Nostrana [21]

Answer:

C) 8

Explanation:

Total number of carbon atoms = 3

Number of single bonds = 3

So, each carbon is bonded to the next carbon with the single bond, Number of unshared electrons left with the terminate carbons are 3 and with the intermediate carbon is 2.

Thus, the two terminate carbon have 3 hydrogen each and the intermediate will have 2.

<u>Total - 8</u>

6 0
3 years ago
Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles
  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles* 58 g/mole= 116 g
  • O₂: 13 moles* 32 g/mole= 416 g
  • CO₂: 8 moles* 44 g/mole= 352 g
  • H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

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