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liubo4ka [24]
3 years ago
7

What are the isomers of butanol​

Chemistry
1 answer:
Juliette [100K]3 years ago
7 0

There are three other structural isomers of 1-butanol: 2-butanol (sec-butyl alcohol), 2-methyl-1-propanol (isobutyl alcohol), and 2-methyl-2-propanol (tert-butyl alcohol). 2-Butanol, or sec-butanol, or sec-butyl alcohol, or s-butyl alcohol, is a four-carbon chain, with the OH group on the second carbon.

Chemicals of this type: Ethanol

Hope this helps

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How many molecules of SrCrO4 are in a sample of SrCrO4 0.556 moles?
Amiraneli [1.4K]

Answer:

3.35*10^{23}\ SrCrO_4\ molecules

Explanation:

We\ are\ given\ that,\\No.\ of\ moles\ of\ SrCrO_4=0.556\\Hence,\\As\ we\ know\ that,\\No.\ of\ particles=Avagadro's\ Constant*No.\ of\ moles\\We\ already\ know\ that\ Avagadro's\ Constant=6.022*10^{23}\\Here,\\No.\ of\ SrCrO_4\  molecules= 6.022*10^{23}*0.556\\Hence,\\No.\ of\ SrCrO_4\  molecules=3.348*10^{23} molecules\ \approx 3.35*10^{23}\ SrCrO_4\ molecules

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2 years ago
The recommended daily requirement of calcium is1,000 mg. Tap water in a Midwestern city contains approximately 150 mgCa/L. A per
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The answer is D. 30 percent

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3 years ago
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AnnZ [28]
True- the finches would eat different sizes nuts and seeds as they would occupy different niches over times which lead to differences in beak sizes within the island
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3 years ago
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Explain the difference between the reacts and products of a chemical reaction
Alika [10]
Reactants are what is changed in the chemical reaction. They go before the arrow when you wrote the equation. Products are what is formed when the chemical reaction occurs. Products always contain the same elements as the reactants, but they might be rearranged because of the reaction. In this picture, you can see that the reactants go before the arrow, and the products go after.

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3 years ago
The benzoate ion, c6h5coo− is a weak base with kb=1.6×10−10. how many moles of sodium benzoate are present in 0.50 l of a soluti
lyudmila [28]

NaC6H5COO \rightarrow Na{^{+}} + C6H5COO^{-}

Here the base is a benzoate ion, which is a weak base and reacts with water.

C6H5COO^{-}(aq) + H2O (l)\leftrightarrow C6H5COOH(aq)+ OH^{-}(aq)

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.

Therefore [OH-] = [C6H5COOH]

In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]

pOH = 14 - pH

pH given = 9.04

pOH = 14-9.04 = 4.96

pOH = -log[OH-] or [OH^{-}] = 10^{^{-pOH}}

[OH^{-}] = 10^{^{-4.96}}

[OH^{-}] = 1.1\times 10^{-5}

The base dissociation equation kb = \frac{Product}{Reactant}

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.

Value of Kb is given = 1.6\times 10^{-10}

And value of [OH-] we have calculated as 1.1\times 10^{-5} and value of C6H5COOH is equal to OH-

Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-

kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}

1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}

[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}

[C6H5COO^{-}] = 0.76 M or 0.76\frac{mol}{L}

So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = 0.76(\frac{mol}{L}) \times (0.50L)

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

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3 years ago
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