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1 year ago

Choose the right reagent or series of reagents from the ones listed below to prepare 2-methyl-3-hexyn-2-ol from acetylene.

Chemistry

1 answer:

Feliz [49]1 year ago

7 0

2-methyl-3-hexyn-2-ol can be prepared from Acetylene by treating Acetylene with NaNH₂ followed by CH₃CH₂Br, then treating the intermediate with NaNH₂ followed by acetone and then doing aqueous workup. NaNH₂ acts as base and abstracts proton from acetylene as the terminal alkynes are acidic in nature. While acetone on reduction gives tertiary alcohol.

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Determine the theoretical yield of H2S (in moles) if 4.0 molAl2S3 and 4.0 mol H2O are reacted according to the followingbalanced

mamaluj [8]

Answer:

The theoretical yield of the hydrogen sulfide is 68.0 grams.

Explanation:

$Al_2S_3(s)+6H_2O(l)\rightarrow 2Al(OH)_3(s)+3H_2S(g)$

Moles of aluminum sulfide = 4.0 mol

Moles of water = 4.0 mol

According to reaction, 6 moles of water reacts with 1 mole of aluminum sulfide,then 4 moles of water will react with :

$\frac{1}{6}\times 4 mol=1.5 mol$ of aluminum sulfide

1.5 moles aluminum sulfide < 4 moles aluminum sulfide

This means that water is present in limiting amount and aluminum sulfide is in excess amount.So, amount of hydrogen sulfide will depend upon moles of water.

According to reaction, 6 moles of water gives with 3 mole of hydrogen sulfide,then 4 moles of water will give :

$\frac{3}{6}\times 4 mol=2.0 mol$ of hydrogen sulfide

Mass of hydrogen sulfide:

2.0 mol × 34 g/mol = 68.0 g

The theoretical yield of the hydrogen sulfide is 68.0 grams.

7 0

1 year ago

Forming a hypothesis is accomplished through _______ reasoning.

vfiekz [6]

Deductive reasoning involves making broad observations and examining possibilities to reach a specific, logical conclusion, which is required in the formation of a hypothesis. Conversely, inductive reasoning involves taking specific observations and making broad generalisations from these. Therefore the answer is A.

Hope this helps!

4 0

1 year ago

Read 2 more answers

If a 300.0 ml sample of a gas is heated at constant pressure from 25.0ºc to 55.0ºc, its new volume is _____ ml?

jekas [21]

The new volume is 330.2 ml

calculation

The new volume is calculated using the Charles law formula

that is V1/T1= V2/T2

where T1= 25.0 c into kelvin = 25 +273 = 298 K

V1= 300.0 ml

T2 = 55.0 c into kelvin = 273 +55 =328 K

V2 = ? ml

make V2 the subject of the formula by multiplying both side by T2

V2= V1T2/ T1

V2 =[ (300.0 ml x 328 k) / 298 k} = 330.2 ml

8 0

2 years ago

You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2

avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

M(TNT) = 0.150 g; mass of the trinitrotoluene.

ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

n(CO₂) = 0.204 g ÷ 44 g/mol.

n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

m(C) = 0.0552 g; mass of the carbon.

n(H₂O) = 0.030 g ÷ 18 g/mol.

n(H₂O) = 0.00166 mol.

n(H) = 2 · n(H₂O) = 0.0033 mol; amount of the hydrogen.

m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

m(O) = 0.06375 g.

n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

n(C) : n(N) : n(O) : n(H) = 0.0046 mol : 0.002 mol : 0.004 mol : 0.0033 mol.

n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

8 0

2 years ago

In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th

zmey [24]

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

$w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms$

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

=0.045 gms

6 0

2 years ago