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2 years ago

What is the purpose of sand bath in preparation of ethane

Chemistry

1 answer:

kykrilka [37]2 years ago

7 0

Answer:

The sand bath spreads the heat out so that the flask is heated evenly. This reduces the chance of the flask breaking and ensures that there are no hot spots in the reaction mixture which could lead to excessive charring,

Explanation:

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A solution containing 75.0 mL of 0.150 M strong acid (HCl) is titrated with 75.0 mL of 0.300 M strong base (NaOH). What is the p

Alborosie

Answer:

12.875

Explanation:

Balanced equation of the reaction:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

mole of acid = mole of base

75 × 0.150 = volume of base needed × 0.3

volume of base needed = 37.5 ml

excess OH = 0.3 × 37.5 ml / 150 ml where total volume = 75 ml + 75 ml = 150 ml

excess OH = 0.075 M

pOH = - log (OH⁻) = - log (0.075 M) = 1.125

pH + pOH = 14

pH = 14 - pOH = 14 - 1.125 = 12.875

7 0

2 years ago

How many moles of 2H2 are formed from 3.64 mole of H2O?

tatiyna

Answer: 2 H2 + O2 → 2 H2O, you get the same number of moles of water as H2, as long as you have 1 mole of O2. So, with 3 moles of H2, as long as you have 1.5 moles of O2, you will get 3 moles of H2O. In my opinion.

5 0

2 years ago

How many grams of water will be produced when 1.6 moles of ethanol (CH3CH2OH) are burned completely? Enter a number only (no uni

AfilCa [17]

Answer:

Explanation:

The reaction that takes place is:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

First we convert moles of ethanol to moles of water:

1.6 mol ethanol * $\frac{3molH_2O}{1molEthanol}$ = 4.8 mol H₂O

Then we convert moles of water to grams of water, using its molar mass:

4.8 mol H₂O * 18 g/mol = 86.4 g

So 84.6 grams of water will be produced.

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2 years ago

A buffer was prepared by mixing 1.00 mole of ammonia and 1.00 mole of ammonium chloride to form an aqueous solution with a total

lidiya [134]

Answer:

Idek sorry

Explanation:

So sorry i think its 1.00

3 0

2 years ago

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = 181,66 kJ/mol

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = 0,216 kJ/Kmol

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ 4. +117,1 kJ/mol

I hope it helps!

5 0

3 years ago

What is the purpose of sand bath in preparation of ethane​

What is the purpose of sand bath in preparation of ethane