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MArishka [77]
3 years ago
6

Explain the similarities and differences between putting a beaker of ethanoic acid in the refrigerator and mixing it with sodium

carbonate
Chemistry
1 answer:
frez [133]3 years ago
5 0

Answer:

See explanation

Explanation:

When a beaker of ethanoic acid is placed in the refrigerator, its temperature drops and the vessel feels cool.

Now, when we mix ethanoic acid and sodium carbonate, an endothermic reaction occurs, fizzing is observed as carbon dioxide is given off and heat is lost to the surroundings causing the reaction vessel to feel cool to touch.

The difference between putting ethanoic acid in the refrigerator and adding  sodium carbonate to the solution is that, in the former, no new substance is formed. The substance remains ethanoic acid when retrieved from the refrigerator. In the later case, new substances are formed. The substance is no more ethanoic acid because a chemical reaction has taken place.

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Calculate the no. of each atoms presents in 10 grams of calcium carbonate.
raketka [301]

Molar Mass of Calcium carbonate:-

\\ \sf\longmapsto CaCO_3

\\ \sf\longmapsto 40u+12u+3(16u)

\\ \sf\longmapsto 52u+48u

\\ \sf\longmapsto 100u

\\ \sf\longmapsto 100g/mol

  • Given Mass=10g

\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}

\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}

\\ \sf\longmapsto No\:of\:moles=0.1mol

Now

\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}

\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}

\\ \sf\longmapsto 6.023\times 10^{22}molecules

8 0
3 years ago
What reaction is depicted by the given equation: au3 3e− au
mrs_skeptik [129]
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3 years ago
A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture co
DaniilM [7]

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = <em>20144 mol air/h</em>

<em></em>

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = <em>289 kg O₂</em>

43058 mol air×29g/mol <em>1249 kg air</em>

Percent of oxygen is: \frac{289kg}{1249 kg} =<em>0,231 kg O₂/ kg air</em>

<em></em>

I hope it helps!

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