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Pepsi [2]
3 years ago
5

Which of the following changes can be reversed by changing the temperature: (a) dew condensing on a leaf; (b) an egg turning har

d when it is boiled; (c) ice cream melting; (d) a spoonful of batter cooking on a hot griddle?
Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

(a) dew condensing on a leaf

(c) ice cream melting

Explanation:

The process which can be reversed by changing the temperature , i.e. , by increasing or decreasing the temperature , is called the <u>reversible process</u> .

From the question ,  

  • dew condensing on a leaf is a reversible process ,  

As the water on the leaf can be changed to gaseous state by increasing the temperature .  

The process of condensation and evaporation are reversible of each other .

  • Ice cream melting

The melted ice cream can be solidified again by reducing its temperature , hence , it is a reversible process .

  • an egg turning hard when boiled and the batter cooking on a hot griddle

They both are irreversible process , and by changing the temperature , we can not get back the original substance .

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Shkiper50 [21]

Answer:

I  can answer some of these.

Explanation:

Q 36.  B. Is false.

Q. 37  C. is false.

Q. 28.

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5 0
3 years ago
Bromine (Br2) is produced by reacting HBr with O2, with water as a byproduct. The O2 is part of an air (21 mol % O2, 79 mol % N2
Karolina [17]

Answer:

The mole fractions:

x_{HBr}=\frac{100mol}{318.5}=0.314

x_{Br_2}=\frac{78mol}{318.5}=0.245

x_{H_2O}=\frac{78mol}{318.5}=0.245

x_{O_2}=\frac{62.5mol}{318.5}=0.196

Explanation:

The reaction described is:

2 HBr (g) + 1/2 O_2 (g) \longrightarrow Br_2 (g) + H_2O (g)

The limiting reactant is the HBr (oxygen is in excess).

a) The mass (in moles) balance for this sistem:

n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *n_{HBr}*0.78

(the 0.78 is because of the fractional conversion)

n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *n_{HBr}*1.25

(the 1.25 is because of the oxygen excess)

n_{H_2O}=\frac{ 1 mol H_2O}{1 mol Br_2} *n_{Br_2}

There is only one degree of freedom in this sistem, you can either deffine the moles of HBr you have or the moles of Br2 you want to produce. The other variables are all linked by the equations above.

b) Base of calculation 100 mol of HBr:

nn_{HBr}=100 mol HBr

n_{Br_2}=\frac{ 1 mol Br_2}{1 mol HBr} *100mol HBr*0.78

n_{Br_2}=78 mol Br2

n_{O_2}=\frac{ 0.5 mol O_2}{1 mol HBr} *100 mol HBr*1.25

n_{O_2}=62.5 mol O_2

n_{H_2O}=n_{Br_2}= 78 mol

n_{total}=(78+78+100+62.5)mol= 318.5mol

The mole fractions:

x_{HBr}=\frac{100mol}{318.5}=0.314

x_{Br_2}=\frac{78mol}{318.5}=0.245

x_{H_2O}=\frac{78mol}{318.5}=0.245

x_{O_2}=\frac{62.5mol}{318.5}=0.196

4 0
3 years ago
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Gemiola [76]

Answer:

I am pretty sure that is a : Catalyst but if i'm not correct sorry

Explanation:

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Answer:

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Answer:

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