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Vladimir79 [104]
3 years ago
11

12. How is the melting point of a substance related to its freezing point? Both indicate the temperature at which the solid and

liquid states of a substance are in equilibrium. The melting point of a substance is a slightly higher temperature than its freezing point. The melting point of a substance is a slightly lower temperature than its freezing point.
Chemistry
2 answers:
krok68 [10]3 years ago
8 0
Both indicate the temperature at which the solid and liquid states of a substance are in equilibrium would be your answer.

This is beacause the melting point of a substance is the same as the freezing point of a substance. At this particular temp, the substance can be either a solid or a liquid. 

hope this helps!
olga nikolaevna [1]3 years ago
7 0

The correct answer is both indicate the temperature at which the solid and liquid states of a substance are in equilibrium  

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FromTheMoon [43]

Answer:

a. NH3 is limiting reactant.

b. 44g of NO

c. 40g of H2O

Explanation:

Based on the reaction:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)

4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.

To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:

<em>a. </em><em>Moles NH3 -Molar mass. 17.031g/mol-</em>

25g NH3*(1mol/17.031g) = 1.47moles NH3

Moles O2 = 4 moles

For a complete reaction of 4 moles of O2 are required:

4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.

As there are just 1.47 moles, NH3 is limiting reactant

b. Moles NO:

1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO

Mass NO -Molar mass: 30.01g/mol-

1.47mol NO * (30.01g/mol) = 44g of NO

c. Moles H2O:

1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O

Mass H2O -Molar mass: 18.01g/mol-

2.205mol H2O * (18.01g/mol) = 40g of H2O

3 0
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Andrew [12]

Given :

2NOBr(g) - -> 2NO(g) + Br2(g)

Initial pressure of NOBr , 1 atm .

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To Find :

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Solution :

             2NOBr(g) - -> 2NO(g) + Br2(g)

t=0 s           1 atm                 0             0

t=t_{eqb}       1( 1-2x)               2x           x

So ,

1-2x=0.82\\\\x=0.09

At equilibrium :

K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm

Hence , this is the required solution .

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