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1 year ago

you are riding in a train that is traveling at a speed of 120km/h how long will it take to travel 950km

Chemistry

1 answer:

muminat1 year ago

4 0

The time take for the train to travel 950 Km is 7.92 h

<h3>What is speed? </h3>

Speed is the distance travelled per unit. Mathematically, it can be expressed as:

Speed = distance / time

With the above formula, we can determine the time. Details below:

<h3>How to determine the time</h3>

The time taken for the train t travel 950 Km can be obtained as follow:

Speed = 120 Km/h
Distance = 950 Km
Time =?

Speed = distance / time

120= 950 / time

Cross multiply

120 × time = 950

Divide both side by 120

Time = 950 / 120

Time = 7.92 h

Learn more about speed:

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The net equation will be as follows.

$K(s) + Cl_{2}(g) \rightarrow KCl(s)$

So, we are required to find $\Delta H_{formation}$ for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1: Convert K from solid state to gaseous state

$K(s) \rightarrow K(g)$ , $\Delta H_{1}$ = 89 kJ

Step 2: Ionization of gaseous K

$K(g) \rightarrow K^{+}(g) + e^{-}$ , $H_{2}$ = 418 KJ

Step 3: Dissociation of $Cl_{2}$ gas into chlorine atom .

$\frac{1}{2} Cl_{2}(g) \rightarrow Cl(g)$ , $\Delta H_{3} = \frac{244}{2}$ = 122 KJ

Step 4: Iozination of chlorine atom.

$Cl(g) + e^{-} \rightarro Cl^{-}(g)$ , $H_{4}$ = -349 KJ

Step 5: Add $K^{+}$ ion and $Cl^{-}$ ion formed above to get KCl .

$K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s)$ , $H_{5}$ = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

$\Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}$

= 89 + 418 + 122 - 349 - 717

= - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

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