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Alex Ar [27]
1 year ago
6

A jogger runs a mile in 8.92 min. Calculate the speed in:

Chemistry
1 answer:
Reil [10]1 year ago
3 0

A. The speed (m/s) of the car is 3 m/s

B. The speed (m/min) of the car is 180.4 m/min

A. The speed (Km/h) of the car is 10.8 Km/h

<h3>What is speed? </h3>

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

Speed = distance / time

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

A. The speed (m/s) can be obtained as follow:

  • Distance = 1 mile = 1609.34 m
  • Time = 8.92 mins = 8.92 × 60 = 535.2 s
  • Speed (m/s) =?

Speed = distance / time

Speed = 1609.34 / 535.2

Speed (m/s) = 3 m/s

B. The speed (m/min) can be obtained as follow:

  • Distance = 1 mile = 1609.34 m
  • Time = 8.92 mins
  • Speed (m/min) =?

Speed = distance / time

Speed = 1609.34 / 8.92

Speed (m/min) = 180.4 m/min

C. The speed (Km/h) can be obtained as follow:

  • Distance = 1 mile = 1.60934 Km
  • Time = 8.92 mins = 8.92 / 60 = 0.149 h
  • Speed (Km/h) =?

Speed = distance / time

Speed = 1.60934 / 0.149

Speed (Km/h) = 10.8 Km/h

Learn more about speed:

brainly.com/question/680492

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Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

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