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3 years ago

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A student dissolves equal amounts of salt in equal amounts of warm water, room-temperature water, and ice water. Which result is

most likely?
The salt dissolved fastest in the warm water.

The salt dissolved fastest in the ice water.

The salt dissolved in the same amount of time at all three temperatures.

The salt dissolved fastest in the room-temperature water.

I chose the first one but can someone double check for me? Thx

2 answers:

Tanya [424]3 years ago

6 0

doesnt salt desolve ice? so wouldn't the salt dissolve in the ice water?

ziro4ka [17]3 years ago

3 0

Answer:

The salt dissolved fastest in the warm water.

Step-by-step explanation:

You were correct.

The salt dissolved faster in the warm water, because the water molecules have more kinetic energy to dislodge the salt particles from the surface of the crystal.

The colder the water, the less kinetic energy the water molecules will have, and the more slowly the salt will dissolve.

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

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Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

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Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how

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Answer:

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Explanation:

1. The balanced chemical equation is the following:

$Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)$

2. Use the molar mass of the $H_{2}O$ , the molar mass of the $NH_{3}$ and the stoichiometry of the balanced chemical reaction to find how many grams of $NH_{3}$ are produced:

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$36.0gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesNH_{3}}{6molesH_{2}O}*\frac{17gNH_{3}}{1molNH_{3}}=11.3gNH_{3}$

Therefore 11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

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