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Julli [10]
3 years ago
13

A tank originally contains 100 gallons of fresh water. Then, water containing 0.5 lb of salt per gallon is poured into the tank

at a rate of 2 gallons per minute, and the mixture is allowed to leave at the same rate. After 10 minutes, the process is stopped and fresh water is poured in the tank at a rate of 9 gallons per minute, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 minutes.
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
7 0

Answer:

The amount of salt in the tank at the end of an additional 10 minutes are 4.38lb.

Explanation:

<u>Situation 1:</u>

Tank with 100 gallons of fresh water

<u>Situation 2:</u>

Tank with 100 gallons of fresh water + water with 0.5lb of salt per gallon

After 10 minutes, as the rate in which the new water is poured is 2 gallons per minute, the result is 20 gallons added (2×10=20) . And taking in account that the water contains 0.5 lb of salt per gallon the amount of salt added is 20×0.5= 10lb of salt.

That amount of salt is now in all the water inside the tank which is 100 gallons+ 20 gallons= 120 gallons. <em>That means that in situation 2 we have 10lb of salt in 120 gallons of water.</em>

That mixture is allowed to leave the tank at a rate of 2 gallons per minute so we will have after 10 minutes: 120 gallons- (2×10) gallons= 100 gallons remaining in the tank. And the amount of salt if we remember that we had 10lb in 120 gallons, now in 100 gallons we will have: (100 gallons × 10lb of salt)/  120 gallons= 8.33 lb of salt.

<u>Situation 3:</u>

Tank with 100 gallons of water with 8.33lb of salt.

After 10 minutes in which fresh water is poured in the tank at a rate of 9 gallons per minute, the result is: 9×10= 90 gallons added to the tank. So now we have 100+90=190 gallons of water in  the tank. <em>That means in situation 3 we have 8.33 lb of salt in 190 gallons of water. </em>

That mixture is leaving the tank at a rate of 9 gallons per minute so we have after 10 minutes: (190- (9×10))= 100 gallons of mixture remaining in the tank.

And the amount of salt if we remember that we had 8.33lb in 190 gallons, now in 100 gallons we will have: (100 gallons × 8.33lb of salt)/  190 gallons= 4.38 lb of salt.

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The system described by the reaction CO(g)+Cl2(g)⇌COCl2(g) is at equilibrium at a given temperature when PCO= 0.26 atm , PCl2= 0
givi [52]

<u>Answer:</u> The new pressure of CO is 0.09 atm

<u>Explanation:</u>

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CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}             .......(1)

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Putting values in above equation, we get:

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Now, the equilibrium is re-established:

                    CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial:            0.26     0.15             0.66

At eqllm:     0.26-x   0.54            0.66+x

Putting values in expression 1, we get:

16.92=\frac{(0.66+x)}{(0.26-x)\times 0.54}\\\\x=0.17

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4 0
3 years ago
Calculate Delta H in KJ for the following reactions using heats of formation:
lozanna [386]

Answer:

<h3>(a)</h3>

\Delta H\textdegree = -2856.8\;\text{kJ} per mole reaction.

<h3>(b)</h3>

\Delta H\textdegree = -22.3\;\text{kJ} per mole reaction.

Explanation:

What is the standard enthalpy of formation \Delta H_f\textdegree{} of a substance? \Delta H_f\textdegree{} the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, \Delta H_f\textdegree{} = 0 for the most stable allotrope of each element under standard conditions. For example, oxygen \text{O}_2 (not ozone \text{O}_3) is the most stable allotrope of oxygen. Also, under STP \text{O}_2  is a gas. Forming \text{O}_2\;(g) from itself does not involve any chemical or physical change. As a result, \Delta H_f\textdegree{} = 0 for \text{O}_2\;(g).

Look up standard enthalpy of formation \Delta H_f\textdegree{} data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam \text{H}_2\text{O} in particular) and the sign of the enthalpy changes.

  • \text{C}_2\text{H}_6\;(g): -84.0\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{CO}_2\;(g): -393.5\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{H}_2\text{O}\;{\bf (g)}: -241.8\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{PbO}\;(s): -217.9\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{PbO}_2\;(s): -276.6\;\text{kJ}\cdot\text{mol}^{-1};
  • \text{Pb}_3\text{O}_4\;(s): -734.7\;\text{kJ}\cdot\text{mol}^{-1}

How to calculate the enthalpy change of a reaction \Delta H_\text{rxn} (or simply \Delta H from enthalpies of formation?

  • Multiply the enthalpy of formation of each product by its coefficient in the equation.
  • Find the sum of these values. Label the sum \Sigma (n\cdot \Delta_f(\text{Reactants})) to show that this value takes the coefficients into account.
  • Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
  • Find the sum of these values. Label the sum \Sigma (n\cdot \Delta_f(\text{Products})) to show that this value takes the coefficient into account.
  • Change = Final - Initial. So is the case with enthalpy changes. \Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants})).

For the first reaction:

  • \Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1};
  • \Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1};
  • \begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}.

Try these steps for the second reaction:

\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}.

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3 years ago
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