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natulia [17]
3 years ago
14

Calculate the freezing temperature of the following solution of 0.50 M glucose (a covalent compound). Assume that the molality o

f the solution is 0.50 m. (The molar and molal concentrations of dilute aqueous solutions are often identical to two significant figures.) Enter your answer in the provided box. 0.50 m glucose (a covalent compound) °C
Chemistry
1 answer:
kirza4 [7]3 years ago
4 0

Answer:

-0.93 °C

Explanation:

Hello,

The freezing-point depression is given by:

T_f-T_f^*=-iK_{solvent}m_{solute}

Whereas T_f is the freezing temperature of the solution, T_f^* is the freezing temperature of the pure solvent (0 °C since it is water), i the Van't Hoff factor (1 since the solute is covalent), K_{f,solvent} the solvent's freezing point depression point constant (in this case 1.86 C\frac{kg}{mol}) and m_{solute} the molality of the glucose.

As long as the unknown is T_f, solving for it:

T_f=T_f^*-iK_fm\\T_f=0C-1*1.86C\frac{kg}{mol}*0.5\frac{mol}{kg}  \\T_f=-0.93C

Best regards.

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Answer:

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Explanation:

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7 0
3 years ago
7. If you start with 3.20 mol of Li3N and 9.0 mol water, how many moles of NH3 will be formed?
pickupchik [31]

Taking into account the reaction stoichiometry, 3 moles of NH₃ will be formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Li₃N + 3 H₂O → NH₃ + 3 LiOH

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Li₃N: 1 mole
  • H₂O: 3 moles
  • NH₃: 1 mole
  • LiOH: 3 moles

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Li₃N reacts with 3 moles of H₂O, 3.20 moles of Li₃N reacts with how many moles of H₂O?

moles of H_{2} O=\frac{3.2 moles of Li_{3}Nx 3 moles of  H_{2} O}{1 mole of Li_{3}N}

<u><em>moles of H₂O= 9.6 moles</em></u>

But 9.6 moles of H₂O are not available, 9 moles are available. Since you have less moles than you need to react with 3.20 moles of Li₃N, H₂O will be the limiting reagent.

<h3>Mass of NH₃ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of H₂O form 1 mole of NH₃, 9 moles of H₂O form how many moles of NH₃?

molesof NH_{3} =\frac{9 moles H_{2} Ox1 mole of NH_{3}}{3 moles H_{2} O}

<u><em>moles of NH₃= 3 moles</em></u>

Finally, 3 moles of NH₃ will be formed.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

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Explanation:

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For the reactions below, describe the reactor system and conditions you suggest to maximize the selectivity to make the desired
alexandr1967 [171]

Answer:

hello your question lacks the required reaction pairs below are the missing pairs

Reaction system 1 :

A + B ⇒ D  -r_{1A}  = 10exp[-8000K/T]C_{A}C_{B}

A + B ⇒ U -r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}

Reaction system 2

A + B ⇒ D  -r_{1A} = 10exp( -1000K/T)C_{A}C_{B}

B + D ⇒ U  -r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}

Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

reaction 2 :

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

Explanation:

Reaction system 1 :

A + B ⇒ D  -r_{1A}  = 10exp[-8000K/T]C_{A}C_{B}

A + B ⇒ U -r_{2a} = 100exp(-1000K/T)C_{A} ^\frac{1}{2}C_{B} ^\frac{3}{2}

the selectivity of D is represented  using the relationship below

S_{DU} = \frac{-r1A}{-r2A}

hence SDu = 1/10 * \frac{exp(-800K/T)}{exp(-1000K/T)} * C_{A} ^{0.5} C_{B} ^{-0.5}

description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1

condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2

Reaction system 2

A + B ⇒ D  -r_{1A} = 10exp( -1000K/T)C_{A}C_{B}

B + D ⇒ U  -r_{2B} = 10^9exp(-10000 K/T) C_{B}C_{D}

selectivity of D

S_{DU} = \frac{-r1A}{-r2A}

hence Sdu = 1/10^7  *  \frac{exp(-1000K/T)}{exp(-10000K/T)} *\frac{C_{A} }{C_{D} }

description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1

condition to maximize selectivity:

to increase selectivity the concentration of D should be minimal

3 0
3 years ago
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