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Marta_Voda [28]
3 years ago
7

Un estudiante debe preparar una disolución 2M de NaCl (58,44 g/mol) en un recipiente de 7L. ?Cuántos gramos de NaCl debe agregar

?
Chemistry
1 answer:
Gnesinka [82]3 years ago
3 0

Answer:

818.2 g.

Explanation:

  • Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of NaCl)/(Volume of the solution (L))</em>

<em></em>

M = 2.0 M.

no. of moles of NaCl = ??? mol,

Volume of the solution = 7.0 L.

∴ (2.0 M) = (no. of moles of NaCl)/(7.0 L)

∴ (no. of moles of NaCl) = (2.0 M)*(7.0 L) = 14.0 mol.

  • To find the mass of NaCl, we can use the relation:

<em>no. of moles of NaCl = mass/molar mass</em>

<em></em>

<em>∴ mass of NaCl = (no. of moles of NaCl)*(molar mass) =</em> (14.0 mol)*(58.44 g/mol) = <em>818.2 g.</em>

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3 0
3 years ago
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
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The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

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Which section of the reaction represents the reactants
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A should be the products and D should be the reactants. So D should be the answer.
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