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Viefleur [7K]
3 years ago
15

How many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?

Chemistry
2 answers:
miv72 [106K]3 years ago
6 0

Answer: 6.75 moles

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n}{V_s}

Given;

Molarity = 4.5 M

n= moles of solute  (glucose) = ?

 V_s = volume of solution in L = 1.5 L

Putting in the values we get,

4.5=\frac{n}{1.5}

n=6.75moles

Thus moles of glucose are 6.75 moles in 1.5 liters of a 4.5 M C_6H_{12}O_6 solution.

zepelin [54]3 years ago
3 0
For the answer to the question above asking, h<span>ow many moles of glucose (C6H12O6) are in 1.5 liters of a 4.5 M C6H12O6 solution?
The answer to your question is the the third one among the given choices which is 6.8 mol.
</span><span>moles glucose = 1.5 x 4.5 = 6.8 </span>
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The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate
vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

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

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

            k = Ae^-(Ea/RT)

where,   k = rate constant

              A = pre-exponential factor

          Ea  = activation energy

             R = gas constant

              T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

         

      T1= (244 + 273.15) K = 517.15 K

      T2= (324+ 273.15) K =597.15 K

    K1  = 6.7 ,     K2 = ?

         R = 8.314 J/mol K

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Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

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Answer:

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Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

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Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

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