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Andreyy89
3 years ago
11

Without doing any calculations, determine which mass is closest to the atomic mass of carbon. without doing any calculations, de

termine which mass is closest to the atomic mass of carbon. 12.00 amu 12.50 amu 13.00 amu
Chemistry
1 answer:
yarga [219]3 years ago
7 0
The mass closest to the mass of carbon is 12.00 amu

This is because carbon is used as a standard compound in chemistry, and its mass is defined as being 12.00 amu per mole of carbon. Its definition as a standard is such that the mole is defined as, "the amount of substance containing the same number of atoms as 12 grams of carbon-12".
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Citrate synthase catalyzes the reaction: ????x????????o????c???????????????????? + ????c????????y???? − ????o???? → c???????????
Nady [450]

The given question is incomplete. The complete question is as follows.

Citrate synthase catalyzes the reaction

Oxaloacetate + acetyl-CoA \rightarrow citrate + HS-CoA

The standard free energy change for the reaction is -31.5 kJ*mol^-1

( a) Calculate the equilibrium constant for this reaction a 37degrees C

Explanation:

(a).  It is known that , relation between change in free energy (\Delta G) of a reaction and equilibrium constant (K) is as follows.

             \Delta G = -RT \times ln K  

where,  T = temperature in Kelvin

The given data is as follows.

         T = 310 K,       \Delta G = -31.5 kJ /mol = -31500 J/mol  (as 1 kJ = 1000 J)

Now, putting the given values into the above formula as follows.

     ln K = \frac{-(\Delta G)}{RT}

            = \frac{31500}{8.314 \times 310}

      ln K = 12.22

         K = antilog (12.22)

           = 2.1 \times 10^{5}

Therefore, we can conclude that value of equilibrium constant for the given reaction is 2.1 \times 10^{5}.

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Determine the number of moles of iodine that reacts with 50g of aluminum​
frozen [14]

Answer: The correct answer is option B i.e., 2.78 mol

Explanation:

Aluminium reacts with iodine to form Aluminium iodide

$2Al(s)+3I_{2}(s) \to Al_{2}I_{6}(s)$

From the equation, it is clear that 3 moles of iodine reacts with 2 moles of Aluminium to form Aluminium iodide

We know $No.\,of\,moles=\frac{weight}{molecular\,weight} \\

For 2 moles of Aluminium, $2=\frac{weight}{27}$\\ $weight=2 \times 27=54g$

3 moles of Iodine reacts with 54 g of Aluminium

? moles of iodine react with 50 g of Aluminium

$=\frac{3 \times 50}{54} =2.78 mol$

8 0
3 years ago
Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a
nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

               Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

  • n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

                             =  (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

                                                                                           = 75.3%

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4 years ago
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