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Angelina_Jolie [31]
3 years ago
8

Find the mass in grams of 2.00x10^23 molecules of F2

Chemistry
1 answer:
earnstyle [38]3 years ago
4 0
Hello!

To find the mass in grams of 2.00 * 10²³ molecules of F₂ you'll need to apply the Avogadro's number, which tells the number of molecules in a mol of a substance, and the Molar Mass, which tells the weight of a mole of a substance. These two quantities are used in the following conversion factor:

2,00*10^{23} moleculesF_2* \frac{1 mol F_2}{6,02*10^{23} moleculesF_2 }* \frac{37,997 g F_2}{1 mol F_2} \\  \\ =12,51 g F_2

So, the mass of 2.00 * 10²³ molecules of F₂ is 12,51 g

Have a nice day!
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The recommended daily intake of potassium ( K ) is 4.725 g . The average raisin contains 3.513 mg K . Fill in the denominators o
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Explanation:

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             \frac{1 g}{1000 mg} or \frac{1000 mg}{1 g}

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The value of Ka for nitrous acid (HNO2) at 25 ∘C is 4.5×10−4 .a. Write the chemical equation for the equilibrium that correspond
kvv77 [185]

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

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   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

6 0
4 years ago
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hope that helps :)
8 0
3 years ago
Read 2 more answers
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