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SVETLANKA909090 [29]
3 years ago
10

How many grams of silver are present in 12.044 x 1023 atoms of Ag?

Chemistry
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

The correct answer is 201.7 g

Explanation:

Silver is a chemical element which symbol is Ag. From the Periodic Table, the molar mass of Ag is 100.86 g/mol. Also, we know that 1 mol of atoms is equal to 6,022 x 10²³atoms (Avogadro's number). So, we have the following relations (convertion factors):

1 mol Ag atoms = 6,022 x 10²³ atoms

1 mol Ag = 100.86 g

We use this factors and we calculate the grams og Ag as follows:

g Ag = 12.044  x 10^{23} x \frac{1 mol Ag}{6.022 x 10^{23} atoms } x \frac{100.86 g}{1 mol Ag} = 201.7 g

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Which of the following is an example of a safe laboratory procedure?
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The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5
Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

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