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ioda
3 years ago
7

Znco3 has a molar mass of 125.40 g/mol. which of the conversion factors shown below correctly expresses this relationship as wel

l as the relationship between molar mass and the number of formula units of the compound? avogadro's number is 6.022 × 1023.
Chemistry
2 answers:
Naddika [18.5K]3 years ago
8 0

Answer:

the correct expression is 125.4 g ZnCO3 * 1 mol ZnCO3/6.022x10^23 formula units ZnCO3

Explanation:

The conversion factor that correctly expresses the relationship between the molar mass and the number of formula units of the compound is as follows.

Avogadro's number is defined as 6.022x10^23 particles/mole, therefore the conversion would be equal to:

125.4 g ZnCO3 * 1 mol ZnCO3/6.022x10^23 formula units ZnCO3

Airida [17]3 years ago
4 0
Answer : ZnCO_{3} will be 1 mole of ZnCO_{3} / 125. 40 g of ZnCO_{3}

Explanation : The molar mass and the formula units are the same thing in any compound so when given a molar mass of compound then one mole of that compound is divided by its molar mass. Where one mole will be equal to avogadro's number which is 6.022 X 10^{23}.
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Complete Question:

A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.

Answer:

13 mol/L

Explanation:

The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:

M = n/V

The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:

n = 134/207.319

n = 0.646 mol

So, for a volume of 50 mL (0.05 L), the concentration is:

M = 0.646/0.05

M = 12.92 mol/L

Rounded to 2 significant digits, M = 13 mol/L

7 0
3 years ago
Nucleophilicity is a kinetic property. A higher nucleophilicity indicates that the nucleophile will easily donate its electrons
lord [1]

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K = A(e^(-Ea/RT))

Taking natural log of both sides

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Comparing this to the equation of a straight line; y = mx + c

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c) K = A(e^(-Ea/RT))

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K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

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