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ivann1987 [24]
3 years ago
6

How many formula units of CaCO3 are present in 4.72 n of calcium carbonate?

Chemistry
1 answer:
arlik [135]3 years ago
4 0

The number of formula units is calculated with the use of Avogadro's constant. That is, on the basis of Avogadro's law:

1 mole = 6.02 × 10²³ formula units

4.72 moles = x formula units

By cross multiplication, we get:

4.72 moles × 6.02 × 10²³ formula units / 1 mole = 2.841 × 10²⁴ formula units.

Thus, the number of formula units of CaCo₃ that are found in 4.72 moles calcium carbonate is 2.841 × 10²⁴.

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What allowed Mendeleev to make predictions of undiscovered elements
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Answer:

he predicted the properties from known elements above and belws the unknown in the same group

Explanation:

What allowed Mendeleev to make predictions of undiscovered elements

He realized that an element on this table with one known element above it and one known  element below it had to have properties between the two known elements

How did Mendeleev predict gallium and germanium?

Based on gaps in the periodic table Mendeleev deduced that in these gaps belonged elements yet to be discovered. Based on other elements below and above in the same group he predicted the existence of eka-aluminum, eka-boron, and eka-silicon, later to be named gallium (Ga), scandium (Sc), and germanium (Ge).

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2 years ago
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Answer:

Divergent plate boundary

Explanation:

7 0
3 years ago
a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh
Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

Molarity = M = \frac{1,22 mol solute}{lts solution}

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution

With the molecular weight of solute (<em>Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol</em>), we can obtain the mass of solute:

1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: <em>Moles solution = moles solute + moles solvent.</em> First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent  

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03

6 0
3 years ago
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Explanation:

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timofeeve [1]
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