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anyanavicka [17]
3 years ago
7

How many moles is 50g of CaCO3?

Chemistry
1 answer:
Neko [114]3 years ago
3 0

\huge\underline\mathbb\pink{♡Your Answer♡}

Formula mass of CaCO3 is

40 + 12 + 3-100amu....

100g CaCO3 = 1 mole..

50g of CaCO3 = 1÷ 100x 5 = 0.5mole...

<h3>Hence ,answer is 0.5mole...</h3>

Hope it helps you..

Thanks...

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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

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Read 2 more answers
_Fe2O3 + 2CO —&gt; _Fe + _CO2
soldi70 [24.7K]
<h3>Answer:</h3>

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

<h3>Explanation:</h3>

Concept tested: Balancing of chemical equations

  • A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
  • Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
  • In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
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      Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

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The art of compounding employs advanced medicine while still sticking to the roots of the profession of pharmacy. Compounding pharmacies can create unparalleled dosage forms based on patient preferences. Compounded medicines are usually reliable and sufficient for most victims who have spoken to their physician about their wellness.

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