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2 years ago

8

with strong heating calcium carbonate undergoes thermal decomposition how many mole of CaCO3 are there in 50g of calcium carbona

te

2 answers:

aev [14]2 years ago

6 0

Ca=40
C=12
O=16
1 mole of CaCO3 has 100 grams
So 50 grams is 0.5 mole

devlian [24]2 years ago

5 0

Answer:

a)CaCO 3→CaO+CO 2CaCO3→CaO+CO2

b) Moles of CaCO3=4.37moles

c) I) Mass of calcium oxide =28g

ii)Mass of carbon dioxide=16.7g

iii)Volume=203cm3

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Two identical bottles at the same temperature contain the same gas. If bottle B has twice the volume and contains half the numbe

yuradex [85]

Answer:

The ratio of pressure in bottle B to that of bottle A is 1 : 4

Explanation:

We'll be by calculating the pressure in both bottles. This is illustrated below below:

For A:

Temperature (T) = T

Volume (V) = V

Number of mole (n) = n

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =...?

PV = nRT

PV = n x 0.0821 x T

Divide both side by V

P = nT0.0821/V

Therefore, the pressure, in bottle A is

PA = nT0.0821/V

For B:

Temperature (T) = the same as that of A = T

Volume (V) = twice that of A = 2V

Number of mole (n) = half that of A = ½n

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =...?

PV = nRT

P x 2V = ½n x 0.0821 x T

Divide both side by 2V

P = ½n x 0.0821 x T/2V

P = nT0.0821/4V

Therefore, the pressure in bottle B is:

PB = nT0.0821/4V

Now, we can obtain the ratio of pressure in bottle B to that of bottle A as follow:

Pressure in bottle A (PA) = nT0.0821/V

Pressure in bottle B (PB) = nT0.0821/4V

PB/PA = nT0.0821/4V ÷ nT0.0821/V

PB/PA = nT0.0821/4V x V/nT0.0821

PB/PA = 1/4

Therefore, the ratio of pressure in bottle B to that of bottle A is 1 : 4.

3 0

3 years ago

What happens to an object if the total of the forces acting on it are greater than zero?

d1i1m1o1n [39]

The objects motion will change

5 0

3 years ago

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In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide Al2O3 dissolved in molten c

gtnhenbr [62]

Answer:

1.13 × 10⁶ g

Explanation:

Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.

Al³⁺ + 3 e⁻ → Al

We can establish the following relations:

1 Ampere = 1 Coulomb / second
The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
1 mole of Al is produced when 3 moles of electrons circulate
The molar mass of Al is 26.98 g/mol.

The mass of aluminum produced under these conditions is:

$90.0 s \times \frac{1s}{620c} \times \frac{96,468c}{1mole^{-} } \times \frac{3mole^{-}}{1molAl} \times \frac{26.98gAl}{1molAl} =1.13 \times 10^{6} g Al$

6 0

4 years ago

Which of the following is a lanthanide

Gelneren [198K]

Answer:

Hope it helps.

Explanation:

The lanthanide (/ˈlænθənaɪd/) or lanthanoid (/ˈlænθənɔɪd/) series of chemical elements[1] comprises the 15 metallic chemical elements with atomic numbers 57–71, from lanthanum through lutetium.[2][3][4] These elements, along with the chemically similar elements scandium and yttrium, are often collectively known as the rare earth elements.

The informal chemical symbol Ln is used in general discussions of lanthanide chemistry to refer to any lanthanide. All but one of the lanthanides are f-block elements, corresponding to the filling of the 4f electron shell; depending on the source, either lanthanum or lutetium is considered a d-block element, but is included due to its chemical similarities with the other 14.[5] All lanthanide elements form trivalent cations, Ln3+, whose chemistry is largely determined by the ionic radius, which decreases steadily from lanthanum to lutetium.

They are called lanthanides because the elements in the series are chemically similar to lanthanum. Both lanthanum and lutetium have been labeled as group 3 elements, because they have a single valence electron in the 5d shell. However, both elements are often included in discussions of the chemistry of lanthanide elements. Lanthanum is the more often omitted of the two, because its placement as a group 3 element is somewhat more common in texts and for semantic reasons: since "lanthanide" means "like lanthanum", it has been argued that lanthanum cannot logically be a lanthanide, but IUPAC acknowledges its inclusion based on common usage.[6]

In presentations of the periodic table, the lanthanides and the actinides are customarily shown as two additional rows below the main body of the table,[2] with placeholders or else a selected single element of each series (either lanthanum and actinium, or lutetium and lawrencium) shown in a single cell of the main table, between barium and hafnium, and radium and rutherfordium, respectively. This convention is entirely a matter of aesthetics and formatting practicality; a rarely used wide-formatted periodic table inserts the lanthanide and actinide series in their proper places, as parts of the table's sixth and seventh rows (periods).

The 1985 International Union of Pure and Applied Chemistry “Red Book” (p. 45) recommends that "lanthanoid" is used rather than "lanthanide". The ending “-ide” normally indicates a negative ion. However, owing to wide current use, “lanthanide” is still allowed.

6 0

3 years ago

Read 2 more answers

How much faster does helium escape through a porous container than ozone?

otez555 [7]

According to Graham's Law ," the rates of effusion or diffusion of two gases are inversely proportional to the square root of their molecular masses at given pressure and temperature".

r₁ / r₂ =   $\sqrt{M2 / M1}$ ---- (1)

r₁ = Rate of effusion of He

r₂ = Rate of Effusion of O₃

M₁ = Molecular Mass of He = 4 g/mol

M₂ = Molecular Mass of O₃ = 48 g/mol

Putting values in eq. 1,

r₁ / r₂ =   $\sqrt{48 / 4}$

r₁ / r₂ =   $\sqrt{12}$

r₁ / r₂ = 3.46

Result:
Therefore, Helium will effuse 3.46 times more faster than Ozone.

4 0

3 years ago