What element is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 4f10

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na3(CO3)(HCO3)·2H2O. 2Na3(CO3)(HCO

zhenek [66]

Percentage yield = (actual yield / theoretical yield) x 100%

The balanced equation for the decomposition is,
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂<span>O(g)

The stoichiometric ratio between </span>Na₃(CO₃)(HCO₃)·2H₂O(s) and Na₂CO₃(s) is 2 : 3

The decomposed mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 1000 kg
= 1000 x 10³ g

Molar mass of Na₃(CO₃)(HCO₃)·2H₂O(s) = 226 g mol⁻¹
moles of Na₃(CO₃)(HCO₃)·2H₂O(s) = mass / molar mass
= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol

Hence, moles of Na₂CO₃ formed = 4424.78 mol x $\frac{3}{2}$
= 6637.17 mol

Molar mass of Na₂CO₃ = 106 g mol⁻¹

Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg

Hence, the theoretical yield of Na₂CO₃ = 703.540 kg
Actual yield of Na₂CO₃ = 650 kg

Percentage yield = (650 kg / 703.540 kg) x 100%
= 92.34%

7 0

3 years ago

A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.

adell [148]

Answer:

a) $V=4.3x10^3mL$

b) $V=4.3x10^6mm^3$

c) $\rho=1.5x10^5g/L$

Explanation:

Hello,

a) In this case, the given height in cm is:

$h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm$

And the radius in cm is:

$r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm$

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

$V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL$

b) In this case, the given height in mm is:

$h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm$

And the radius in mm is:

$r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm$

Thus, the volume in cubic millimeters is:

$V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3$

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

$\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\ \\\rho=1.5x10^5g/L$

Best regards.

8 0

3 years ago

The shells further away from the nucleus are larger/smaller and can hold more/less electrons.

Sergeu [11.5K]

The shells further away from the nucleus are LARGER and can hold MORE electrons

8 0

2 years ago

The molar mass of HgO is 216.59 g/mol. The molar mass of O2 is 32.00 g/mol. How many moles of HgO are needed to produce 250.0 g

Irina-Kira [14]

A reaction in which Oxygen (O₂) is produced from Mercury Oxide (HgO) would be a decomposition reaction.
2HgO → 2Hg + O₂

If 250g of O₂ is needed to be produced,
then the moles of oxygen needed to be produced = 250g ÷ 32 g/mol
= 7.8125 mol

Now, the mole ratio of Oxygen to Mercury Oxide is 1 : 2
∴ if the moles of oxygen = 7.8125 mol
then the moles of mercury oxide = 7.8125 mol × 2
= 15.625 mol

Thus the number moles of HgO needed to produce 250.0 g of O₂ is 15.625 mol

6 0

3 years ago

How much energy is wasted as heat by an incandescent light globe

muminat

Electric current passes through a filament of an incandescent bulb, thereby increasing it temperature. When current flows, it contains electrons through the filament to produce light. Typically, incandescent light bulb consists of a glass enclosure that contains tungsten filament. The glass enclosure contains either a vacuum or an inert gas that serves as the filament protection from evaporating. Incandescent light bulbs contain a stem attached at to its base to allow the electrical contacts to run through the envelope without gas or air leaks.

8 0

2 years ago