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LuckyWell [14K]
3 years ago
12

Argon, Ar, is a noble gas. How many miles of argon are in 200 gram sample of argon.? (Answer rounded to the ones place)

Chemistry
1 answer:
tatuchka [14]3 years ago
3 0

The answer is 39.948 g/mol so rounded to the nearest ones would be 40.00 g/mol.

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Photosynthesis by land plants leads to the fixation each year of about 1 kg of carbon on the average for each square meter of an
Rama09 [41]

Answer:

a) mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

Explanation:

first we calculate the moles of carbon

moles = mass/molar mass

= 1kg/12gmol⁻¹

= 1000g/12gmol⁻¹

= 83.33 mol

now using the ideal gas equation

we find the volume of co₂required based on 83.33 moles

PVco₂ = nRT

Vco₂ = nRT/P

Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm

Vco₂ = 2083.73 L

so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon

therefore

Vair × 0.0390/100 = 2038.73L

Vair = (2038.73L × 100) / 0.0390

Vair = 5.23 × 10⁶L

therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon

a)

Here we calculate the mass of air over 1 square meter of surface.

Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²

NOW

mass of air = 1.020×10⁴kgm⁻² × 1m²

= 1.020×10⁴kg

= 1.020×10⁷g    [1kg = 10³g]

we now find the moles of air associated with it

moles = mass/molar mass

= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)

= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)

= 1.020 × 10⁷g / 28.8 gmol⁻¹

= 354166.67mol

so based on the question, for each mole (air), there is 0.0390% of CO₂

now to calculate the moles of CO₂ we say;

MolesCo₂ = 0.0390/100 × 354166.67mol

= 138.125 moles

Now we calculate mass of CO₂ from the above findings

Moles = mass/molar mass

mass = moles × molar mass

= 138.125 moles × 12gmol⁻¹

= 1657.5g

we covert to KG

= 1657.5g / 1000

mass = 1.65kg

therfore mass of carbon directly above 1 ( each)  square meter of the earth is 1.65kg

b)

to find the number years required to use up all the CO₂, WE SAY

Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year

Number of years = 1.65kgm⁻² / 1kg²year⁻¹

Number of years = 1.65 years

Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years

6 0
3 years ago
Balance the equation <br> Ca(s) + H3PO4(aq) ----&gt;Ca3(PO4)2(s) + H2(g)
vovikov84 [41]
Answer: CO2x+O4H I don’t know if this is the right answer
3 0
2 years ago
) If you started with 2.3182 g of 3-nitrophthalic acid and had a 93% yield of 3-nitrophthalhydrazide, how many grams of 3-nitrop
MArishka [77]

Answer: 1.8124 g of 3-nitrophthalhydrazide were recovered.

Explanation:

The balanced chemical reaction will be :

C_8H_5NO_6+N_2H_4\rightarrow C_8H_5N_3O_4+2H_2O

moles of 3-nitrophthalic acid  = \frac{\text {given mass}}{\text {molar mass}}=\frac{2.3182g}{211.13g/mol}=0.0110mol

As 1 mole of 3-nitrophthalic acid gives = 1 mole of 3-nitrophthalhydrazide

0.0110 moles of 3-nitrophthalic acid gives = \frac{1}{1}\times 0.0110=0.0110 mole of 3-nitrophthalhydrazide

mass of 3-nitrophthalhydrazide = moles\times {\text {molar mass}}=0.0110mol\times 177.16g/mol=1.9488g

As the percentage yield is 93% , the mass of 3-nitrophthalhydrazide recovered = \frac{1.9488\times 93}{100}=1.8124g

Therefore 1.8124 g of 3-nitrophthalhydrazide were recovered.

4 0
3 years ago
For the 52nd electron must pair with one of the<br> electrons in the 5p orbital.
Umnica [9.8K]

Answer:

This question is incomplete

Explanation:

This question is incomplete, however, the element that has 52 electrons only is Tellurium (Te) and when the electronic configuration of elements with more than 52 electrons are written, the 52nd electron is indicated/paired the same way the 52nd electron of Te is indicated/paired. Hence, while writing the electronic configuration of Te, it is written as

[Kr] 4d¹⁰ 5s² 5p⁴ where [Kr] is the electronic configuration of krypton. Based on this, we can deduce that the 52nd electron will be in the first orbital of the P subshell (as attached in the picture). This is because when indicating the electrons in the subshell, one electron will be spread across each orbital and if any electron is still remaining, it will be added starting from to the first orbital of the subshell, however no two electrons in an orbital in a subshell can have the same spin and hence must face opposite direction based on pauli's exclusion principle (as seen in attached); thus for the 5p-orbital of elements with 52 or more electrons, when one electron each is represented in each box (3 boxes in total) in the 5p-orbital, the remaining electron is paired with the the first electron in the first box of the 5p-orbital

5 0
3 years ago
The solubility of gases in liquids The solubility of gases in liquids increases as temperature increases and increases as pressu
eimsori [14]

Answer:

As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases.

Explanation:

As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases

3 0
3 years ago
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