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artcher [175]
4 years ago
7

Can someone answer theses problems? Only do the Even number problems,Im in a rush so please help me.

Chemistry
1 answer:
8090 [49]4 years ago
7 0
The answer to the questions. All 15 of them.

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The lithosphere, hydrosphere, and ______________ are the abiotic parts of the planet.
VladimirAG [237]
The lithosphere, hydrosphere, and atmosphere are the abiotic parts of the planet.
8 0
3 years ago
I NEED HELP PLEASE, THANKS!
ira [324]

Answer:

About 0.652

Explanation:

Because the reaction is balanced, we can go straight to the next step. The molar mass of potassium is about 39.098, while the molar mass of hydrogen gas is 2 and the molar mass of water is 18. Therefore, 25.5g of potassium would be about 0.652 moles, and 220 grams of water would be about 12.222 moles, making potassium the limiting reactant. Since there is a single unit of each compound on both sides of the equation, there would be an equal amount of moles of potassium and hydrogen, and therefore about 0.652 moles of hydrogen gas would be produced. Hope this helps!

6 0
3 years ago
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
4 years ago
Can matter undergo chemical reactions but still keep the same physical properties?
Nastasia [14]
Yes it can!!!!!!!!!!
6 0
3 years ago
Sulfur dioxide gas (SO2) reacts with excess oxygen gas (O2) and excess liquid water (H2O) to form liquid sulfuric acid (H2SO4).
inessss [21]
A.2SO₂ + O₂ + 2H₂O → 2H₂SO₄
B.Moles of SO₂ = 67.2 / 22.4Moles of SO₂ = 3 molesMoles of H₂SO₄ = 3 molesMass of H₂SO₄ = 3 x 98Mass of H₂SO₄ = 294 grams
Assuming sulfuric acid to have the same density as water,density = 1000 g / LVolume = mass / densityVolume = 294 / 1000Volume = 0.29 liters of sulfuric acid

6 0
3 years ago
Read 2 more answers
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