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Butoxors [25]
3 years ago
13

How can a gardener remedy soil that is overly acidic?

Chemistry
1 answer:
DochEvi [55]3 years ago
4 0
A gardener can fix acidic soil by adding lime over time.
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Which of these is an example of a physical change? Question options: melting a substance breaking chemical bonds forming chemica
kherson [118]
Your answer is Breaking chemical bonds. Hope this helps
4 0
3 years ago
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What is the volume of 0.4 mole of gas at stp? 30 points!
iragen [17]

Answer:

89.6 liters

Explanation:

A STP (standard temperature and pressure) ONE mole of any idea gas will occupies 22.4 liters. So,...

4 moles x 22.4 L/mol = 89.6 liters

5 0
2 years ago
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14.2 grams of Na2SO4 is dissolved in water to make a 2.50 L
Trava [24]

Answer:

0.04 M

Explanation:

Given data:

Mass of Na₂SO₄= 14.2 g

Volume of solution = 2.50 L

Molarity of solution = ?

Solution:

Number of moles of Na₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 14.2 g/ 142.04 g/mol

Number of moles = 0.1 mol

Molarity :

Molarity = number of moles of solute / volume of solution in L

Molarity = 0.1 mol / 2.50 L

Molarity = 0.04 M

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
Is PH3 polar or nonpolar
Andru [333]

Answer:

polar

Explanation:

8 0
3 years ago
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