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Ierofanga [76]
3 years ago
5

Private land is generally used for _______. a. homes b. businesses c. factories d. all of the above Please select the best answe

r from the choices provided A B C D
Chemistry
2 answers:
alekssr [168]3 years ago
5 0

HM, I think the answer would be D. This is just a guess, so please use it if ou want to answer D it's ok :D

kumpel [21]3 years ago
5 0

Answer: d. all of the above

Explanation:

Private land is the land which is legally occupied by a single ownership. The government has no right over it. The single owner can use the land for the purpose of self benefit. A private land owner can use the land for the purpose of construction of homes, factories and use the land for the commercial businesses like fishing, farming and for other needs.

On the basis of the above explanation, d. all of the above is the correct option.

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8. Which of the following have bacteria in their root nodules that fix nitrogen?
Natalka [10]

Answer:

Option A

Explanation:

Leguminous plants like pulses etc. have root nodules comprising of rhizobacterium which live in a symbiotic relationship with the roots of the plant and in turn fix the nitrogen in the soil in the roots of the leguminous plants.

Hence, option A is correct

8 0
2 years ago
Why is granite better then marble for a statue in a city centre
nasty-shy [4]
Granite is stronger, and it doesn't grow mold in the rain. It also looks better than marble and is easier to carve. 
5 0
3 years ago
Read 2 more answers
Consider the reaction: CaCO3(s) à CaO(s) + CO2(g)
Vsevolod [243]

Answer:

131.5 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

First, we will calculate the standard enthalpy of the reaction (ΔH°).

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g) ) - 1 mol × ΔH°f(CaCO₃(s) )

ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)

ΔH° = 179.2 kJ

Then, we calculate the standard entropy of the reaction (ΔS°).

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g) ) - 1 mol × S°(CaCO₃(s) )

ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)

ΔS° = 160.2 J/K = 0.1602 kJ/K

Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.

ΔG° = ΔH° - T × ΔS°

ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K

ΔG° = 131.5 kJ

6 0
3 years ago
Which base has the lowest ionization constant (Kb)?
inysia [295]

Answer:D.Blood

Explanation:

4 0
3 years ago
What is an element in the alkali metal group
Tema [17]

Answer:

Group of highly-reactive chemical elements. The alkali metals are a group (column) in the periodic table consisting of the chemical elements lithium (Li), sodium (Na), potassium (K), rubidium (Rb), caesium (Cs), and francium (Fr).

8 0
3 years ago
Read 2 more answers
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