Explanation:
prevent becoz crop rotation means using it by planting the crop according to nuetrients so ur soil is prevented
Answer:
HCl
Explanation:
The best solvent for NaF is a polar liquid. The only liquid having a significant dipole moment among the options is HCl due to the large electro negativity difference between hydrogen and chlorine.
The polar solvent can interact with the NaF via its dipoles such that the NaF dissolves due to ion-dipole interaction.
Answer:
Explanation:
1)
Given data:
Initial volume of balloon = 0.8 L
Initial temperature = 12°C ( 12+273= 285 K)
Final temperature = 300°C (300+273 = 573 K)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 0.8 L .573 K / 285 K
V₂ = 458.4 L / 285
V₂ = 1.61 L
2)
Initial pressure = 204 kpa
Initial temperature = 29°C ( 29 + 273 = 302 K)
Final temperature = ?
Final pressure = 300 kpa
Solution:
P₁/T₁ = P₂/T₂
T₂ = T₁P₂/P₁
T₂ = 302 K . 300 kpa / 204 kpa
T₂ = 90600 K/ 204
T₂ = 444.12 K
3)
Given data:
Initial volume = 14 L
Initial pressure = 2.1 atm
Initial temperature = 100 K
Final temperature = 450 K
Final volume = ?
Final pressure = 1.2 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm
V₂ = 13230 L / 120
V₂ = 110.25 L
Answer:
Explanation:
stoichiometry of C₂H₂ to H₂O is 2:2.
Number of moles of C₂H₂ = molar mass of C₂H₂
Since the molar mass of C₂H₂ is 26 g/mol.
Number of C₂H₂ moles reacted = 64.0 g / 26 g/mol = 2.46 mol.
according to a molar ratio of 2:2.
the number of H₂O moles formed = a number of C₂H₂ moles reacted.
Therefore the number of H₂O moles produced = 2.46 mol
Question:
A chemistry student needs of 10 g isopropenylbenzene for an experiment. He has available 120 g of a 42.7% w/w solution of isopropenylbenzene in acetone. Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button.
Answer:
The answer to the question is as follows
The mass of solution the student should use is 23.42 g.
Explanation:
To solve the question we note the following
A solution containing 42.7 % w/w of isopropenylbenzene in acetone has 42.7 g of isopropenylbenzene in 100 grams of the solution
Therefore we have 10 g of isopropenylbenzene contained in
100 g * 10 g/ 42.7 g = 23.42 g of solution
Available solution = 120 g
Therefore the quantity to used from the available solution = 23.42 g of the isopropenylbenzene in acetone solution.
