Answer:
potassium, nitrogen and phosphorous cycle
Explanation:
A fertilizer is a substance which is applied on the plants by farmers to increase the supply of nutrients for the plants. Fertilizers have known to be toxic in many ways such as they alter the potassium, nitrogen and phosphorus cycles. Nitrogen, potassium and phosphorus are present in abundant amounts in the fertilizers. Draining of these fertilizers into rivers and ponds is toxic for the aquatic life. Hence, the use of fertilizers disrupts the natural cycles and is toxic for many aquatic plants and animals.
Answer:
28 g/mol, N2
Explanation:
Given data:
Volume of gas = 5.0 L
Mass of gas = 6.3 g
Pressure = 1 atm
Temperature = 273 K
Molar mass of gas = ?
Solution:
We will calculate the density first.
d = mass/ volume
d = 6.3 g/ 5.0 L
d = 1.26 g/L
Molar mass:
d = PM/RT
M = dRT/P
M = 1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm
M = 28 g/mol
Molar mass of N₂ is 28 g/mol thus given gas is N₂.
Answer:
B. Decreasing the pressure applied to the gas molecules
Explanation:
According to Boyle's Law, the pressure of the gas is inversely proportional to the volume of the gas. So, the option B is correctly implied to it.
Other values such as Temperature, Number of molecules are inversely proportional to the volume of the gas.
Size of the gas molecules is negligible as compared to volume.
The balanced equation for the above reaction is
2K₃PO₄ + 3NiCl₂ ---> 6KCl + Ni₃(PO₄)₂
stoichiometry of K₃PO₄ to NiCl₂ is 2:3
the number of NiCl₂ moles reacted - 0.0110 mol/L x 0.154 L = 1.69 x 10⁻³ mol
if 3 mol of NiCl₂ reacts with - 2 mol of K₃PO₄
then 1.69 x 10⁻³ mol of NiCl₂ reacts with - 2/3 x 1.69 x 10⁻³ = 1.13 x 10⁻³ mol of K₃PO₄
molarity of K₃PO₄ solution given - 0.205 M
there are 0.205 mol in 1 L
therefore 1.13 x 10⁻³ mol are in - 1.13 x 10⁻³ mol / 0.205 mol/L = 5.51 mL
volume of K₃PO₄ required - 5.51 mL
Answer:
Mass = 40.4 g
Explanation:
Given data:
Mass in gram = ?
Volume of SO₂ = 14.2 L
Temperature = standard = 273 K
Pressure = standard = 1 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × 14.2 L = n × 0.0821 atm.L/ mol.K × 273 K
14.2 atm.L = n × 22.41 atm.L/ mol
n = 14.2 atm.L/22.41 atm.L/ mol
n = 0.63 mol
Mass of sulfur dioxide:
Mass = number of moles × molar mass
Mass = 0.63 mol × 64.1 g/mol
Mass = 40.4 g
