Your question isn't quite clear, but if you're wondering if a chemical is polar or non-polar, you simply draw a VSEPR sketch and draw arrows where the bonds are. Only draw arrows between atoms, NOT between an atom and a lone pair of electrons. The arrow should point to the most electronegative atom (you should be given an electronegativity scale). Afterwards, you add up the arrows as vectors, and look at the sum of the vectors. If the sum is zero (CH4 is a good example), the chemical is non-polar. If the sum is a vector, the chemical is polar (H2O, or water, is polar).
Answer:
V = 81.14 L
Explanation:
Given data:
Volume of gas = ?
Number of moles = 3.30 mol
Temperature of gas = 25°C
Pressure of gas = 0.995 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
Now we will convert the temperature.
25+273 = 298 K
now we will put the values in formula:
V = 3.30 mol 0.0821 atm.L/ mol.K 298 K / 0.995 atm
V = 80.74 L. atm / 0.995 atm
V = 81.14 L
Answer:
Option B. 4.25×10¯¹⁹ J
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 6.42×10¹⁴ Hz
Energy (E) =?
Energy and frequency are related by the following equation:
Energy (E) = Planck's constant (h) × frequency (f)
E = hf
With the above formula, we can obtain the energy of the photon as follow:
Frequency (f) = 6.42×10¹⁴ Hz
Planck's constant (h) = 6.63×10¯³⁴ Js
Energy (E) =?
E = hf
E = 6.63×10¯³⁴ × 6.42×10¹⁴
E = 4.25×10¯¹⁹ J
Thus, the energy of the photon is 4.25×10¯¹⁹ J
Answer:
True
Explanation:
Stars are bright and enormous astronomical bodies that are located at a large distance from the earth. They are made up of the lightest gaseous elements namely hydrogen and helium. These hydrogen atoms undergo the reaction of nuclear fusion, which fuels the energy of the star and allows the star to shine, releasing a large amount of energy. They are located about trillions of miles away from the location of the earth, due to which they appear smaller in size.
Thus, the above-given statement is True.
The answer is a conjugate acid.