1. Dry Ice (solid carbon dioxide)
2. Iodine
3. Arsenic
4. Naphthalene
Explanation:
Matter also exhibits physical properties. Physical properties are used to observe and describe matter. Physical properties can be observed or measured without changing the composition of matter. These are properties such as mass, weight, volume, and density.
Answer:
The given atom is of Ca.
Explanation:
Given data:
Speed of atom = 1% of speed of light
De-broglie wavelength = 3.31×10⁻³ pm (3.31×10⁻³ / 10¹² = 3.31×10⁻¹⁵ m)
What is element = ?
Solution:
Formula:
m = h/λv
m = mass of particle
h = planks constant
v = speed of particle
λ = wavelength
Now we will put the values in formula.
m = h/λv
m = 6.63×10⁻³⁴kg. m².s⁻¹/3.31×10⁻¹⁵ m ×( 1/100)×3×10⁸ m/s
m = 6.63×10⁻³⁴kg. m².s⁻¹/ 0.099×10⁻⁷m²/s
m = 66.97×10⁻²⁷ Kg/atom
or
6.69×10⁻²⁶ Kg/atom
Now here we will use the Avogadro number.
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
Now in given problem,
6.69×10⁻²⁶ Kg/atom × 6.022 × 10²³ atoms/ mol × 1000 g/ 1kg
40.3×10⁻³×10³g/mol
40.3 g/mol
So the given atom is of Ca.
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
Answer:
The u (amu is the old unit name) is 1/12 of the weight of an 12C atom. The way the u is chosen ensures that all core and atom masses are multiples of 1(±0.1) u.
Explanation:
Further explanation if needed...
Carbon 12 was chosen because the chemical atomic weights based on C12 are almost identical to the chemical atomic weights based on the natural mix of oxygen. Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.
