Answer:
Most radio waves have wavelengths between 1 mm and 100 km.
A cooling curve shows A. how the temperature of a substance falls as heat is removed.
Explanation:
<em>Radio waves</em> are the longest of all the waves in the electromagnetic spectrum.
Most have wavelengths between 1 mm and 100 km, although there is no upper limit.
Some radio waves have wavelengths of 10 000 km.
A <em>cooling curve</em> (see image below) shows how the temperature of a substance falls as it is cooled.
In Option E., a decrease in temperature would cause an energy <em>loss</em>.
Options B., C., and D. involve the <em>addition of heat</em>.
Answer:
The correct answer is 5.447 × 10⁻⁵ vacancies per atom.
Explanation:
Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,
Fv = Nv/N------ (i)
Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,
N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,
N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol
N = 5.14073 × 10²⁸ atoms/m³
Now putting the values of Nv and N in the equation (i) we get,
Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3
Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.
Answer:
The value of an integer x in the hydrate is 10.
Explanation:
Molarity of the solution = 0.0366 M
Volume of the solution = 5.00 L
Moles of hydrated sodium carbonate = n
Mass of hydrated sodium carbonate = n= 52.2 g
Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol
Solving for x, we get:
x = 9.95 ≈ 10
The value of an integer x in the hydrate is 10.
The correct answer is (3)
I-131 and P-32
The explanation:
according to attached table:
- we can see that the half life of p 32 is 14.28d (more than one hour)
- and the half life of I-131 is 8.021 d
(more than one hour)
and They both have β- decay mode and with half-lives greater than hour.
Answer:
the Glancing angle is the angle between the incident ray and plane mirror which is 90o in the given case. The angle between the direction of the incident ray and the reflected ray is the angle of deviation. Since the angle of deviation for a plane mirror is twice the glancing angle, the angle of deviation is 1800.
