Explanation:
Higher the frequency smaller will be the wavelength. Higher frequency have shorter wavelength and lower frequency waves have larger wavelength. Also, Beats are formed by the superposition of two waves with slightly different frequencies but with similar amplitudes. In time, waves switch between constructive interference and disruptive interference, giving the resultant wave a time-varying amplitude.
Answer:
The first thing that you need to do here is to figure out the mass of the sample.
To do that, you can use its volume and the fact that aluminium is said to have a density of
2.702 g cm
−
3
, which implies that every
1 cm
3
of aluminium has a mass of
2.702 g
.
Explanation:
No because dew is water droplets that come from condensation rather than from rain coming at night
Answer:
1.155 moles of potassium nitrate are required to make 550 mL of a 2.1M solution.
Explanation:
In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes.
Molarity is a unit of concentration of a solution and indicates the amount of moles of solute that appear dissolved in each liter of the mixture. In other words, the Molarity (M) or Molar Concentration is the number of moles of solute that are dissolved in a given volume.
The Molarity of a solution is determined by the following expression:

Molarity is expressed in units (
).
In this case:
- Molarity= 2.1 M
- number of moles of solute= ?
- Volume= 550 mL= 0.550 L (being 1L=1000 mL)
Replacing:

Solving:
number of moles of solute= 2.1 M* 0.550 L
number of moles of solute= 1.155 moles
1.155 moles of potassium nitrate are required to make 550 mL of a 2.1M solution.
Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:

Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:

The expression for equilibrium constant is:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
Now put all the given values in this expression, we get:

x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M