Given :
2NOBr(g) - -> 2NO(g) + Br2(g)
Initial pressure of NOBr , 1 atm .
At equilibrium, the partial pressure of NOBr is 0.82 atm.
To Find :
The equilibrium constant for the reaction .
Solution :
2NOBr(g) - -> 2NO(g) + Br2(g)
t=0 s 1 atm 0 0
1( 1-2x) 2x x
So ,
At equilibrium :
![K_{eq}=\dfrac{[NO]^2[br_2]}{[NOBr]^2}\\\\K_{eq}=\dfrac{0.18^2\times 0.9}{0.82^2}\\\\K_{eq}=0.043\ atm](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cdfrac%7B%5BNO%5D%5E2%5Bbr_2%5D%7D%7B%5BNOBr%5D%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D%5Cdfrac%7B0.18%5E2%5Ctimes%200.9%7D%7B0.82%5E2%7D%5C%5C%5C%5CK_%7Beq%7D%3D0.043%5C%20atm)
Hence , this is the required solution .
Answer:
The number of copper atoms 12.405 ×10²³ atoms.
The number of silver atoms 13.13 ×10²³ atoms.
Beaker B have large number of atoms.
Explanation:
Given data:
In beaker A
Number of moles of copper = 2.06 mol
Number of atoms of copper = ?
In beaker B
Mass of silver = 222 g
Number of atoms of silver = ?
Solution:
For beaker A.
we will solve this problem by using Avogadro number.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms in one mole of substance.
While we have to find the copper atoms in 2.06 moles.
So,
63.546 g = 1 mole = 6.022×10²³ atoms
For 2.06 moles.
2.06 × 6.022×10²³ atoms
The number of copper atoms 12.405 ×10²³ atoms.
For beaker B:
107.87 g = 1 mole = 6.022×10²³ atoms
For 222 g
222 g / 101.87 g/mol = 2.18 moles
2.18 mol × 6.022×10²³ atoms = 13.13 ×10²³ atoms
Answer:
Sorry mate! I can't understand this language...
Answer:
Amplitude ----> C.) The distance from a crest or trough to the rest position on the horizontal axis
Crest ---> A.) The highest point of a wave
Destructive interference ---> E.) A situation in which the crest of one wave and the trough of another overlap, resulting in a wave that has a smaller amplitude than the original waves
Sound Wave ---> B.) A vibration transmitted through an elastic medium, such as a gas, liquid, or solid
Trough ---> D.) The lowest point of a wave
V ( NaOH ) = mL ?
M ( NaOH ) = 0.100 M
V ( HCl ) = 9.00 mL / 1000 => 0.009 L
M ( HCl ) = 0.0500 M
number of moles HCl:
n = M x V
n = 0.009 x 0.0500 => 0.00045 moles HCl
mole ratio:
<span>HCl + NaOH = NaCl + H2O
</span>
1 mole HCl ---------------- 1 mole NaOH
0.00045 moles HCl ----- ??
0.00045 x 1 / 1 => 0.00045 moles of NaOH
M = n / V
0.100 = 0.00045 / V
V = 0.00045 / 0.100
V = 0.0045 L
1 L ------------ 1000 mL
0.0045 L ----- ??
0.0045 x 1000 / 1 => 4.5 mL of NaOH
