What change in volume results if 60 mL of a gas is cooled from 33 C to 5 C?
1 answer:
Answer:
Change in volume on changing temperature from 33 to 5
is 5.49 mL
Explanation:
Initial volume of gas = V = 60 mL
Assuming final volume of gas to be V' mL
Initial temperature = T = 33 = 306 K
Final temperature = T' = 5 = 278 K
The relationship between volume and temperature of gas at constant pressure is shown below
Change in volume on changing temperature = 5.49 mL
You might be interested in
Answer:
a) [Ag+]dilute = 6.363 × 10⁻¹⁶ M
b) 1.273 × 10⁻¹⁶
c) 2.629×10⁻¹⁹ M Thus; the value for [Ag+ ]dilute will be too low
Explanation:
In an Ag | Ag+ concentration cell ,
The anode reaction can be written as :
Ag ----> Ag+(dilute) + e- &:
The cathode reaction can be written as:
Ag+(concentrated) + e- ----> Ag
The Overall Reaction : is
Ag+(concentrated) -----> Ag+(dilute)
However, the Standard Reduction potential of cell = E°cell = 0
( since both cathode and anode have same Ag+║Ag )
Also , given that the theoretical slope is - 0.0591 V
Therefore; the reduction potential of cell ; i.e
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963
[Ag+]dilute = × 1.0 × 10⁻¹ M
[Ag+]dilute = 6.363 × 10⁻¹⁶ M
b)
AgI ----> Ag + (dilute) + I⁻
So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]
= 6.363 × 10⁻¹⁶ M × 0.20 M
= 1.273 × 10⁻¹⁶
c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804
[Ag+]dilute = × 1.0 × 10⁻¹ M
[Ag+]dilute = 2.629×10⁻¹⁹ M
Thus, the value for [Ag+ ]dilute will be too low
Answer:
1s² 2s²2p³
Explanation:
If the atom has seven electrons, it is Element 7 (nitrogen).
In the Periodic Table, you count the electrons in all the subshells up to
No. 7.
In the first Period, you have filled the 1s level (2 electrons).
In the second Period, you have filled the 2s subshell (2 electrons) and put three electrons in the 2p subshell.
Thus, the electron configuration is
1s² 2s²2p³
Note how the superscripts tell you the number of electrons in each subshell: <em>2 + 2 + 3 = 7</em>.
