<u>Answer:</u> The phase change process in which solids gets converted to gases is sublimation.
<u>Explanation:</u>
For the given options:
<u>Option a:</u> Condensation
It is a type of process in which phase change occurs from gaseous state to liquid state at constant temperature.
<u>Option b:</u> Melting
It is a type of process in which phase change occurs from solid state to liquid state at constant temperature.
<u>Option c:</u> Sublimation
It is a type of process in which phase change occurs from solid state to gaseous state without passing through the liquid state at constant temperature.
<u>Option d:</u> Deposition
It is a type of process in which phase change occurs from gaseous state to solid state without passing through the liquid state at constant temperature.
Hence, the phase change process in which solids gets converted to gases is sublimation.
Answer:
1 mole of platinum
Explanation:
To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.
The empirical formula for the compound can be obtained as follow:
Platinum (Pt) = 117.4 g
Carbon (C) = 28.91 g
Nitrogen (N) = 33.71 g
Divide by their molar mass
Pt = 117.4 / 195 = 0.602
C = 28.91 / 12 = 2.409
N = 33.71 / 14 = 2.408
Divide by the smallest
Pt = 0.602 / 0.602 = 1
C = 2.409 / 0.602 = 4
N = 2.408 / 0.602 = 4
The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄
From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.
Answer:
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Explanation:
Answer:
A) 22.4L
Explanation:
we know, ideal gas law states
PV=nRT
V=nRT/P
At STP,
T= 273.15K P=1atm R=0.082L.atm/mol/K n=1 mole
V=(1*0.082*273.15)/ 1
V=22.4L
Answer:
See explaination
Explanation:
1)
we know that
half cell with higher reduction potential is cathode
so
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
anode :
Cr(s) ---> Cr+3 + 3e-
so
overall reaction is
3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3
now
Eo cell = Eo cathode - Eo anode
so
EO cell = 1.77 + 0.74
Eo cell = 2.51 V
now
in this case
oxidizing agents are N20 and Cr+3
reducing agents are Cr and N2
higher the reduction potential , stronger the oxidizing agent
lower the reduction potential , stronger the reducing agent
so
oxidzing agents
N20 > Cr+3
reducing agents
Cr > N2
2)
cathode :
Au+ + e- --> Au
anode :
Cr ---> Cr+3 + 3e-
overall reaction
3Au+ + Cr ---> 3Au + Cr+3
Eo cell = 1.69 + 0.74
Eo cell = 2.43
now
oxidizing agents :
Au+ > Cr+3
reducing agents :
Cr > Au
3)
cathode :
N20 + 2H+ + 2e- ---> N2 + H20
andoe :
Au ---> Au+ + e-
overall
2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20
Eo cell = 1.77 - 1.69
Eo cell = 0.08
oxidizing agents
N20 > Au+
reducing agents
Au > N2
