Answer:
M.Mass = 3.66 g/mol
Data Given:
M.Mass = M = ??
Density = d = 0.1633 g/L
Temperature = T = 273.15 K (Standard)
Pressure = P = 1 atm (standard)
Solution:
Let us suppose that the gas is an ideal gas. Therefore, we will apply Ideal Gas equation i.e.
P V = n R T ---- (1)
Also, we know that;
Moles = n = mass / M.Mass
Or, n = m / M
Substituting n in Eq. 1.
P V = m/M R T --- (2)
Rearranging Eq.2 i.e.
P M = m/V R T --- (3)
As,
Mass / Volume = m/V = Density = d
So, Eq. 3 can be written as,
P M = d R T
Solving for M.Mass i.e.
M = d R T / P
Putting values,
M = 0.1633 g/L × 0.08205 L.atm.K⁻¹.mol⁻¹ × 273.15 K / 1 atm
M = 3.66 g/mol
<span>Ammonia (NH3) is the combination of Nitrogen and Hydrogen
elements.
=> N2 + 3H2 => 2NH3
Ammonia is basically used as a fertilizer. It is a gas composed of nitrogen and
hydrogen. It is colorless with strong odor. Here are some other uses of Ammonia
aside from fertilizer:
=> used by manufacturer to produce synthetic fiber
=> Used in metallurgical process
Ammonia can be decomposed easily and it produce hydrogen that is very
convenient in welding.
Ammonia’s boiling point is -28.03 F and freezing point is -107.8F.
</span>
The first option, collapsed in on itself.
The star's core mass becomes so dense that the resulting gravity implodes the star.
Interesting enough, the third option is kindof true too...some large and tenacious black holes that absorb other stars will form incredibly bright accretion disks around their perimeter before filling absorbing the star.
Answer:
1.5055×10²⁴ molecules
Explanation:
From the question given above, the following data were obtained:
Number of mole CO₂ = 2.5 moles
Number of molecules CO₂ =?
The number of molecules present in 2.5 moles CO₂ can be obtained as:
From Avogadro's hypothesis,
1 mole of CO₂ = 6.022×10²³ molecules
Therefore,
2.5 mole of CO₂ = 2.5 × 6.022×10²³
2.5 mole of CO₂ = 1.5055×10²⁴ molecules
Thus, 1.5055×10²⁴ molecules are present in 2.5 moles CO₂
Balanced Half reactions are:
At anode 2
==> Cl₂+ 
+ H₂O ==> 
+ 2
+
At Cathode: 2 + ==> H₂
Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, , 
and
Now at Anode reaction will occur as given:
2 ==> Cl₂+ + H₂O ==> + 2 + (will occur)
At Cathode:
2 + ==> H₂ (will occur)
At Cathode:
+ ==> Mn (This reaction will not occur)
The deposition of solid Mn will not occur because in aqueous solution, will be reduced before .
The reduction potentials for is zero whereas reduction potential for is - 1.18V.
The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.
To learn more about the half reaction please click on the link brainly.com/question/13186640
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