Answer:
The compound is N2O4
Explanation:
We have certain important pieces of information about the compound;
1) it is an oxide (a binary compound of nitrogen and oxygen)
2) there are no N-N bonds present
3) there are no O-O bonds present
Since it contains only nitrogen and oxygen then nitrogen accounts for 25.9% of the molecule by mass then oxygen should account for (100-25.9) = 74.1% oxygen
Relative atomic mass of oxygen = 16
Relative atomic mass of nitrogen = 14
We now deduce the empirical formula
Nitrogen. Oxygen
25.9/14. 74.1/16
1.85/1.85. 4.6/1.85 (divide through by the lowest ratio)
1 2
Empirical formula is NO2
To find the molecular formula
(NO2)n = 108
(14+2(16))n= 108
46n=108
n= 108/46
n= 2
Therefore molecular formula= N2O4
Formula: NA2S2O3. Valency: 2
Find your answer in the explanation below.
Explanation:
PV = nRT is called the ideal gas equation and its a combination of 3 laws; Charles' law, Boyle's law and Avogadro's law.
According to Boyle's law, at constant temperature, the volume of a gas is inversely proportional to the pressure. i.e V = 1/P
From, Charles' law, we have that volume is directly proportional to the absolute temperature of the gas at constant pressure. i.e V = T
Avogadro's law finally states that equal volume of all gases at the same temperature and pressure contain the same number of molecules. i.e V = n
Combining the 3 Laws together i.e equating volume in all 3 laws, we have
V = nT/P,
V = constant nT/P
(constant = general gas constant = R)
V = RnT/P
by bringing P to the LHS, we have,
PV = nRT.
Q.E.D
Answer:
Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.
Explanation:
The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:
.
Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both
and
will increase if the pressure is increased through compression. However, because
and
have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient
.
As a result, the increase in pressure will have no impact on the value of
. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.
Answer:
yes
Explanation:
same properties different size
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