Use IUPAC rules to name this compound. A) nitrogen iodide B) nitrogen triiodide C) mononitrogen iodide D) mononitrogen triiodide
2 answers:
Answer:
Name of the compound is Nitrogen triiodide.
Explanation:
According to the IUPAC rules, to naming of the compound the following formula can be applied.
Prefix + Name of first element + Base name element of second element + Suffix.
The given compound -
Name of first element- Nitrogen
Base name element of second element - Iodine
Suffix = 3 = tri
Here, iodine is in ionic form therefore, it becomes iodide. and then suffix will be added in front of the halogen.
Therefore, name of the compound will be Nitrogen triiodide..
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Answer:
<h2>2.49 g/cm³</h2>Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>2.49 g/cm³</h3>Hope this helps you
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .
will all the alcohol evaporate? or none at all?
Answer:
Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.
Explanation:
Given that:
The volume of alcohol which is placed in a small laboratory = 1.0 L
Vapor pressure of ethyl alcohol at 25 ° C = 59 mmHg
Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;
Then, we have:
=
= 0.078 atm
Temperature = 25 ° C
= ( 25 + 273 K)
= 298 K.
Density of the ethanol = 0.785 g/cm³
The volume of laboratory = l × b × h
= 3.0 m × 2.0 m × 2.5 m
= 15 m³
Converting the volume of laboratory to liter;
since 1 m³ = 100 L; Then, we have:
15 × 1000 = 15,000 L
Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:
PV = nRT
Making n the subject of the formula; we have:
n = 47. 88 mol of ethanol
Moles of ethanol in 1.0 L bottle can be calculated as follows:
Since numbers of moles =
and mass = density × vollume
Then; we can say ;
number of moles =
number of moles =
number of moles =
number of moles = 17.039 mol
Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.