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:
-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.
Explanation:
Mass of nitrogen triiodide = 20.0 g
Moles of nitrogen triiodide = 

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.
Answer:
i also had this question:P
Explanation:
Answer:
See attached picture.
Explanation:
Hello,
In this case, for the given name, you can verify the structure on the attached picture, wherein you can see verify the presence of both the ethyl and methyl radicals at the third carbon as well as the triple bond at the first carbon.
Best regards.