These problems are a bit interesting. :)
First let's write the molecular formula for ammonium carbonate.
NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)
17.6 gNH4CO3
Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.
17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)
Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)
NH4 = 14.01 + 4*1.01 = 18.05 g/mol
NH4CO3 = 14.01 + 4*1.01 + 12.01 + 3*16.00 = 78.06 g/mol
17.6 gNH4CO3 * (18.05 molNH4 / 78.06 molNH4CO3)
= 4.07 gNH4
Now just take the molar mass we found to convert that amount into moles!
4.07 gNH4 * (1 molNH4 / 18.05 gNH4) = 0.225 molNH4
Answer:
It's obviously true
Explanation:
As we have evolved over the years we have become more advanced
Answer: Option (C) is the correct answer.
Explanation:
Chemical formula of a secondary amide is R'-CONH-R, where R and R' can be same of different alkyl or aryl groups. Here, the hydrogen atom of amide is attached to more electronegative oxygen atom of the C=O group.
Therefore, the hydrogen atom will be more strongly held by the electronegative oxygen atom. As a result, there will be strongly hydrogen bonded in the liquid phase of secondary amide.
Whereas chemical formula of nitriles is RCN, ester is RCOOR' and acid chlorides are RCOCl. As no hydrogen bonding occurs in any of these compounds because hydrogen atom is not being attached to an electronegative atom.
Thus, we can conclude that secondary amides are strongly hydrogen bonded in the liquid phase.
Answer:
1120 gm
Explanation:
6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
(a) Balance the equation.
(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete
combustion reaction?
FIRST, CORRECT THE EQUATION THEN BALANCE
2C2H6(G) + 7O2------------> 4CO2 + 6H2O
so for 10 moles of ethane, we need
7 X 5 = 35 MOLES O2
=35 MOLES O2
O2 HAS A MOLAR MASS OF 2X16 = 32 gm
35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm
Answer: El carbono, que en estado sólido, puede adoptar muchas formas alotrópicas, siendo las más comunes el diamante (red tridimensional) y el grafito (láminas), aunque también puede formar nanoestructuras en forma de balón de fútbol (fullerenos) o tubos diminutos (nanotubos de carbono), entre otras posibilidades.
Explanation:
