Answer:
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
Explanation:
Step 1: Data given
tetraphosphorus decoxide = P4O10
Molar mass of P4O10 = 283.89 g/mol
Mass of P4O10 = 4.5 grams
Number of Avogadro = 6.022 * 10^23 / mol
Step 2: Calculate moles of P4O10
Moles P4O10 = mass P4O10 / molar mass P4O10
Moles P4O10 = 4.5 grams / 283.89 g/mol
Moles = 0.016 moles
Step 3: Calculate moles of P
For 1 mol P4O10 we have 4 moles of phosphorus
For 0.016 moles P4O10 we have 4*0.016 = 0.064 moles P
Step 4: Calculate number of P atoms
Number of P atoms = moles P * number of Avogadro
Number of P atoms = 0.064 moles * 6.022*10^23
Number of P atoms = 3.85 * 10^22 atoms
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
Explanation:
The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.
Positive entropy means the system has increased its degree of disorderness.
<em>Paper chromatography is especially useful in characterizing amino acids. The different amino acids move at differing rates on the paper because of differences in their R groups.</em>
Answer:
64J of energy must have been released.
Explanation:
Step 1: Data given
One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.
The product contains 435 J of chemical energy.
Step 2:
Since the energy is conserved
Sum of energy of Reactants = Energy of Products
Sum of energy of Reactants = 346 J + 153 J = 499 J
The energy of the product = 435 J
435 < 499
This means energy must have been lost as heat.
Step 3: Calculate heat released
499 J - 435 J = 64 J
64J of energy must have been released.
Answer:
Shown below
Explanation:
a) for BrN3
80+3(14)=122amu
b) forC2H6
2(12) + 6(1) = 30amu
C) for NF2
14+2(19) = 52amu
D) Al2S3
2(27) + 3(32)= 150amu
E) for Fe(NO3)3
56 + 3 [14+3(16)] =242amu
F) Mg3N2
3(24) + 2(14)= 100amu
G) for (NH4)2CO3
2[14 +4(1)] +12 +3(16)=96amu
