The following are the answers to the different questions:
<span>The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?
A. 101.53° C
Which of the following solutions will have the lowest freezing point?
D. 1.0 mol/kg magnesium fluoride (MgF2)
Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?
A. sodium iodide (NaI)
Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?
B. 96.3 kPa
Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?
D. 108.1°C</span>
In a chemical reaction,
the limiting reagent is the chemical being used up while the excess reactant is
the chemical left after the reaction process.
Before calculating the limiting
and excess reactant, it is important to balance the equation first by stoichiometry.
C25N3H30Cl + NaOH = C25N3H30OH + NaCl
Since the reaction is already balanced, we can now identify which
is the limiting and excess reagent.
First, we need to determine the number of moles of each chemical
in the equation. This is crucial for determining the limiting and excess reagent.
<span>Assuming that there is the
same amount of solution X for each reactant</span>
1.0 M NaOH ( X ) = 1.0
moles NaOH
1.00 x 10-5 M C25N3H30Cl
( X ) = 1.00 x 10-5 moles C25N3H30Cl
<span>The result showed that the
crystal violet has lesser amount than NaOH. Thus, the limiting reactant in this
chemical reaction is crystal violet and the excess reactant is NaOH.</span>
Answer:
2.4 ×10^24 molecules of the herbicide.
Explanation:
We must first obtain the molar mass of the compound as follows;
C3H8NO5P= [3(12) + 8(1) + 14 +5(16) +31] = [36 + 8 + 14 + 80 + 31]= 169 gmol-1
We know that one mole of a compound contains the Avogadro's number of molecules.
Hence;
169 g of the herbicide contains 6.02×10^23 molecules
Therefore 669.1 g of the herbicide contains 669.1 × 6.02×10^23/ 169 = 2.4 ×10^24 molecules of the herbicide.
1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm
