We calculate for the amount of chromium metal in the reactant by,
= 350 x (mass of Cr2/mass of Cr2O3)
= 350 x (104/152)
= 239.47 grams
The amount of Cr metal in the product is only 213.2 grams. Thus, the percent yield.
percent yield = (213.2 grams/239.47) x 100%
= 89%
<span>Displaced volume :
</span>Final volume - <span>Initial volume
</span>13.45 mL - 12.00 mL => 1.45 mL
Mass = 4.50 g
Therefore:
density = mass / volume
D = 4.50 / 1.45
<span>D = 3.103 g/mL </span>
Answer:
0.258 mg of iron remains.
Explanation:
To solve this problem we can use the formula
M₂ = M₀ * 
Where M₂ is the mass remaining, M₀ is the initial mass, and t is time in days.
Using the data given by the problem:
M₂ = 2.000 mg * 
M₂ = 0.258 mg
Answer:
333.7 g.
Explanation:
- The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>
Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).
Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>
molaity (m) is the no. of moles of solute per kg of the solvent.
∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))
∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.
∵ no. of moles = mass/molar mass.
<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>
The trends and exceptions to the trends in ionization energy observed includes;
B, ionization energy tends to increase across a period because the nuclear charge increases.
C, ionization energy tends to increase across a period because electrons are added to the same main energy level.
E, The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron, in the p orbital from the nucleus.
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example to remove an electron from a neutral fluorine atom to form a positively charged ion. <span />
