Let <span>A= ebc </span>
<span> 0.765 = 86,300*1.3*c </span>
<span>
Solve for c = approximately 7E-6 Molar
= mols/L soln. </span>
<span>g = 7E-6*893.49
= about 0.006 g chlorophyll/L soln. </span>
<span>
1000 x 0.785 = 785 g ethanol. </span>
<span>Conc. = about 0.006g chlorophyll/785 g soln. </span>
<span>
Change that to ppm. by using formula:
(0.006/785)*1E6</span>
Well they help your body in lots of different ways so you could say that they help you in your senses if im crrect
Answer:
The dipole moment of H-C bond will be smaller than that of an H-I bond.
Explanation:
The electronegativity of iodine is greater than that of hydrogen.As a result the iodine atom tends to attract the bond electron pair of H-I bond towards itself creating a bond dipole which does not occur in case of H-C bond as the electronegativity of carbon and hydrogen are almost same.
That"s why dipole moment of H-C bond is smaller than that of H-I bond.
Answer:
-152.92°C
Explanation:
Initial volume = 0.750 mL
Initial temperature = -32.7 °C (-32.7 + 273.15 K = 240.45 k)
Final temperature = ?
Final volume = 0.750 mL / 2= 0.375 mL
Solution:
The given problem will be solve through the Charles Law.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = T₁V₂ / V₁
T₂ = 240.45 k × 0.375 mL / 0.750 mL
T₂ = 90.17 mL.K / 0.750 mL
T₂ = 120.23 K
Temperature in celsius:
120.23 K - 273.15 = -152.92°C
Answer:
MoO₃
Explanation:
To solve this question we must find the moles of molybdenum in Mo2O3. The moles of Mo remain constant in the new oxide. With the differences in masses we can find the mass of oxygen and its moles obtaining the empirical formula as follows:
<em>Moles Mo2O3 -Molar mass: 239,878g/mol-</em>
11.79g * (1mol / 239.878g) = 0.04915 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.09830 moles Mo
<em>Mass Mo in the oxides:</em>
0.09830 moles Mo * (95.95g/mol) = 9.432g Mo
<em>Mass oxygen in the new oxide:</em>
14.151g - 9.432g = 4.719g oxygen
<em>Moles Oxygen:</em>
4.719g oxygen * (1mol/16g) = 0.2949 moles O
The ratio of moles of O/Mo:
0.2949molO / 0.09830mol Mo = 3
That means there are 3 moles of oxygen per mole of Molybdenum and the empirical formula is:
<h3>MoO₃</h3>
