The concentration of the reactants and products remain constant. Because the rates of the forward and reverse reaction are equal there is no net change to the amount of reactants or products produced.May 19, 2011
Answer:
<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>
Explanation:
This question is about solubility.
Regarding solubility, the solutions may be classified as:
- Unsaturated: the concentration is below the maximum concentration permited at the given temperature.
- Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.
- Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.
Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.
- In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.
- In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g of water.
- Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ
115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O ⇒ x = 57.5 g NaNO₃
- <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>
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Answer:
4.52 x 10¹⁴ cycles/s
Explanation:
From c = f·λ => f = c/λ = (3.0 x 10⁸ m/s)/(6.63 x 10⁻⁷m) = 4.52 x 10¹⁴ cycles/s.
f = frequency = ?
λ = wavelength = 6.63 x 10⁻⁷ meter
c = speed of light in vacuum = 3.0 x 10⁸ meters/s
Absorbed photon energy
Ea = hc/λ.. (Planck's equation)
Ea = hc / 92.05^-9m
<span>Energy emitted
Ee = hc/ 1736^-9m </span>
Energy retained ..
∆E = Ea - Ee = hc(1/92.05<span>^-9 - 1/1736^-9) </span>
<span>∆E = (6.625^-34)(3.0^8) (1.028^7)
∆E = 2.04^-18 J </span>
<span>Converting J to eV (1.60^-19 J/eV)
∆E = 2.04^-18 / 1.60^-19
∆E = 12.70 eV </span>
<span>Ground state (n=1) energy for Hydrogen = - 13.60eV </span>
<span>New energy state = (-13.60 + 12.70)eV = -0.85 eV </span>
<span>Energy states for Hydrogen
En = - (13.60 / n²) </span>
n² = -13.60 / -0.85 = 16
n = 4
Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.
Explanation:
Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.
A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.
A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.
B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.
B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.
Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.
Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.
