Answer:
A spectral line is a dark or bright line in an otherwise uniform and continuous spectrum, resulting from emission or absorption of light in a narrow frequency range compared with the nearby frequency. spectral lines are often more used to identify atoms and molecules.
Explanation:
hope this helps
Answer:
0.422 mol/L.
Explanation:
- <em>Molarity (M)</em><em> is the no. of moles of solute dissolved in a 1.0 L of the solution.</em>
<em />
M = (no. of moles of glucose)/(volume of the solution (L))
M = (mass/molar mass)of glucose / (volume of the solution (L)).
∴ M = (mass/molar mass)of glucose / (volume of the solution (L)) = (289 g/342.2965 g/mol)/(2.0 L) = 0.422 mol/L.
Answer: The number of moles in 250.0 L of He at STP is 11.0 mole.
Explanation:
- It is known that 1.0 mole of a gas at STP conditions will occupy 22.7 L.
- To show this information: STP means that T = 0.0 °C = 273.15 K and P = 1.0 kPa = (100/101.325) = 0.9869 atm.
- From the ideal gas law: PV = nRT.
- Where, P is the pressure in atm <em>(P = 1.0 atm at STP).</em>
- n is the number of moles (n = 1.0 mole).
- R is the general gas constant (R = 0.0821 L.atm/mol.K).
- T is the temperature in K (T = 273.15 K at STP).
- and now we can get the volume of 1.0 mole at STP: V = nRT/P
- V = (1.0 mole x 0.0821 L.atm/mol.K x 273.15 K) / (0.9869 atm) = 22.7 L.
- Now, we can get the number of moles of 250.0 L of He at STP:
<em>Using cross multiplication:</em>
1.0 mole → 22.7 L
??? mole → 250.0 L
- The number of moles in 250.0 L of He at STP = (250.0 L x 1.0 mole) / (22.7 L) = 11.01 mole ≅ 11.0 mole.
Answer: The concentration of hydrogen ions is 
and the concentration of hydroxide ions is 
. Thus the concentration of hydroxide ions is more than the concentration of hydrogen ions.
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
Putting in the values:
![11=-\log[H^+]](https://tex.z-dn.net/?f=11%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-11}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-11%7DM)
![3=-log[OH^-]](https://tex.z-dn.net/?f=3%3D-log%5BOH%5E-%5D)
![[OH^-]=10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-3%7DM)
Thus the concentration of hydroxide ions is more than the concentration of hydrogen ions.
Answer:
Option (B) is correct
Explanation:
Oxidation: 
Reduction: 
---------------------------------------------------------------------------
Overall: 
Nernst equation for this cell reaction at 
:
![E_{cell}=(E_{Sn^{2+}\mid Sn}^{0}-E_{Zn^{2+}\mid Zn}^{0})-\frac{0.059}{n}log\frac{[Zn^{2+}]}{[Sn^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D%28E_%7BSn%5E%7B2%2B%7D%5Cmid%20Sn%7D%5E%7B0%7D-E_%7BZn%5E%7B2%2B%7D%5Cmid%20Zn%7D%5E%7B0%7D%29-%5Cfrac%7B0.059%7D%7Bn%7Dlog%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BSn%5E%7B2%2B%7D%5D%7D)
Where, n is number of electron exchanged and species inside third bracket represents concentrations in molarity.
Here, n = 2, 
and ![[Zn^{2+}]=2.5\times 10^{-3}M](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D%3D2.5%5Ctimes%2010%5E%7B-3%7DM)
So, plug in all the given values into above equation:
![0.660V=(-0.136V+0.76V)-\frac{0.059}{2}log\frac{2.5\times 10^{-3}M}{[Sn^{2+}]}](https://tex.z-dn.net/?f=0.660V%3D%28-0.136V%2B0.76V%29-%5Cfrac%7B0.059%7D%7B2%7Dlog%5Cfrac%7B2.5%5Ctimes%2010%5E%7B-3%7DM%7D%7B%5BSn%5E%7B2%2B%7D%5D%7D)
So, ![[Sn^{2+}]=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BSn%5E%7B2%2B%7D%5D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
As the value "0.059" varies from literature to literature and 
is most closest to 
therefore option (B) is correct.
