Different isotopes of the same element emit light at slightly different wavelengths, the minimum number of slits is mathematically given as
N=1820slits
<h3>What minimum number of slits is required to resolve these two wavelengths in second-order?</h3>
Generally, the equation for the wave is mathematically given as

Where the chromatic resolving power (R) is defined by

R = nN,
Therefore


and


In conclusion, the minimum number of slits is required to resolve these two wavelengths in second-order

Therefore

N=1820slits
Read more about slits
brainly.com/question/24305019
#SPJ1
Answer:
202 L
Explanation:
Step 1: Write the balanced equation
C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)
Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.16 g/mol.
270 g × 1 mol/180.16 g = 1.50 mol
Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose
The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol
Step 4: Calculate the volume of 9.00 moles of CO₂ at STP
The volume of 1 mole of an ideal gas at STP is 22.4 L.
9.00 mol × 22.4 L/mol = 202 L
Condensation is the change of water from its gaseous form (water vapor) into liquid water. Condensation generally occurs in the atmosphere when warm air rises, cools and looses its capacity to hold water vapor. As a result, excess water vapor condenses to form cloud droplets.
Molar mass of NH_3



We know.
No of moles=Given mass/Molar mass


Now
Lets write the balanced equation

- There is 2moles of Ammonia
- 3moles of H_2
- 1mole of N_2
Now

For Hydrogen



For Ammonia



For Nitrogen

