E = hc/(lamda)
The lamda symbol is wavelength, which this site does not have. I can represent it with an "x" instead.
Plancks constant, h = 6.626×10^-32 J·s
Speed of light, c = 3.00×10^8 m/s
The energy must be greater than or equal to 1×10^-18 J
1×10^-18 J ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / x
x ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / (1×10^-18 J)
x ≤ 1.99×10^-7 m or 199 nm
The wavelength of light must be greater than or equal to 199 nm
Answer:
A. 1350
You multiply 18.21HNO3* 1mol MgN2O6 * 148.30MgN2O6
Then divide it by the 2mol HNO3 to get 1350
Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of fusion = 3.17 kJ/mol
As we know that:
= freezing point temperature = 
Now put all the given values in the above formula, we get:
Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K
The answer is letter A definitively .
2.4212 X 10^ 7
How I at least figure this problem out is I take a pencil and start on the right side of the 0 and make a loop to the left for each number and count until I get to the first two numbers that are between 1-9 when reading from left to right. This is where you put the decimal point. Some teachers rather you keep the 0's there, while others prefer one to get rid of them. Anyways with that new decimal number, you multiply the decimal by ten to what ever number you counted, which was 7.
