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: ≈ 67.4 g
Explanation:
A(t) = amount remaining in t years = A0ekt, where A0 is the initial amount and k is a constant to be determined. Since A(1690) = (1/2)A0 and A0 = 80, we have 40 = 80e1690k 1/2 = e1690k ln(1/2) = 1690k k = -0.0004 So, A(t) = 80e-0.0004t Therefore, A(430) = 80e-0.0004(430) = 80e-0.172 ≈ 67.4 g
Answer:
5.112 grams of simple sugar glucose is produced by the reaction of 7.50 g of carbon dioxide.
Explanation:
![6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2](https://tex.z-dn.net/?f=6CO_2%2B6H_2O%5Crightarrow%20C_6H_%7B12%7DO_6%2B6O_2)
Moles of carbon dioxide = ![\frac{7.50 g}{44 g/mol}=0.1704 mol](https://tex.z-dn.net/?f=%5Cfrac%7B7.50%20g%7D%7B44%20g%2Fmol%7D%3D0.1704%20mol)
According to reaction 6 moles of carbon dioxide gives 1 mole of glucose.
Then 0.1704 moles of carbon dioxide will give:
of glucose
Mass of 0.0284 moles of glucose:
0.0284 mol 180 g/mol =5.112 g[/tex]
5.112 grams of simple sugar glucose is produced by the reaction of 7.50 g of carbon dioxide.
Ethane can be turned into its carboxylic acid form called the ethanoic acid by a series of steps. First chlirinate it in the presence of light. Then, add potassium hydroxide to get an alcohol. Finally, adding KMnO4 to oxidize it to a carboxylic acid. The equations are as follows:
C2H6 + Cl2 ---> C2H5Cl + HCl
C2H5Cl+KOH-> C2H5OH+ KCl.
C2H5OH ----KMnO4 ---> CH3COOH.