I wanna say 3,398.08 but let me know if that sounds very off
In order to find the answer you must plug in the information to the frequency equation,

. C is the speed of light, 3.0 x 10^8. By plugging in the information you receive a frequency of
3.0 x 10^7 hz.
Answer:
219.15 grams
Explanation:
What is the mass of 3.75 moles of NaCI? ( Na= 22.99g/mol, CI= 35.45 g/mol)
Mole of Na = 22.99g
Mole of Cl = 35.45g
For NaCl we have ratio of 1:1, so we have 1 Na for every Cl
So we just add the two together to get the molar mass of NaCl which is
22.99 + 35.45 = 58.44g/mol
And we know we have 3.75 moles of NaCl so we multiply that by the molar mass of NaCl to get our answer
3.75 x 58.44 = 219.15grams
The question is incomplete, complete question is :
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?
Structure is given in an image?
Answer:
There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Explanation:
Total numbers of carbon = 10
Number of primary carbons that is carbon joined to just single carbon atom = 6
Number of secondary carbons that is carbon joined to two carbon atoms = 1
Number of tertiary carbons that is carbon joined to three carbon atoms = 2
Number of quartenary carbons that is carbon joined to four carbon atoms = 1
So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.
Answer:
N₂O₄ + 14 kcal ⇄ 2NO₂.
Explanation:
Since the sign of ΔH determines either the reaction is exothermic or endothermic:
+ve, the reaction is endothermic.
-ve, the reaction is exothermic.
∵ The change of enthalpy of this reaction when proceeding left to right is + 14 kcal (+ ve sign).
∴ The reaction is endothermic, the heat is a part of the reacatnts in the reaction.
So, the reaction is:
N₂O₄ + 14 kcal ⇄ 2NO₂.