The correct answer that would best complete the given statement above would be option 2. <span>The relationship between molecular velocities and temperature is a direct relationship. In other words, their relationship is directly proportional. Hope that this is the answer that you are looking for. </span>
Answer:
Six C atoms (C₆); five H atoms (H₅); one N atom (N); no O atoms
Explanation:
The rule of 13 states that the formula of a compound is a multiple n of 13 (the molar mass of CH) plus a remainder r.
MF = CₙHₙ₊ᵣ
Y has a molecular mass of 91 u
91/13 =7r0
The formula can't be C₇H₇ because a hydrocarbon must have an even number of H atoms,
The odd mass and the odd number of H atoms make it reasonable to add an N atom and subtract CH₂ (CH₂ = 14):
C₇H₇ + N - CH₂ = C₆H₅N
Check:
6C = 6 × 12.000 = 72.000 u
5H = 5 × 1.008 = 5.040
1N = 1 × 14.003 = <u>14.003 </u>
TOTAL = 91.043 u
This is excellent agreement with the observed mass of 91.0425 u.
There are six C atoms (C₆)
There are five H atoms (H₅)
There is one N atom (N)
There are no O atoms.
The stoichiometry of the reaction gives the molar ratio in which the reactants react with each other and the ratio in which products are formed.
The coefficients of the reactants in the reaction follow the stoichiometry
the balanced chemical equation for the reaction is as follows;
2C₃H₆(g) + 9O₂(g) ---> 6CO₂(g) + 6H₂O(l)
Answer:
The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.
Explanation:
To know the mass in grams of 0.280 moles of sample of sodium hydroxide NaOH, you must know the molar mass of the compound, that is, the mass of one mole of a substance, which can be an element or a compound.
So you know:
- Na: 23 g/mole
- O: 16 g/mole
- H: 1 g/mole
So, the molar mass of NaOH is:
NaOH= 23 g/mole + 16 g/mole+ 1 g/mole= 40 g/mole
Then the following rule of three can be applied: if in 1 mole of sodium hydroxide there are 40 grams, in 0.280 moles how much mass is there?
mass= 11.2 grams
<u><em>The mass of 0.280 mole sample of sodium hydroxide NaOH is 11.2 grams.</em></u>
