Answer:
option A
an increase in entropy and a decrease in enthalpy
pls mark brainliest
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.
Your answer is probably
Vaporization point
<span>D.) Oxygen would acquire a stable arrangement of electrons by bonding with two atoms of "Magnesium"
[ As Mg has 2 extra electrons & their size are quite similar ]
Hope this helps!</span>
Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, 
Explanation:
The reaction equation for given reaction is as follows.
Here, 1 mole of reacts with 3 moles of 
.
As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide 
is 152 g/mol.
Number of moles is the mass of substance divided by its molar mass. So, moles of is calculated as follows.
Now, moles of .given by 0.5 mol of is calculated as follows.
As molar mass of is 2.016 g/mol. Therefore, mass of is calculated as follows.
Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, .
