The formula is m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
1 mole = 6.022 E23 atoms
0.97 mole=?
6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms
Answer:
The answer is: <u>Al2O3</u>
Explanation:
The data they give us is:
To find the empirical formula without knowing the grams of the compound, we find it per mole:
- 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
- 0.485 g O * 1 mol O / 16 g O = 0.03 mol O
Then we must divide the results obtained by the lowest result, which in this case is 0.02:
- 0.02 mol Al / 0.02 = 1 Al
- 0.03 mol O / 0.02 = 1.5 O
Since both numbers have to give an integer, multiply by 2 until both remain integers:
Now the answer is given correctly:
<span>Answer:
H-C-N H-N-C C-H-N
Notice that C-H-N is the same as N-H-C just written backwards. ( i.e. they have the same connectivtiy.) You can exclude the last one with H in the middle since H has two bonds and 4 electrons around it. At this point you couldn't differentiate between the first two, so I would give you the connectivity in such a problem, which in this case is H-C-N.</span>
Answer:
V CH4(g) = 190.6 L
Explanation:
assuming ideal gas:
∴ STP: T =298 K and P = 1 atm
∴ R = 0.082 atm.L/K.mol
∴ moles (n) = 7.80 mol CH4(g)
∴ Volume CH4(g) = ?
⇒ V = RTn/P
⇒ V CH4(g) = ((0.082 atm.L/K.mol)×(298 K)×(7.80 mol)) / (1 atm)
⇒ V CH4(g) = 190.6 L
