Answer:
The answer to your question is C₂HO₃
Explanation:
Data
Hydrogen = 3.25%
Carbon = 19.36%
Oxygen = 77.39%
Process
1.- Write the percent as grams
Hydrogen = 3.25 g
Carbon = 19.36 g
Oxygen = 77.39 g
2.- Convert the grams to moles
1 g of H ----------------- 1 mol
3,25 g of H ------------- x
x = (3.25 x 1) / 1
x = 3.25 moles
12 g of C ---------------- 1 mol
19.36 g of C ---------- x
x = (19.36 x 1) / 12
x = 1.61 moles
16g of O --------------- 1 mol
77.39 g of O --------- x
x = (77.39 x 1)/16
x = 4.83
3.- Divide by the lowest number of moles
Carbon = 3.25/1.61 = 2
Hydrogen = 1.61/1.61 = 1
Oxygen = 4.83/1.61 = 3
4.- Write the empirical formula
C₂HO₃
Answer:
See explanation
Explanation:
All molecules possess the London dispersion forces. However London dispersion forces is the only kind of intermolecular interaction that exists in nonpolar substances.
So, the only kind of intermolecular interaction that exists in dimethyl ether is London dispersion forces.
As for ethyl alcohol, the molecule is polar due to the presence of polar O-H bond. In addition to London dispersion forces, dipole-dipole interactions and specifically hydrogen bonding also occurs between the molecules.
Because ethyl alcohol is polar, it is more soluble in water than dimethyl ether.
Answer:
[C₆H₁₂O₆] = 0.139 M
Explanation:
Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.
We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.
For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)
(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL
We determine moles, we convert them to mmoles, we divide by mL
M = 0.139 M
Moles = 3.95 g . 1mol / 180g → 0.0219 mol
We convert mL to L → 158 mL . 1L/1000mL = 0.158L
M = 0.0219 mol / 0.158L = 0.139 M
Sb has the largest atomic radius by 206
Initial concentration of magnesium nitrate M1 = 2.13 M
Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L
Final concentration of MgNO3, M2 = 1.60 M
Let the final volume of MgNO3 upon dilution be V2
Formula to use:
M1*V1 = M2*V2
V2 = M1*V1/M2
= 2.13 M * 1.24 L/1.60 M = 1.65 L
Thus, the final volume of magnesium nitrate solution upon dilution is 1.65 L
