Which drawing best accounts for the polarity of methanol, CH3OH, and the bond polarities that make a major contribution to the overall molecular polarity?

A) drawing (1) B) drawing (2)

D) drawing (4) C) drawing (3)

Answer:

B) drawing (2)

Explanation:

In Chemistry, the direction of dipole is shown from positive end to negative end.

The image that contains the options in the question asked has been attached.

We can see in image 2 that the oxygen atom was correctly designated as the negative end of the dipole while the carbon and hydrogen atoms were each designated as positive ends of the dipole in accordance with the magnitude of electronegativity difference between the two atoms. The net dipole moment is now taken in the direction shown in image 2. This is the correct answer.

3 0

3 years ago

What is the empirical formula of a compound composed of 3.25% hydrogen (H), 19.36% carbon (C), and 77.39% oxygen (O) by mass?

blsea [12.9K]

CH2O3

C=19.36\12 = 1.61
H=3.25\1.008 =3.22
O=77.39\16=4.83

then dived the moles with the smallest mole

C=1.61\1.61=1
H=3.22\1.61=2
O=4.83\1.61=3

E.F
CH2O3

8 0

3 years ago

Read 2 more answers

The standard molar enthalpy of formation of NH3(g) is -45.9 kJ/mol. What is the enthalpy change if 9.51 g N2(g) and 1.96 g H2(g)

Vladimir [108]

Answer:

$\Delta H=-29.7kJ$

Explanation:

Hello!

In this case, since the undergoing chemical reaction is:

$N_2+3H_2\rightarrow 2NH_3$

We first need to identify the limiting reactant given the masses of nitrogen and hydrogen:

$n_{NH_3}^{by\ H_2}=1.96gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.647molNH_3\\\\ n_{NH_3}^{by\ N_2}=9.51gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.679molNH_3$

It means that only 0.647 moles of ammonia are yielded, so the resulting enthalpy change is:

$\Delta H=0.647molNH_3*\frac{-45.9kJ}{1molNH_3}\\\\ \Delta H=-29.7kJ$

Best regards!

3 0

3 years ago