What is the term?

Identify the functional groups attached to the benzene ring as either, being electron withdrawing, electron donating, or neither

Dominik [7]

-OH is elctron donating -C=-N is electron withdrawing -O-CO-CH3 is electron withdrawing -N(CH3)2 is electron donating -C(CH3)3 is electron donating -CO-O-CH3 is electron withdrawing -CH(CH3)2 is electron donating -NO2 is electrong withdrawing -CH2

8 0

3 years ago

What is the limitation of the kinetic molecular theory

Minchanka [31]

Answer:

1)The molecules themselves take up some volume.

2)There is some attractions between the molecules.

( in ideal gas the molecules volume is neglected

and it's assume there is no attractions between the molecules)

Explanation:

3 0

3 years ago

What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 15.0% NaOH by mass?

GalinKa [24]

Hello!

a) The mole fraction of solute of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams of water. We can calculate the number of moles of each one in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH \\ \\ moles H_2O=85 gH_2O* \frac{1 mol H_2O}{18 g H_2O}=4,7222 moles H_2O$

To finish, we calculate the mole fraction by dividing the moles of NaOH between the total moles:

$X_{NaOH}= \frac{moles NaOH}{total moles}= \frac{0,3750 moles NaOH}{0,3750 moles NaOH+4,7222 molesH_2O} =0,073$

So, the mole fraction of NaOH is 0,073

b) The molality (moles NaOH/ kg of solvent) of a 15% NaOH aqueous solution can be calculated in the following way:

First, we have to assume that we have 100 grams of solution. This will simplify the calculations.

Now, we know that this solution has 15 grams of NaOH and 85 grams (0,085 kg) of water. We can calculate the moles of NaOH in the following way:

$molesNaOH= 15 g NaOH*\frac{1molNaOH}{39,997 g NaOH}=0,3750 moles NaOH$

Now, we apply the definition of molality to calculate the molality of the solution:

$mNaOH= \frac{moles NaOH}{kg_{solvent}}= \frac{0,3750 moles NaOH}{0,085 kg H_2O}=4,41 m$

So, the molality of this solution is 4,41 m

Have a nice day!

4 0

3 years ago

96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight?

g100num [7]

Answer:

I hope this helps 52.2 g/mol

Explanation:

1) Solve for the moles using PV = nRT:

n = PV / RT

n = [(700.0 mmHg / 760.0 mmHg atm¯1) (48.0 L)] / [(0.08206 L atm mol¯1 K¯1) (293.0 K)]

n = 1.8388 mol

2) Divide the grams given (96.0) by the moles just calculated above:

96.0 g / 1.8388 mol = 52.2 g/mol

7 0

2 years ago

The empirical formula of hydrazine, molecular mass 32 grams, is NH2. What is the molecular formula of hydrazine?

fgiga [73]

Answer:

C) I think..

Explanation:

7 0

2 years ago