What is the combining number of lead?

g A 500. mL solution contains 0.665 M NaC2H3O2 and 0.475 M HC2H3O2. What mass of HCl in grams needs to be added for the solution

77julia77 [94]

Answer:

7.38g HCl

Explanation:

Using H-H equation for acetic buffer:

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

Where pKa is -log Ka = 4.74 and [] could be taken as moles of each compound.

The initial moles of each specie is:

[NaC2H3O2]:

0.500L * (0.665mol/L) = 0.3325moles

[HC2H3O2]:

0.500L * (0.475mol/L) = 0.2375 moles

That means total moles are:

[NaC2H3O2] + [HC2H3O2] = 0.57 moles (1)

And solving H-H equation for a pH of 4.21:

4.21 = 4.74 + log [NaC2H3O2] / [HC2H3O2]

0.29512 = [NaC2H3O2] / [HC2H3O2] (2)

Replacing (1) in (2):

0.29512 = 0.57mol - [HC2H3O2] / [HC2H3O2]

0.29512 [HC2H3O2] = 0.57mol - [HC2H3O2]

1.29512 [HC2H3O2] = 0.57mol

[HC2H3O2] = 0.44 moles

The HCl reacts with NaC2H3O2 producing HC2H3O2, that means you need to add:

0.44 moles - 0.2375 moles =

0.2025 moles of HCl

Using molar mass of HCl (36.45g/mol), to convert these moles to grams:

0.2025 moles * (36.45g/mol) =

8 0

3 years ago

Question 3 0

Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

5 0

3 years ago

Phosgene is formed from carbon monoxide and chlorine. At 611 oC, Kc is approximately 16.5 M-1 or L/mol. Calculate Kp (in atm-1)

defon

Answer:

Kp is 0.228/atm

Explanation:

This is the reaction:

CO + Cl₂ → CCl₂O

1 mol of carbon monoxide and 1 mol of chlorine produce 1 mol of phosgene.

Formula for Kp which derivates from Kc is:

Kp = Kc (R.T)ⁿᵇ ⁻ ⁿᵃ

Δп = nb (moles in the products) - nₐ(moles in the reactants)

Δп = 1 - 2 = -1

T is T° in K → T°C + 273 = 611°C +273 = 884K

R → Universal constant gas → 0.082 L.atm/mol.K

We replace the data: Kp = 16.5 L/mol (0.082 . 884K)⁻¹ → 0.228/atm

5 0

3 years ago

Identify the balanced single displacement reaction.

tekilochka [14]

CuCl2 + Fe ⟶ Cu + FeCI2

3 0

3 years ago

Read 2 more answers

The vapor pressure of benzene is 73.03 mm Hg at 25°C. How many grams of estrogen (estradiol), C18H24O2, a nonvolatile, nonelectr

alexira [117]

Answer:

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

Explanation:

The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

Where:

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of the solution

$n_1$ = Number of moles of solvent

$n_2$ = Number of moles of solute

$p_o = 73.03 mmHg$

$p_s= 71.61 mmHg$

$n_1=\frac{216.7 g}{78.12 g/mol}=2.7739 mol$

$\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}$

$n_2=0.05499 mol$

Mass of 0.05499 moles of estrogen :

= 0.05499 mol × 272.4 g/mol = 14.9802 g

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

6 0

3 years ago