Besides the concentration of the acid and its conjugate base, what else does one need to calculate the ph of a weak acid

Calculate the molarity of a solution obtained dissolving 10.0 g of cobalt(Ⅱ) bromide tetrahydrate in enough water to make 450 mL

Vladimir [108]

Answer:

The molarity of the CoBr2•4H2O solution is 7.64 × 10-2 M

Explanation:

Cobalt (II) bromide tetrahydrate

• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+

• Bromide - anion from group 7A → -1 charge → symbol: Br-

• Tetrahydrate- tetra- means 4 and hydrate is H2O

The chemical formula of the compound is: CoBr2•4H2O

We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.

We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.

•The molar mass of CoBr2•4H2O is:

CoBr2•4H2O

1 Co x 58.93 g/mol Co = 58.93 g/mol

2 Br x 79.90 g/mol Br = 159.80 g/mol

8 H x 1.008 g/mol H = 8.064 g/mol

4 O x 16.00 g/mol O = 64.00 g/mol

________________________________________

Sum = 290.79 g/ mo

The moles of CoBr2•4H2O is:

= 10.0 g CoBr2•4H2O x $\frac{ 1 mol CoBr_2 . 4H_2O}{290.79 g CoBr_2 . 4H_2O}$

= 0.0344 mol CoBr2•4H

We know that the volume of the solution is 450 mL.

We can now calculate for molarity:

Convert mL to L → 1 mL = 10-3 L

Formula:

Molarity (M)= Mole of solute / Liters of solution

= 0.0344 mol CoBr2•4H / 450 mL x 1 ml / 10^ -3 L

= 0.0764

= 7.64 × 10-2 mol/L

8 0

3 years ago

1500 millimeters to km

gizmo_the_mogwai [7]

0.0015 kilometers is for sure the answer!

4 0

3 years ago

1. Consider the decomposition reaction of sodium chlorate. There are 100 grams of

IRINA_888 [86]

A. NaCl(s) and O2(g)

B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)

C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3

D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)

E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl

F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2

G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2

H. Percent yield = 10/45.1 • 100% = 22.2% yield

6 0

3 years ago

A 1.0 g sample of hydrogen reacts completely with 19.0 g of fluorine to form a compound of hydrogen and fluorine. a. What is the

Rashid [163]

Explanation:tr

a) Molar mass of HF = 20 g/mol

Atomic mass of hydrogen = 1 g/mol

Atomic mass of fluorine = 19 g/mol

Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{Molar mass of compound }}\times 100$