What new information can you add to your definition of chemical change

g Calculate the [OH-] from the results of your titrations. Explain your calculations2. Calculate the [Ca2+]. Explain your calcul

RideAnS [48]

Answer:

Explanation:

*Since the titration is between the strong acid HCl and the strong base Ca(OH)2, the pH at the equivalent point should be 7. On interpolation, we will obtain that 9.50mL and 9.82 mL of HCl is required to completely neutralized the given Ca(OH)2 solution.

*pH at the equivalence point =7

we know that pH + pOH = 14

Hence pOH= 14-7=7

pOH= -log(OH-)

The concentration of OH-= 10-pH= 1X10-7 M

One reason for the low solubility may be the higher reaction temperature, Another reason is the common ion effect.

3 0

3 years ago

Name the following lonic Compounds using the lonic naming rules. Remember, place the metal's name

liq [111]

1.Calcium Chloride
2.Lithium Bromide
3.Beryllium Sulfide
4.Lithium Fluoride
5. Potassium hydroselenide
6. Strontium phosphide
7.Barium Chloride
8.Iron Oxide
9.Iron
10.?
11.Copper Nitride

6 0

3 years ago

If 100% is not passed from one trophic level to the next, where does this energy go?

Sonja [21]

D is the answer to question

6 0

3 years ago

Read 2 more answers

What type(s) of intermolecular forces are expected between ch3ch2ch2ch2ch2oh molecules?

notka56 [123]

Answer : Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.

Explanation :

The given molecule is, $CH_3CH_2CH_2CH_2CH_2OH$

Three types of inter-molecular forces are present in this molecule which are Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.

Hydrogen-bonding : when the partial positive end of hydrogen is bonded with the partial negative end of another molecule like, oxygen, nitrogen, etc.

Dipole-dipole attraction : When the partial positively charged part of the molecule is interact with the partial negatively charged part of the molecule. For example : In case of HCl.

London-dispersion force : This force is present in all type of molecule whether it is a polar or non-polar, ionic or covalent. For example : In case of Br-Br , F-F, etc

Hydrogen-bonding is present between the oxygen and hydrogen molecule.

Dipole-dipole forces is present between the carbon and oxygen molecule.

London-dispersion forces is present between the carbon and carbon molecule.

5 0

4 years ago

The K a of propanoic acid ( C 2 H 5 COOH ) is 1.34 × 10 − 5 . Calculate the pH of the solution and the concentrations of C 2 H 5

Zigmanuir [339]

Answer:

2.62.

Explanation:

Okay let us first write the parameters in the question in question above out. We are given the ka value of propanoic acid, C2H5COOH to be equals to 1.34 × 10^- 5. Also, we are given the value for the initial concentration of propanoic acid to be 0.441 M.

So, let us delve right into the solution to the question and we will be starting by writting the equation below;

C2H5COOH <--------> H^+ + C2H5COO^-.

Please note that this Reaction is a reversible Reaction.

Therefore, the basic things about acid is its great tendency to release Hydrogen ion in an aqeous solution.

So, we will be taken equation above and correspond it with the time and Concentration.

C2H5COOH <----> H^+ C2H5COO^-.

Initial concentration of the C2H5COOH = 0.441 M and the initial concentration of H^+ and C2H5COO^- are both zero.

So, after a time, t, concentration of C2H5COOH= 0.441 - x and at that time the concentration of H^+ and C2H5COO^- are both x and x respectively.

Hence, Ka = [C2H5COO^-] [H^+]/ C2H5COOH. -----------------------(**).

Therefore, slotting in the values from above into equation (**), we have;

1.34 × 10^-5 = [x] [x]/ [0.441 - x].

1.34 × 10^-5= x^2/ [0.441 - x].

x^2 = 1.34 × 10^-5(0.441) - 1.34 × 10^-5x.

x^2 + 1.34 × 10^-5x - 5.91× 10^-6.

x = 2.4×10^-3.

Hence, the concentration of the propanoic acid at time, t= 0.441 - 2.4 ×10^-3.

==> 0.44 M.

pH = -log [H^+].

Then, we have; pH= - log[2.4× 10^-3].

pH= 2.62.

4 0

3 years ago