What is the difference between the small and large intestines?

Use the molar solubility, 1.08×10−5m, in pure water to calculate ksp for bacro4.

Dmitriy789 [7]

The solubility constant or Ksp is calculated by the product of the concentration of the ions raise to the stoichiometric coefficient. We calculate as follows:

BaCrO4 = Ba2+ + CrO42-

Ksp = [Ba2+][CrO4]
Ksp = [1.08×10−5<span> ] [1.08×10−5] = 1.1664x10^-10</span>

8 0

4 years ago

Which requires more heat to warm from 22.0°C and 85.0°C, 45.0 g of water or 200. g of aluminum metal?

stich3 [128]

Answer:

The answer to your question is Aluminum

Explanation:

We need to warm from 22°C to 85°C

a) 45 g of water

b) 200 g of aluminum Cp = 0.905 J/cal°C

a) Water

Q = mCpΔT

Q = (45)(1)(85 - 22) = 45(63) = 2835 cal

b) Aluminum

Q = (200)(0.905)(85 - 22) = 114030 cal

5 0

3 years ago

Elements with similar properties [A] sodium and bromine [B] lithium and sodium [C] carbon and oxygen [D] potassium and helium

Shkiper50 [21]

Usually, elements which share the same group, they have similar properties, here only Lithium and sodium share same group as they both are alkali metals.

So, In short, Your Answer would be option B) Lithium and Sodium

Hope this helps!

8 0

3 years ago

A certain half-reaction has a standard reduction potential E0red = +0.13V . An engineer proposes using this half-reaction at the

Ivan

Answer:

a. 1.23 V

b. No maximum

Explanation:

Required:

a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?

b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

If E°cell must be at least 1.10 V (E°cell > 1.10 V),

E°red, cat - E°red, an > 1.10 V

E°red, cat - 0.13V > 1.10 V

E°red, cat > 1.23 V

The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.

4 0

3 years ago

alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decima

svetlana [45]

A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$ = 0.10 M

The chemical equation for this reaction is :

$\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}$

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is = $\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}$

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = $\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}$ = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

$\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}$

Initial 0.10 0.035 0.10 -

Change -0.035 -0.035 + 0.035 -

Equilibrium 0.065 0 0.135 -

By using Henderson-Hasselbalch equation:

The pH of this solution = pKa + log $\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}$

The pH of this solution = 4.74 + log $\mathtt{\dfrac{0.135}{0.065}}$

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06 to two decimal places

7 0

3 years ago