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jolli1 [7]
3 years ago
12

Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [

Chemistry
1 answer:
diamong [38]3 years ago
8 0

Answer: pH=2.38

Explanation:

To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.

            HCOOH ⇄ H⁺ + HCOO⁻

I               1.0M          0          0

C              -x            +x        +x

E            1.0-x            x          x

<u>For the steps below, refer to the ICE chart above.</u>

1. Since we were given the initial of HCOOH, we can fill this into the chart.

2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.

3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.

4. We were given the Kₐ of the solution. We know K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}.

5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.

1.8*10^-^4 =\frac{x^2}{0.1-x}

6. Once we plug this into the quadratic equation, we get x=0.00415.

7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).

-log(0.00415)=2.38

Our pH for the weak acid solution is 2.38.

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using information in titles of appendices b and c, calculate the minimum grams of propane, C3H8 (g), that must be combusted to p
Lubov Fominskaja [6]

<u>Given:</u>

Mass of ice = mass of water = 5.50 kg = 5500 g

Temperature of ice = -20 C

Temperature of water = 75 C

<u>To determine:</u>

Mass of propane required

<u>Explanation:</u>

Heat required to change from ice to water under the specified conditions is:-

q = q(-20 C to 0 C) + q(fusion) + q (0 C to 75 C)

  = m*c(ice)*ΔT(ice) + m*ΔHfusion + m*c(water)*ΔT(water)

  = 5500[2.10(0-(-20)) + 334 + 4.18(75-0)] = 3792 kJ

The enthalpy change for the combustion of propane is -2220 kJ/mol

Therefore, the number of moles of propane corresponding to the required energy of 3792 kJ = 1 mole * 3792 kJ/2220 kJ = 1.708 moles of propane

Molar mass of propane = 44 g/mol

Mass of propane required = 1.708 moles * 44 g/mol = 75.15 g

Ans: 75.15 grams of propane must be combusted.



4 0
4 years ago
A 7.337 gram sample of chromium is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 9.595 g. Dete
Thepotemich [5.8K]
<h3>Answer:</h3>

Empirical formula is CrO

<h3>Explanation:</h3>

<u>We are given;</u>

  • Mass of sample of Chromium as 7.337 gram
  • Mass of the metal oxide formed as 9.595 g

We are required to determine the empirical formula of the metal oxide.

<h3>Step 1 ; Determine the mass of oxygen used </h3>

Mass of oxygen = Mass of the metal oxide - mass of the metal

                          = 9.595 g - 7.337 g

                         = 2.258 g

<h3>Step 2: Determine the moles of chromium and oxygen</h3>

Moles of chromium metal

Molar mass of chromium = 51.996 g/mol

Moles of Chromium = 7.337 g ÷ 51.996 g/mol

                                 = 0.141 moles

Moles of oxygen

Molar mass of oxygen = 16.0 g/mol

Moles of Oxygen = 2.258 g ÷ 16.0 g/mol

                            = 0.141 moles

<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>

Mole ratio of Chromium to Oxygen

          Cr : O

0.141 mol : 0.141 mol

             1 : 1

Empirical formula is the simplest whole number ratio of elements in a compound.

Thus the empirical formula of the metal oxide is CrO

6 0
3 years ago
An unknown compound, X is thought to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 an
Westkost [7]

Answer:

7.3

Explanation:

By Henderson Hasselbalch equation we can calculate the pH or the pOH of a solution by its pKa. Remember that pH = -log[H^{+}], and pKa = -logKa. Ka is the equilibrium constant of the acid.

Henderson Hasselbalch equation :

pH = pKa - log \frac{[HA]}{[A^{-}]}

Where [HA] is the concentration of the acid, and [A^{-}] is the concentration of the anion which forms the acid.

So, acid X, has two ionic forms, the carboxyl group and the other one. First, we have 0.1 mol/L of the acid, in 100 mL, so the number of moles of X

n1 = (0.1 mol/L)x(0.1 L) = 0.01 mol

When it dissociates, it forms 0.005 mol of the carboxyl group and 0.005 mol of the other group. Assuming same  stoichiometry.

Adding NaOH, with 0.1 mol/L and 75 mL, the number of moles of OH^- will be

n2 = (0.1 mol/L)x(0.075 L) = 0.0075 mol

So, the 0.0075 mol of OH^- reacts with 0.005 mol of carboxyl, remaining 0.0025 mol of OH^-, which will react with the 0.005 mol of the other group. So, it will remain 0.0025 mol of the other group.

The final volume of the solution will be 175 mL, but both concentrations (the acid form and ionic form) have the same volume, so we can use the number of mol in the equation.

Note that, the number of moles of the acid form is still 0.01 mol because it doesn't react!

So,

6.72 = pKa - log \frac{0.01}{0.0025}

6.72 = pKa - log 4

pKa - log4 = 6.72

pKa = 6.72 + log4

pKa = 6.72 + 0.6

pKa = 7.3

8 0
4 years ago
What is the three-dimensional shape of the molecule with this Lewis structure.
atroni [7]

I would have to say B. Linear

I hope this helps!

Cheers, July.


7 0
3 years ago
Read 2 more answers
Put the steps in the process of photosynthesis in order
shepuryov [24]

Step 1-Light Dependent

CO2 and H2O enter the leaf

Step 2- Light Dependent

Light hits the pigment in the membrane of a thylakoid, splitting the H2O into O2

Step 3- Light Dependent

The electrons move down to enzymes

Step 4-Light Dependent

Sunlight hits the second pigment molecule allowing the enzymes to convert ADP to ATP and NADP+ gets converted to NADPH

Step 5-Light independent

The ATP and NADPH is used by the calvin cycle as a power source for converting carbon dioxide from the atmosphere into simple sugar glucose.

Step 6-Light independent

The calvin cycle converts 3CO2 molecules from the atmosphere to glucose

calvin cycle

The second of two major stages in photosynthesis (following the light reactions), involving atmospheric CO2 fixation and reduction of the fixed carbon into carbohydrate.

4 0
4 years ago
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