Answer:
pH = 5.54
Explanation:
The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:
- pH = pKa + log
![\frac{[CH_3COO^-]}{[CH_3COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_3COO%5E-%5D%7D%7B%5BCH_3COOH%5D%7D)
For acetic acid, pKa = 4.75.
We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:
- CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
- CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH
The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.
Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:
- pH = 4.75 + log
= 5.54
Answer:
39.1 °C
Explanation:
Recall the equation for specific heat:
Where q is the heat, m is the mass, c is the specific heat of the substance (in this case water), and delta T is the change in temperature.
You should know that the specific heat of water is 1 cal/g/C.
Using the information in the question:
The final temperature is about 39.1 °C.
The answer is "elements" :)
DE = dH - PdV
<span>2 H2O(g) → 2 H2(g) + O2(g) </span>
<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>
<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>
<span>Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas </span>
<span>2 moles reacts to form 3 moles </span>
<span>The gas equation is </span>
<span>PV = nRT </span>
<span>P = pressure </span>
<span>V = volume (unknown) </span>
<span>n = moles (1) </span>
<span>R = gas constant = 8.314 J K^-1 mol^-1 </span>
<span>- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) </span>
<span>T = temp in Kelvin (kelvin = deg C + 273.15 </span>
<span>So T = 403.15 K </span>
<span>Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say </span>
<span>dE = dH -dnRT (because PV = nRT) </span>
<span>Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules </span>
<span>dH = 483.6 kJ/mol = 483600 Joules/mol </span>
<span>dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) </span>
<span>dE = 483600 J/mol - 3351.77 J </span>
<span>dE = 480248.23 J/mol </span>
<span>dE = 480.2 kJ/mol </span>
Your answer is 1-3-2-4 , grass sand water rock
