Answer:
5118.50 J
Explanation:
pΔv=nRΔT ;
therefore, ΔT=PV/nR
ΔT = (6.4×10^5)(3.2×10^(-3)/1×8.314
ΔT= 2.4633×10^2 = 246.33 K
specific heat at constant pressure is given as:
c_p = 3/2R
c_p = 12.5 J/mol K
Now, substitute in equation (1)
we know that
Q=ΔU+W ;
and
W=pΔV= 6.4×10^5×3.2×10^(-3) = 2048 J
now
ΔU=CvΔT = 12.465×246.33 =3070.50 J ;
therefore
Q=3070.50+2048= 5118.50 J
<span>When the aqueous solution is heated to 85âC, the gas entrapped in the aqueous solution is released since the vapour pressure of the particular entrapped gas is more than the solution. The solid saturated in the aqueous solution returns back to its original form once the aqueous solution evaporates. A good example is that of the salt left behind after heating a brine solution to evaporate.</span>
Answer:
i dont know thank you soo much
Answer:
6,45mmol/L of NaOH you need to add to reach this pH.
Explanation:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ <em>pka = 4,74</em>
Henderson-Hasselbalch equation for acetate buffer is:
5,0 = 4,74 + log₁₀![\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCH_%7B3%7DCOO%5E-%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D)
Solving:
1,82 =
<em>(1)</em>
As total concentration of acetate buffer is:
10 mM = [CH₃COOH] + [CH₃COO⁻] <em>(2)</em>
Replacing (2) in (1)
<em>[CH₃COOH] = 3,55 mM</em>
And
<em>[CH₃COO⁻] = 6,45 mM</em>
Knowing that:
<em>CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O</em>
Having in the first 10mmol/L of CH₃COOH, you need to add <em>6,45 mmol/L of NaOH. </em>to obtain in the last 6,45mmol/L of CH₃COO⁻ and 3,55mmol/L of CH₃<em>COOH </em>.
I hope it helps!
Answer: 1 grams FeCl3 = 0.0061650760770388 mole using the molecular weight calculator and the molar mass of FeCl3.
Explanation: