<span>During neutralization reaction
Milliequivalent of acid=milliequivalent of base.
Milliequivalent of HCl=0.100*12*1
Milliequivalent of NaOH=M*10*1
So 0.100*12*1=M*10*1
Solve for M.</span>
Answer:
Q = 270 Joules (2 sig. figs. as based on temperature change.)
Explanation:
Heat Transfer Equation of pure condensed phase substance => Q = mcΔT
Mixed phase (s ⇄ l melting/freezing, or l ⇄ g boiling/condensation) heat transfer equation => Q = m∙ΔHₓ; ΔHₓ = phase transition constant
Since this is a pure condensed phase (or, single phase) form of lead (Pb°(s)) and not melting/freezing or boiling/condensation, one should use
Q = m·c·ΔT
m = mass of lead = 35.0g
c = specific heat of lead = 0.16J/g°C
ΔT = Temp change = 74°C - 25°C = 49°C
Q = (35.0g)(0.16J/g·°C )(49°C) = 274.4 Joules ≅ 270 Joules (2 sig. figs. as based on temperature change.)
D. 0.2 M
The concentration of a solution is basically the ratio of the solute present to the solvent in the solution. This is an intrinsic property, independent of the amount of solution that is present. A similar example is that of density. No matter the size of a sample, the density and concentration of that sample remain constant.
Answer:
Explanation:
Electrons are transferred from atoms of sodium to atoms of phosophorus. This transfer makes the sodium atoms positive and the phosphorus atoms negative. As a result, the sodium and phosphorus atoms strongly attract each other.
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4
And
![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
