Answer:
pH = 11.7
Explanation:
pOH= -log [OH]=-log[0.05]
=2.3
pOH+pH= 14
pH= 14-2.3= 11.7
The values of the coefficients would be 4, 5, 4, and 6 respectively.
<h3>Balancing chemical equations</h3>
The equation of the reaction can be represented by the following chemical equation:
ammonia (g) + oxygen (g) ---> nitrogen monoxide (g) + water (g)
+ 
---> 
+ 
Thus, the coefficient of ammonia will be 4, that of oxygen will be 5, that of nitrogen monoxide will be 4, and that of water will be 6.
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The experimental value of Avogadro's number will be ±0.050.
<h3>What is the degree of accuracy?</h3>
The degree of accuracy refers to the extent to which a measurement is correct. Now we know that the measurement obtained by an instrument is as correct as its degree of accuracy.
Hence, given that the margin of error is about 0.050, the experimental value of Avogadro's number will be ±0.050.
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Answer:
a) [H3O+] = 1.00 E-10 M ⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M ⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M ⇒ [OH-] = 1.0 E-9 M
Explanation:
- 14 = pH + pOH
- pH = - Log [H3O+]
a) [H3O+] = 1.00 E-10 M
⇒ pH = - Log(1.00 E-10) = 10
⇒ pOH = 14 - 10 = 4
⇒ 4 = - Log[OH-]
⇒ [OH-] = 1.0 E-4 M
b) [H3O+] = 1.00 E-4 M
⇒ pH = 4
⇒ pOH = 10
⇒ [OH-] = 1.0 E-10 M
c) [H3O+] = 9.90 E-6 M
⇒ pH = 5
⇒ pOH = 9
⇒ [OH-] = 1.0 E-9 M
Answer:
At the second equivalent point 200 mL of NaOH is required.
Explanation:
at the first equivalent point:
H2A + OH- = HA- + H2O
initial mmoles y*100 y*100 - -
final mmoles 0 0 y*100 y*100
at the second equivalent point:
HA- + OH- = A2- + H2O
initial mmoles y*100 y*100 - -
final mmoles - - y*100 y*100
at the second equivalent point we have that y*100 mmoles of NaOH or 100 mL of NaOH ir required, thus:
at the second equivalent point 200 mL of NaOH is required.
