Answer is: <span>the percent ionization is 0,19%.
</span>Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.
Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
Answer:
does your son ever talk to strangers?
Explanation:
Answer: C = 0.014M
Explanation:
From n= m/M= CV
m =43.5 M= 148, V=850ml
43.5/148= C× 0.85
C= 0.35M
Applying dilution formula
C1V1=C2V2
C1= 0.35, V1= 25ml, C2=?, V2= 600ml
0.35× 25 = C2× 600
C2= 0.014M
Answer:
Option B:Publishing scientific journals
Explanation:
We are told that Lindsey is trying to gain credibility for her studies.
Since she completed her experiment and discussed her finding with colleagues, the most logical next step would be to publish scientific journals. This is because the other options given are not steps that should be taken because she has completed the research and therefore has no need to speak at a conference next nor even create new charts which they must have done during the research. No need for her to make sure the topic is popular.
Option B is correct
