Answer:
D. The presence of a base catalyst.
E. The presence of an acid catalyst. and
F. The presence of heat.
Explanation:
Answer:
The correct answer to the following question will be "Particles".
Explanation:
- A particle seems to be a little component of something, it's little. When you're talking about a subatomic particle, that would be a structured user likely won't see because it's quite unbelievably thin, but it has a tiny mass as well as structural integrity. Such particles seem to be tinier than that of the particles or atoms.
- Such that the light which shines on the bit of metal could dissipate electrons, the particles seem to be more compatible with the light.
Answer:
The answer is D. gamma rays
Explanation:
A radioactive atom can have three different types of emission:
alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.
Beta particles (β) = they have 5x
amu and they have an intermediate penetrating power
Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher
For this reason Gamma emissions (γ) has the smallest mass value.
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
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