C. chlorophyll
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<span>Pitch is sometimes defined as the fundamental frequency of a sound wave (i.e. generally, the lowest frequency in a given sound wave). For most practical purposes, this is fine, and pitch and frequency can be thought of as equivalent. On the other hand, for most practical purposes, amplitude can be thought of as volume.However, technically, pitch (and volume) are human perceptions. Thus, our perception of pitch and volume are not solely based on frequency and amplitude respectively, but are based on a combination of both (and even other factors). Frequency overwhelming dictates perceived pitch, but amplitude also does have some small, small effect on our pitch perception, especially when it is very large. For example, a very loud sound can have a different <span>perceived </span>pitch than you would predict from its frequency alone.That all being said, usually these effects are negligible, and pitch can be thought of as equivalent to fundamental frequency.
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The answer would be D because from my research it's the only one that didn't have a catalyst
Answer:
The concentration of acetic acid is 8.36 M
Explanation:
Step 1: Data given
Volume of acetic acid = 1.00 mL = 0.001 L
Volume of NaOH = 32.40 mL = 0.03240 L
Molarity of NaOH = 0.258 M
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate the concentration of the acetic acid
b*Ca*Va = a*Cb*Vb
⇒with b = the coefficient of NaOH = 1
⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED
⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L
⇒with a = the coefficient of CH3COOH = 1
⇒with Cb = the concentration of NaOH = 0.258 M
⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L
Ca * 0.001 L = 0.258 * 0.03240
Ca = 8.36 M
The concentration of acetic acid is 8.36 M
We will use Arrehenius equation
lnK = lnA -( Ea / RT)
R = gas constant = 8.314 J / mol K
T = temperature = 25 C = 298 K
A = frequency factor
ln A = ln (1.5×10 ^11) = 25.73
Ea = activation energy = 56.9 kj/mol = 56900 J / mol
lnK = 25.73 - (56900 / 8.314 X 298) = 2.76
Taking antilog
K = 15.8
