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aliya0001 [1]
3 years ago
9

What is the period of a wave with a frequancy of 3 Hz and an amplitude off 0.01

Chemistry
1 answer:
lara31 [8.8K]3 years ago
4 0

0.33 seconds is the period of a wave with a frequency of 3 Hz and an amplitude off 0.01 .

<u>Explanation:</u>

We have , period of a wave with a frequency of 3 Hz and an amplitude off 0.01 . We know that period of a wave is amount of time needed to complete one oscillation . In order to calculate period of wave we use frequency and the formula use is period = \frac{1}{frequency} . We are given that frequency = 3 Hz:

period = \frac{1}{frequency}

⇒ period = \frac{1}{frequency}

⇒ period = \frac{1}{3 Hz}

⇒ period = 0.33 seconds

Therefore, 0.33 seconds is the period of a wave with a frequency of 3 Hz and an amplitude off 0.01 .

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Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

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