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2 years ago

The period of an osciller is 2.34 minutes. What id the frequency of this oscillator in hertz?

Chemistry

1 answer:

AleksAgata [21]2 years ago

5 0

Explanation:

Given parameters

Period = 2.34minutes

Unknown:

Frequency of the oscillator = ?

Solution:

Frequency is the number of waves that passes through a point at a given time. It is the inverse of period which is the time it takes for a wave to pass through a medium:

F = $\frac{1}{T}$

F is the frequency

T is the period

To solve this, convert the period to seconds:

2.34minutes to seconds:

1 min = 60seconds

2.34 = 60 x 2.34 = 140.4s

F = $\frac{1}{140.4}$ = 0.007hz or 0.007s⁻¹

Learn more:

frequency brainly.com/question/5882803

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2 years ago

Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.

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(a) $pH = -Log (0.1M) = 1$

(b) $pH = -Log (10^{-13}M) = 13$

(d) $pH = -Log (4.93x10^{-10}M) = 9.3$

(e) $pH = -Log (5^{-7}M) = 6.3$

Explanation:

To calculate de pH of an acid solution the formula is:

$pH = -Log ([H^{+}]) = 1$

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

$K_{w} =[H^{+} ][OH^{-}]=10^{-14}$