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3 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]3 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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A geologist is comparing the properties of different minerals. He rubs each one on a piece of white tile and observes the color

Lynna [10]

Scratch Test is your answer

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3 years ago

Read 2 more answers

What is the ph of a buffer consisting of 0.200 m hc2h3o2 and 0.200 m kc2h3o2? the k a for hc2h3o2 is 1.8×10−5. view available hi

AURORKA [14]

PH = pKa + log $\frac{[base]}{[Acid]}$
Acid is HC₂H₃O₂ and conjugate base is KC₂H₃O₂,
pKa = - log Ka = - log (1.8 x 10⁻⁵) = 4.74
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3 years ago

A scientist measures the standard enthalpy change for the following reaction to be 81.1 kJ :2CO2(g) + 5 H2(g)C2H2(g) + 4 H2O(g)B

posledela

<u>Answer:</u> The enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.Z

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

For the given chemical reaction:

$2CO_2(g)+5H_2(g)\rightarrow C_2H_2(g)+4H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]$

We are given:

$\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=226.7kJ/mol\\\Delta H^o_{rxn}=81.1kJ$

Putting values in above equation, we get:

$81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol$

Hence, the enthalpy of the formation of $CO_2(g)$ is coming out to be -410.8 kJ/mol.

8 0

3 years ago

In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5

Rufina [12.5K]

Answer:

The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.

Explanation:

The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:

MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol

If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:

0.070 mol x 152 g/mol = 10.64 g

Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.

7 0

3 years ago

Air enters a diffuser witha velocity of 400 m/s, a pressure of 1 bar and temperature of 25 C. It exits with a temperature of 100

ANEK [815]

Answer:

Exit velocity of air is 96.43 m/s.

Explanation:

Given that

$V_1=400\ m/s$