Exam number: 007170RR

Which equation shows how wavelength is related to velocity and frequency?

Nataly_w [17]

Answer:

The correct option is the option;

$\lambda = \dfrac{v}{f}$

Explanation:

The wavelength of a wave is the distance between two successive crests of the wave

Therefore, the wavelength, λ, is given by the fraction of the velocity, <em>v</em>, of the wave divided by the frequency, <em>f</em>, (the number of cycles that pass through a point) of the wave

Mathematically, we have;

$The \ wavelength, \ \lambda = \dfrac{The \ wave \ velocity, \ v}{The \ wave \ frequency, \ f}$

$\lambda = \dfrac{v}{f}$

4 0

3 years ago

Which would have the lowest input force?

Afina-wow [57]

Answer:

A long lever with the fulcrum as close as possible to the load

Explanation:

If F be the effort , W be the weight , L₁ be the distance of load from fulcrum and L₂ be the distance of effort from the fulcrum ,

Taking moment of force about the fulcrum , we have

W x L₁ = F x L₂

F = W x ( L₁ / L₂ )

F will be minimum when L₁ will be minimum .

Hence fulcrum should be as close as possible to the load.

5 0

3 years ago

An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx

Anna [14]

Answer:

$M=235.42g/mol$

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

$2n_{acid}=n_{base}$

It means that the moles of acid can be computed given the volume and concentration of NaOH:

$n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol$

It means that the approximate molar mass of the acid is:

$M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol$

Best regards!

5 0

3 years ago

Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin

Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0

3 years ago

What is the maximum amount in moles of P2O5 that can theoretically be made from 235 g of P4 and excess oxygen?

Lady_Fox [76]

The correct answer would be 3.8mol

5 0

3 years ago