Exercise releases chemicals in our bodies called underscore which make us feel good

Gennadij [26K]

Answer:

V = wavelength * frequency = 1.5 * 5.5 = 8.25 m/s

7 0

2 years ago

A piece of bismuth with a mass of 4.06 g 4.06 g gains 423 J 423 J of heat. If the specific heat of bismuth is 0.123 J / ( g ° C

Sholpan [36]

Answer: 846°C

Explanation:

The quantity of Heat Energy (Q) required to heat bismuth depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = 423 joules

Mass of bismuth = 4.06g

C = 0.123 J/(g°C)

Φ = ?

Then, Q = MCΦ

423 J = 4.06g x 0.123 J/(g°C) x Φ

423 J = 0.5J/°C x Φ

Φ = (423J/ 0.5g°C)

Φ = 846°C

Thus, the change in temperature of the sample is 846°C

4 0

3 years ago

How would the products of electrolysis be different is a solution of zinc chloride was used instead of molten zinc chloride?

kozerog [31]

Answer:

The electrolysis of zinc chloride shows how an ionic salt will conduct electricity when molten, but not when solid.

Explanation:

hope this helps!!!! GOOD LUCK!!!!

3 0

2 years ago

Which letter does not show one wavelength?<br><br> A <br> C<br> B

Sloan [31]

Answer:

B

Explanation:

the answer is letter B

5 0

2 years ago

Read 2 more answers

Find the ratio of intensities in 4 different sets of red to violet spectral satellites in Raman scattering spectra of CCl4 molec

Vlad1618 [11]

Complete Question

Find the ratio of intensities in 4 different sets of red to violet spectral satellites in Raman scattering spectra of CCl4 molecules at T=27C temperature if corresponding resonant infrared frequencies (equivalent to frequencies of nuclei vibrations) of CCl4 molecule are 217, 315, 457 and 774 cm-1 . (Note: Wavenumber N in cm-1 is defined as $N=1/\lambda\ cm^{-1}$ )

Answer:

The ratio of intensities is $I_1 : I_2 : I_3 : I_4 = 217 : 315 : 457 : 774$

Explanation:

From the question we are told that

The number of sets of satellite is $n = 4$

The temperature is $T = 27 ^oC$

The resonant infrared frequencies are $f_1 = 217 cm^{-1}$

$f_2 = 315 cm^{-1}$

$f_3 = 457 cm^{-1}$

$f_4 = 774 cm^{-1}$

From the question we see that the wave number also has a unit of $cm^{-1}$ hence the value of the wave numbers of the molecule are

$N_1 = 217 cm^{-1}$

$N_2 = 315 cm^{-1}$

$N_3 = 457 cm^{-1}$

$N_4 = 774 cm^{-1}$

Generally intensity is mathematically represented as

$I = \frac{nhc}{A \lambda }$

Here we see that I varies inversely with wavelength i,.e

$I \ \ \alpha \ \ \frac{1}{\lambda }$

From the question we are told that the wave number is mathematically represented as

$N = \frac{1}{\lambda }$

Therefore

$I \ \ \alpha \ \ N$

This implies that the ratio of intensity in first set to that of second set to that of third set to that of fourth set is equal to the ratio of the wavenumber in the first set to that of the second set to that of third set to that of fourth

This is mathematically represented as

$I_1 : I_2 : I_3 : I_4 = N_1 : N_2 : N_3 : N_4$

Substituting values

$I_1 : I_2 : I_3 : I_4 = 217 : 315 : 457 : 774$

5 0

3 years ago