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2 years ago

What is the molarity of 2 mol of calcium carbonate, CaCl2, dissolved in 1 L of solution?

Physics

1 answer:

Agata [3.3K]2 years ago

6 0

Answer:

2mol/L

Explanation:

Given parameters:

Number of moles of calcium carbonate = 2mol

Volume of solution = 1L

Unknown:

Molarity = ?

Solution:

The molarity of a solution is the number of moles of solute dissolved or contained in a solution:

Molarity = $\frac{number of moles }{volume }$

Molarity = $\frac{2}{1}$ = 2mol/L

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What symbol represents energy that moves from a hot object to a cooler object?

strojnjashka [21]

The correct answer is
<span>C) Q

In fact, the symbol Q represents the heat, which is the form of energy transferred from a hot object to a cooler object. Heat generally refers to the energy related to the motion of the particles, and it is related to the temperature of an object: the higher the temperature of an object, the faster the particles of the object move, and so the object can transfer more energy (as heat) to other objects with lower temperature.</span>

6 0

3 years ago

An observer is standing next to the tracks, watching a train approach. The train travels at 30 m/s and blows its whistle at 8,00

SSSSS [86.1K]

7351.35Hz

f0= v-Vo/v-Vs × FSA

= 340-0 /340+30 ×8000

= 340/370× 8000

= 7351.35hz

7 0

3 years ago

Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op

Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

$f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz$

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0

3 years ago

If the wavelength of a sound wave increases and the frequency of the sound wave does not change what happens to the speed of the

salantis [7]

If the wavelength of a sound wave increases and the frequency of the sound wave does not change, the speed of the wave will increase.

Ans: D

Explanation

The sound wave speed is given by E=fλ, where f indicates its frequency and λ indicates its wavelength.

From the equation, it is evident that the sound speed is proportional to both frequency and wavelength.

Here, as wavelength increases, wave speed increases provided there is no change in frequency.

3 0

3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

2 years ago