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3 years ago

Which of the following statements is true with regard to transverse and longitudinal waves?

Chemistry

2 answers:

Ipatiy [6.2K]3 years ago

5 0

Answer:

D (The last answer)

Explanation:

In a transverse wave, particles oscillate perpendicular to the direction of wave motion.

In a longitudinal wave, the oscillations of particles are parallel to the direction of propagation.

azamat3 years ago

5 0

Answer:

Option (D)

Explanation:

Transverse waves are usually defined as those wave in which the particles within the medium vibrates perpendicular to the direction of wave propagation direction. These are comprised of crest and trough and also comprises the other properties of a wave such as amplitude, wave speed, wavelength, time period and frequency. For example, X rays, lightwave, and microwaves.

On the other hand, longitudinal waves are defined as those waves in which the particles present in the medium vibrate parallel to the wave propagation direction. For example, sound waves.

Thus, the correct answer is option (D).

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2 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

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2 years ago

Help! I need to finish work. Brainlist will be given to the most helpful!

Soloha48 [4]

Answer:

1. matches with elements.

2. matches with compounds.

3. matches with atoms

4. matches with weight

5. matches with gas

6. matches with carbon dioxide

7. matches with Mendeleev (there's an element named after him)

8. matches with IUPAC

Hope that helped :)

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