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3 years ago

11

Explain how manipulation of light waves can cause reflection, refraction, diffusion, and absorption; and describe how different

media (solid, liquid, or gas) affect wave behavior.

1 answer:

Effectus [21]3 years ago

6 0

Answer:is A

Explanation:

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Metallic behavior is generally associated with elements with partially filled p orbitals. elements with unpaired electrons. elem

Mila [183]

Answer:

Elements with low ionization energies.

Explanation:

The ionization energy of an atom reffers to the amount of energy that is required to remove an electron from the gaseous form of that atom or ion.

The greater the ionization energy, the more difficult it is to remove an electron. The ionization energybis one of the indicator that shows the reactivity of an element. Elements with a low ionization energy such as metals are usually reffered to as a reducing agents and form cations, this give metals the tendency to

give away their valence electrons when bonding, whereas non-metals tend to take electrons.

Metallic elements have different properties such as shiny, heat and electricity conductivity . They are malleable and ductile Some metals, such as sodium, are soft and can be cut with a knife. while some are very hard such as iron.

4 0

3 years ago

A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.

Bingel [31]

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

$n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH$

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

$m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO$

And consequently the moles:

$n_O=0.12molO$

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

$C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1$

Thus, the empirical formula, taken the nearest whole number is:

$C_5H_{10}O_2$

Now, if we divide the molar mass of the molecular formula (102.1 g/mol) by that of the empirical formula (102.1 g/mol) we infer they are both the same.

Best regards!

6 0

2 years ago

Which of the following elements is the most reactive?<br> A. Ba<br> B. Cs<br> C. Hf<br> D. Lu

hodyreva [135]

Answer: Cs because its further to the right on the table, meaning its more reactive.

Explanation:

8 0

3 years ago

What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.0×1015Hz?

RideAnS [48]

The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency $1.0 \times {10^{15}}\;{\text{Hz}}$ is $\boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}$

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

${\mathbf{E=h\nu }}$ ......(1)

Here,

${\text{E}}$ is the energy.

$h$ is the Plank’s constant.

$\nu$ is the frequency.

The frequency of UV rays is $1.0 \times {10^{15}}\;{\text{Hz}}$ or $1.0 \times {10^{15}}\;{{\text{s}}^{ - 1}}$

The value of Planck’s constant is $6.626 \times {10^{ - 34}}\;{\text{J}}\cdot{\text{s}}$ .

Substitute these values in equation (1)

$\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}$

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is ${\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}}$ .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

7 0

3 years ago

Read 2 more answers

Is Sr3(PO4)2 ionic or covalent

lana66690 [7]

Sr3(PO4)2 is definitely Ionic

6 0

3 years ago