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3 years ago

Do light waves interact with all mediums the same?

Chemistry

2 answers:

jeka943 years ago

7 0

Answer:

Light can be reflected, transmitted, or absorbed. You have read that EM waves can interact with a material medium in the same ways that mechanical waves do.

Explanation:

tankabanditka [31]3 years ago

6 0

The interactions occur when waves pass from one medium to another. Besides bouncing back like an echo, waves may bend or spread out when they strike a new medium. These three ways that waves may interact with matter are called reflection, refraction, and diffraction

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Un gas a una temperatura de 38°C, tienen un volumen de 21 Lts Litros. ¿Qué volumen tendrá si la temperatura sube a 67°C?.

djverab [1.8K]

Answer:

. ¿Qué

Explanation:

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3 0

3 years ago

Determine the volume of water to be added to the nitric acid solution at a concentration of 8.61 mol / L to prepare 500 mL of th

Alex_Xolod [135]

Answer:

398 mL

Explanation:

Using the equation for molarity,

C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L

V₂ = V₁ + V' where V' = volume of water added.

So, From C₁V₁ = C₂V₂

V₁ = C₂V₂/C₁

= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L

= 0.875 mol/8.61 mol/L

= 0.102 L

So, V₂ = V₁ + V'

0.5 L = 0.102 L + V'

V' = 0.5 L - 0.102 L

= 0.398 L

= 398 mL

So, we need to add 398 mL of water to the nitric solution.

6 0

3 years ago

What is the pH of a solution in which [H 3O +] = 3.8 × 10 -8 M?

7nadin3 [17]

Answer:

<h3>The answer is 7.42 </h3>

Explanation:

The pH of a solution can be found by using the formula

$pH = - log [ { H_3O}^{+}]$

From the question we have

$pH = - log(3.8 \times {10}^{ - 8} ) \\ = 7.420216...$

We have the final answer as

Hope this helps you

4 0

3 years ago

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=3.095g$

Mass of $H_2O=1.902g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

The molarity of concentrated acetic acid is 17.45 M. When acetic acid is diluted in water it is commonly called vinegar with a m

Sladkaya [172]

M1 = 17.45 M

M2 = 0.83 M

V2 = 250 ml

M1. V1= M2. V2

V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml

7 0

2 years ago