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3 years ago

15

The diagram below shows a longitudinal wave at one instant in its motion. Each oval represents a particle acted on by the ener t

he wave as it moves from left to right. Which line on the diagram indicates a region of rarefaction?

2 answers:

Anton [14]3 years ago

8 0

Answer:

the first line

Explanation:

A rarefaction is a region in a longitudinal wave where the particles are furthest apart. ... The region where the medium is compressed is known as a compression and the region where the medium is spread out is known as a rarefaction.

use this picture as a reference:

GaryK [48]3 years ago

8 0

Answer:

yea the first line

Explanation:

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The correct electron configuration for Mn is

dmitriy555 [2]

1s22s22p63s23p64s23d5 is the correct answer.

4 0

4 years ago

Elvira Walks 4 miles to the west from school and stops at the store. She then walks 3 miles south.

dalvyx [7]

Answer:

distance = 7 miles

displacement = 5 miles

Explanation:

Distance is a scalar quantity as it takes account of magnitude traveled but not the direction traveled from starting point.

The distance traveled is the sum total of distances moved

distance = 4 + 3 = 7 miles

Displacement however, is a vector and measure the shortest possible distance traveled in a given direction from the starting point.

The path of Elvis' walking forms a right-angle triangle with the hypotenuse being the displacement and the other two sides being the distance traveled west and south.

Using Pythagoras' theorem; c² = a² + b²

where c = hypotenuse and a and b are the other two sides

c² = 4² + 3²

c² = 16 + 9 = 25

√c² = √25

c = 5

Therefore, displacement = 5 miles

4 0

3 years ago

Would you expect an organic compound to be soluble in water why

RSB [31]

Answer: All organic compound depends on H-bonding with water. more stronger H-bonding with water more will be soluble.

Explanation:

1. It depends primarily upon the function groups of that compound. It also depends on the size of the compound.

2. some organic compound which soluble in water for example: alcohols, ethers, carboxylic acids. Because of the functional groups attached to the organic structure (the C-H backbone) are what effect the solubilities.Like carboxylic acids and alcohols form hydrogen bonds with the water, helping to solubilize it.

3. Take alcohols for example: methanol, ethanol, and isopropanol are all completely soluble in water. By the time you get to butanol and some of the larger alcohols, including those with more complex structures, they tend to be less soluble.

3 0

3 years ago

The equilibrium constant for the reaction 2x(g)+y(g)=2z(g) is 2.25 . what would be the concentration of y at equilibrium with 2

Troyanec [42]

$[\text{Y}] \approx0.337\;\text{mol}\cdot\text{dm}^{-3}$ at equilibrium.

<h3>Explanation</h3>

Concentration for each of the species:

$[\text{X}] = \dfrac{n}{V} = 2\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Y}] = \dfrac{n}{V} = 0\; \text{mol}\cdot \text{dm}^{-3}$ ;
$[\text{Z}] = \dfrac{n}{V} = 3\; \text{mol}\cdot \text{dm}^{-3}$ .

There was no Y to start with; its concentration could only have increased. Let the change in $[\text{Y}]$ be $+x \; \text{mol}\cdot \text{dm}^{-3}$ .

Make a $\textbf{RICE}$ table.

Two moles of X will be produced and two moles of Z consumed for every one mole of Y produced. As a result, the <em>change</em> in $[\text{X}]$ will be $+2\;x \; \text{mol}\cdot \text{dm}^{-3}$ and the <em>change</em> in $[\text{Z}]$ will be $-2\;x \; \text{mol}\cdot \text{dm}^{-3}$ .

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& & &&&\end{array}$ .

Add the value in the C row to the I row:

$\begin{array}{l|ccccc}\textbf{R}\text{eaction}&2\; \text{X}\; (g) & + &\text{Y}\; (g) & \rightleftharpoons &2 \; \text{Z}\; (g)\\\textbf{I}\text{nitial Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 & &0 & & 3 \\\textbf{C}\text{hange in Concentration}\; (\text{mol}\cdot\text{dm}^{-3})\;& +2\;x & &+x &&-2\;x\\\textbf{E}\text{quilibrium Condition}\; (\text{mol}\cdot\text{dm}^{-3})& 2 + 2\;x & &x&&3-2\;x\end{array}$ .

What's the equation of $K_c$ for this reaction? Raise the concentration of each species to its coefficient. Products go to the numerator and reactants are on the denominator.

$K_c = \dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]}$ .

$K_c = 2.25$ . As a result,

$\dfrac{[\text{Z}]^{2}}{[\text{X}]^{2} \cdot[\text{Y}]} = \dfrac{(3-2x)^{2}}{(2+2x)^{2} \cdot x} = K_c = 2.25$ .

$(3-2\;x)^{2}= 2.25 \cdot(2+2\;x)^{2} \cdot x\\4\;x^{2} - 12 \;x + 9 = 2.25 \;(4\;x^{3} + 8 \;x^{2} + 4 \;x)\\4\;x^{2} - 12\;x + 9 = 9 \;x^{3} + 18\;x^{2} + 9\;x\\9\;x^{3} + 14\;x^{2} + 21\;x - 9 = 0$ .

The degree of this polynomial is three. Plot the equation $y = 9\;x^{3} + 14\;x^{2} + 21\;x - 9$ on a graph and look for any zeros. There's only one zero at $x \approx 0.337$ . All three concentrations end up greater than zero.

Hence the equilibrium concentration of Y: $0.337\;\text{mol}\cdot\text{dm}^{-3}$ .

7 0

3 years ago

Please guysssss

mina [271]

If you would draw the Lewis structures of these atoms, you would see that A has 2 electron pairs and 2 lone electrons (that can bond). For B you’d see that you only have 1 electron that can form a bond. This means that 1 atom of A (2 lone electrons) can bond with 2 atoms of B. To know the kind of bond you have to know wether or not there will be a ‘donation’ of an electron from one atom to another. This happens when the number of electrons on one atoms is equal to the number of electrons another atom needs to reach the noble gas structure. As you can see, this is not the case here. This means that you get an AB2 structure with covalent character.

6 0

3 years ago