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crimeas [40]
3 years ago
5

What is the highest point of the wave a transverse wave called

Chemistry
2 answers:
topjm [15]3 years ago
7 0
<span>The answer to the question "what is the highest point of the transverse wave called" is a crest or peak. A transverse wave is a wave in which the medium of the wave vibrates at 90 degrees to the direction in which the wave is moving at. The lowest points are called the troughs. Examples of transverse waves are light and electromagnetic radiation.</span>
inna [77]3 years ago
3 0

Answer:

Crest

The highest point on a transverse wave is the <u>crest</u>.

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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
the latest weather report includes the following statement: the temperature is 78°f, barometric pressure is 29 and the relative
NeX [460]

Answer:

78

Explanation:

6 0
3 years ago
If aluminum has a mass of 22.3 g, how many liters of oxygen gas are required at STP?
IceJOKER [234]

Answer:

27.8

Explanation:

6 0
3 years ago
Enter ionic and net equations: feso4(aq)+ na3po4(aq) arrow fe3(po4)2(s)+na2so4(aq)
stepan [7]

Answer:

<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

<em> net ionic equation: </em>3Fe(2+)(aq)  + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

Explanation:

The balanced equation is

3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)

<em>Ionic equations: </em>Start with a balanced molecular equation.  Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions . Indicate the correct formula and charge of each ion. Indicate the correct number of each ion . Write (aq) after each ion .Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation

3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

<em>Net ionic equations: </em>Write the balanced molecular equation.  Write the balanced complete ionic equation.  Cross out the spectator ions, it means the repeated ions that are present.  Write the "leftovers" as the net ionic equation.

3Fe(2+)(aq)  + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

6 0
3 years ago
At a given temperature the vapor pressures of hexane and octane are 183 mmhg and 59.2 mmhg, respectively. Calculate the total va
Bas_tet [7]

Total vapor pressure can be calculated using partial vapor pressures and mole fraction as follows:

P=X_{A}P_{A}+X_{B}P_{B}

Here, X_{A} is mole fraction of A, X_{B} is mole fraction of B, P_{A} is partial pressure of A and P_{B} is partial pressure of B.

The mole fraction of A and B are related to each other as follows:

X_{A}+X_{B}=1

In this problem, A is hexane and B is octane, mole fraction of hexane is given 0.580 thus, mole fraction of octane can be calculated as follows:

X_{octane}=1-X_{hexane}=1-0.58=0.42

Partial pressure of hexane and octane is given 183 mmHg and 59.2 mmHg respectively.

Now, vapor pressure can be calculated as follows:

P=X_{hexane}P_{hexane}+X_{octane}P_{octane}

Putting the values,

P=(0.580)(183 mmHg)+(0.420)(59.2 mmHg)=131 mmHg

Therefore, total vapor pressure over the solution of hexane and octane is 131 mmHg.

4 0
3 years ago
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