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3 years ago

How does wavelength differ in high and low frequency waves?

2 answers:

7 0

In high frequency waves, the wavelength is short and appears in many short bursts. Meanwhile, in a low frequency wave, the wavelength is longer since it takes it a longer period of time to occur/pass.

Taya2010 [7]3 years ago

4 0

Waves with higher frequencies have shorter wavelengths, and lower frequencies have longer wavelengths

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Write the balanced neutralization reaction that occurs between H 2 SO 4 and KOH in aqueous solution. Phases are optional. neutra

Mariulka [41]

Answer:

0.168 M

Explanation:

First, this is a reaction between the a strong base and a strong acid, therefore, we do not have to count with the acid constant of equilibrium. This reaction is taking place completely and occurs a neutralization, which is the following reaction:

H₂SO₄(aq) + 2KOH(aq) <------> K₂SO₄(s) + 2H₂O(l)

Now that we have the reaction, we can go to the second part of the question.

To calculate the remaining concentration after neutralization, we need to calculate the moles of the reactants and determine which is the limiting reactant.

The moles of the reactants:

moles A = 0.42 * 0.15 = 0.063 moles

moles B = 0.210 * 0.1 = 0.021 moles

Now that we have the moles, let's calculate the limiting reactant:

H₂SO₄(aq) + 2KOH(aq) <------> K₂SO₄(s) + 2H₂O(l)

If:

1 moles A ---------> 2 moles B

0.063 A -----------> X

X = 0.063 * 2 = 0.126 moles of B

However, we only have 0.021 moles of base, so, this is the limiting reactant.

Now that we know this, let's see the remaining moles of the acid, after the base reacts completely:

moles of A remaining = 0.063 - 0.021 = 0.042 moles

Finally to get the concentration, we have the volume of acid and the base together, so, the final volume is 0.25 L:

C = 0.042 / 0.25 = 0.168 M

This is the final concentration of the acid

3 0

3 years ago

Which tool gives scientists information about the wind speed and humidity of a region?

frez [133]

Answer:

Anemometer. An anemometer is a type of weather instrument that measures wind speed.

6 0

3 years ago

Why can’t cold air hold much water vapor?

Andru [333]

Warm air can “hold” more water vapor than cool air because as the air warms its molecules move farther apart, making room for more molecules. This leads to the idea that as air cools its molecules move closer together, “squeezing” out water vapor.

4 0

3 years ago

Suppose that 100 grams of water at 50.0°C is placed in contact with 200 grams of iron at 30.0°C. The final

wel

Answer:

The answer would be 1.5 kJ.

Explanation:

When you use the equation q = m x c x ∆T you will be able to find the energy gained or lost. The data for the water in this case is just there to distract you so ignore it. :D

4 0

3 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

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6 0

2 years ago