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3 years ago

Which of the following leads to a higher rate of diffusion?

Chemistry

1 answer:

tiny-mole [99]3 years ago

5 0

<h3><u>Answer;</u></h3>

Higher velocity of particles

<h3><u>Explanation;</u></h3>

The diffusion rate is determined by a variety of factors which includes;

Temperature such that the higher the temperature, the more kinetic energy the particles will have, so they will move and mix more quickly and the diffusion rate will be high.
Concentration gradient such that the greater the difference in concentration, the quicker the rate of diffusion.
Higher velocity of particles increases the diffusion rate as this means more kinetic energy by the particles and hence the particles will mix and move faster, thus higher diffusion rate.

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Ham radio operators often broadcast on the 6-meter band. The frequency of this electromagnetic radiation is MHz. [speed of light

Volgvan

Answer:

50 MHz

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength$

c is the speed of light having value $3\times 10^8\ m/s$

Given, Wavelength = 6 m

Thus, Frequency is:

$Frequency=\frac{c}{Wavelength}$

$Frequency=\frac{3\times 10^8}{6}\ Hz$

$Frequency=50\times 10^{6}\ Hz$

Also,

Also, 1 Hz = $10^{-6}$ MHz

So,<u> Frequency = 50 MHz.</u>

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3 years ago

Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying

PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? ( $K_a$ for HF is $6.8\times 10^{-4}$ .)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = $K_a=6.8\times 10^{-4}$

$HF\rightleftharpoons H^++F^-$

Initially

c 0 0

At equilibrium :

(c-x) x x

The expression of disassociation constant is given as:

$K_a=\frac{[H^+][F^-]}{[HF]}$

$K_a=\frac{x\times x}{(c-x)}$

$6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}$

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

$[H^+]=x=0.01346 M$

The pH of the solution is ;

$pH=-\log[H^+]=-\log[0.01346 M]=1.87$

The pH of an 0.280 M HF solution is 1.87.

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