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2 years ago

Compare and contrast the fine and coarse adjustment knobs on a microscope

Chemistry

1 answer:

IceJOKER [234]2 years ago

8 0

The fine adjustment knob focuses an image. The coarse adjustment knod moves the slide up or down.

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

2 years ago

Volcanoes are most closely related to which type

sergeinik [125]

Answer:

Hey mate.....

Explanation:

This is ur answer.....

Volcanoes are closely associated with plate tectonic activity. Most volcanoes, such as those of Japan and Iceland, occur on the margins of the enormous solid rocky plates that make up Earth's surface.

Hope it helps!

Brainliest pls....

Follow me! ;)

4 0

2 years ago

Read 2 more answers

The radioactivity of U-235 is of low intensity. Why then were the people of Hiroshima exposed to high intensity radiation in the

Marina86 [1]

Answer:

Explanation:

What occurred then is as a result of nuclear fission. This occurs as the Uranium-235 split into two smaller nuclei while releasing high energy neutrons. These neutrons bombard existing U-235 in the atmosphere and this reaction continue in a spontaneous manner until a chain reaction is formed of U-235, whose fall out fills the environment. This process was what led to people been exposed to high intensity radiation in the days and months after the atomic bomb was dropped.

7 0

2 years ago

What is melted rock and minerals found beneath earths rust called

satela [25.4K]

Magma- hot fluid or semifluid material below or within the earth's crust from which lava and other igneous rock is formed by cooling. Hope this helps!

3 0

2 years ago

Select the compound that is most likely to increase the solubility of ZnSe when added to water.NaCnMgBr2NaClKClO4

vlada-n [284]

Answer:

NaCl.

Explanation:

In the solution, ZnSe ionizes to $Zn^2^+$ and $Se^2^-$ . Following reaction represents the ionization of ZnSe in solution -

$ZnSe$ ⇄ $Zn^2^+ + Se^2^-$

As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have $Na^+$ and $Cl^-$ in the solution which will be formed by the ionization of NaCl.

Now, $Zn^2^+$ in the solution will react with two $Cl^-$ ions to form $ZnCl_2$ as follows -

$Zn^2^++2Cl^-$ ⇄ $ZnCl_2$

Due to this reaction the concentration of $Zn^2^+$ will decrease in the solution and more ZnSe can be soluble in the solution.

6 0

2 years ago