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3 years ago

The units on the ruler are in nanometers (nm). look at the graduations on the ruler. what does each graduation represent?

2 answers:

8 0

The units on the ruler are inches, centimeters, and millimeters. Nanometers (nm) is equivalent to 10^-9 m. Each graduation on the ruler represents 0.1 millimeters or 0.01 centimeters. The other side also has a graduation of 0.1 inch. The function of the ruler depends on how you use it.

kati45 [8]3 years ago

7 0

The answer is 0.2 nanometers

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What is the specific latent heat of fusion of ice if it takes 863 kJ to convert 4.6 kg of ice into water at 0 C?

allochka39001 [22]

Answer:

$1^{f} =187.7 \frac{J}{kg}$

Explanation:

SO in order to calculate the specific latent heat of fusion, you need to remember the formula:

$1^{f} =\frac{Q}{m}$

Where $1^{f}$ representes the specific latent heart of fusion.

$Q$ represents the heat energy added, usually represented in kJ

$m$ represents the mass of the object, in kg.

Now that we have our formula we just have to put our values into the formula:

$1^{f} =\frac{Q}{m}$

$1^{f} =\frac{863kJ}{4.6kg}$

$1^{f} =187.7 \frac{J}{kg}$

SO our answer would be $1^{f} =187.7 \frac{J}{kg}$

7 0

3 years ago

What happens to air pressure as altitude increases

gulaghasi [49]

Answer:

the air pressure decreases

explanation

as altitude increases air pressure decreases the surface of the Earth is bottom of the ocean of the are the layer on top increase pressure so that sea levels a body is under 14.7 pounds of pressure per square inch moving up in altitude decreases weight of the air that causes the pressure

4 0

3 years ago

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

3 years ago

"Given a block of frozen salt water, how could you separate

cricket20 [7]

Answer: You could put the frozen block of ice on the stove and let it melt and eventually boil out leaving the salt behind

Explanation:

4 0

2 years ago

What would happen if the sand dunes in an area were destroyed?

Komok [63]

Answer:

Our beaches would be unprotected

In the short-term, these artificial sand hills will be destroyed by the elements. Because sand dunes protect inland areas from swells, tides, and winds, they must be protected and defended like national treasures. ... The ocean and the wind can have an unpredictable, destructive force on coastal regions.

- surfertoday

Natural sand dunes play a vital role in protecting our beaches, coastline and coastal developments from coastal hazards such as erosion, coastal flooding and storm damage. Sand dunes protect our shorelines from coastal erosion and provide shelter from the wind and sea spray.

- Waikato Regional Council

3 0

2 years ago

Read 2 more answers