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3 years ago

What ismole used for

Chemistry

1 answer:

nordsb [41]3 years ago

8 0

Answer:

What is mole used for?

Explanation:

The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions.

SI base units: Base unit (Dimensionless)

Symbol: mol

Unit of: Amount of substance

Unit system: SI base unit

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Discuss the significance of assigning an atomic mass of exactly 12 amu to the carbon-12 isotope.

evablogger [386]

Answer:

Cabon-12 has same average atomic weight and mass number.

Explanation:

carbon-12 has average atomic weight 12 amu and mass number of 12.

amu represents average of mass of a nucleon.

As carbon-12 has same average atomic mass and mass number therefore carbon-12 is a good standard to determine average mass of a nucleon.

Again, abundance of carbon-12 isotope is almost equal to 99%. Therefore fluctuation of average atomic weight from 12 amu is very very low.

So, carbon-12 is taken as a standard to determine mass of a nucleon.

Hence atomic mass of carbon-12 is 12 amu.

5 0

3 years ago

Balance Reaction K(s) + H2O(l) —> KOH(aq) + H2

Wewaii [24]

Answer:

Chemistry is a branch of science that involves the study of the composition, structure and properties of matter. Often known as the central science, it is a creative discipline chiefly concerned with atomic and molecular structure and its change, for instance through chemical reactions.Explanation:Chemistry is a branch of science that involves the study of the composition, structure and properties of matter. Often known as the central science, it is a creative discipline chiefly concerned with atomic and molecular structure and its change, for instance through chemical reactions.

4 0

2 years ago

If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit

Blizzard [7]

Answer:

For a: The volume of the box is 217.5 mL

For b: The volume of the box is 0.2175 L

Explanation:

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

$V=10.00\times 7.25\times 3.00=217.5cm^3$

For a:

To convert the volume of cuboid into milliliters, we use the conversion factor:

$1mL=1cm^3$

So,

$\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL$

Hence, the volume of the box is 217.5 mL

For b:

To convert the volume of cuboid into liters, we use the conversion factor:

$1L=1000cm^3$

So,

$\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L$

Hence, the volume of the box is 0.2175 L

3 0

3 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

2 years ago

If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature of 1100˚C for 8 hours, wh

RideAnS [48]

Answer:

8.354 nanometers

Explanation:

To treat a diffusive process in function of time and distance we need to solve 2nd Ficks Law. This a partial differential equation, with certain condition the solution looks like this:

$\frac{C_{s}-C{x}}{C_{s}-C_{o}}=erf(x/2\sqrt{D*t})$

Where Cs is the concentration in the surface of the solid

Cx is the concentration at certain deep X

Co is the initial concentration of solute in the solid

and erf is the error function

Then we solve right side,

$\frac{C_{s}-C{x}}{C_{s}-C_{o}}=\frac{1018atoms/cm3-1016atoms/cm3}{1018atoms/cm3}=0.001964$

And we need to look up the inverse error function of 0.001964 resulting in: 0.00174055

Then we solve for x:

$x=0.00174055*2*\sqrt{D*t} =0.00174055*2*\sqrt{2*10^{-12}cm^{2}/s*8h*3600s/h}=8.35464*10^{-7}cm$

6 0

3 years ago

What ismole used for​

What ismole used for