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kotykmax [81]
2 years ago
12

Really need help on this thing plz someone... help number 2 and 3:(

Chemistry
1 answer:
olasank [31]2 years ago
8 0

Answer:

I don't have one sorry

Explanation:

it's because I can't see the question I don't know it I'm sorry dude

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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
Guys- my science teacher doesn't teach, he just says go on schoology ​
Free_Kalibri [48]
The relative mass of each element can be found from the periodic table (the larger number). For instance, P2O5, P = 31.0 O = 16.0, thus the formula mass (Mr) is 2(31) + 5(16) = 142 amu (atomic mass unit). I used a not really specific periodic table. Maybe your teacher is referring to open Schoology (a website where teacher can post files or announcements to students in his or her class). Sorry if i got something mistaken.
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2 years ago
How many outer planets are there?
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Answer:

four outer planetsThe Outer Planet the four outer planets and the Sun, with sizes to scale. From left to right, the outer planets are Jupiter, Saturn, Uranus, and Neptune. The gas giants are mostly made of hydrogen and helium. These are the same elements that make up most of the Sun.

Explanation:

I hope this helps you

Ari  

mwah


3 0
2 years ago
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