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VMariaS [17]
3 years ago
11

Help!! I need help on these! Please show all work! Will give extra points!

Chemistry
1 answer:
alukav5142 [94]3 years ago
7 0

Answer:

20 times 10 to the third is 77 and a half

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A piece of PVC plumbing pipe displaces 60 mL when placed into a container of water. If the pipe has a mass of 78g, what is the d
babymother [125]
Use the equation d=m/v

your mass or "m" is 78 g

your volume or "v" is 60mL

if you plug those values into the equation it will look like this:

d=78/60
d=1.3g/mL should be what you come up with
3 0
3 years ago
Read 2 more answers
Have an infinite number of significant figures
vovikov84 [41]
The answer is: Non-zero
7 0
3 years ago
What is the amount of space that a gas takes up called?
trasher [3.6K]
The answer is:Matter
5 0
2 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
3 years ago
What volume of water has the same mass as 9.0m3 of ethyl alcohol?
Triss [41]
This question requires the knowledge of density. 

The density of ethyl alcohol = 789 kg m⁻³
The density of water = 1000 kg m⁻³

Density = Mass / Volume 

By applying ethyl alcohol,
    789 kg m⁻³ = Mass / 0.9 m³
         Mass     = 710.1 kg
hence the mass of 0.9 m³ ethyl alcohol is 710.1 kg.

Then by applying water,
     1000 kg m⁻³ = 710.1 kg / Volume
      Volume        = 0.7101 m³
                          = 0.7 m³
hence the equal water volume is 0.7 m³
8 0
3 years ago
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