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2 years ago

In your video respond to the following question/promt. Make sure you state a clear claim and explain with

Chemistry

2 answers:

Pavlova-9 [17]2 years ago

6 0

Answer:

Vvvbn n njjnjj

Explanation:

Natasha_Volkova [10]2 years ago

5 0

Answer:

Huh

Explanation:

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MethaneProteinAlcohol can be found in beer, wine and liquor.

artcher [175]

The correct answer is alcohol. It is the common component in beer, wine and any liquor. Usually, alcohol is produced by fermentation of organic products containing glucose to produce alcohol, specifically ethanol, as the important product and the by-products water and carbon dioxide.

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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

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

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To how much water should 100 ml of 18m sulfuric acid be added to prepare a 1.5m solution

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Take 100 ml of a 18 molar solution. The total number of moles is (1 liter/1000 ml) 100 ml 18 moles is 1.8 moles.

1.5 moles in 1 liter so If 1.1 liters of water is added, the total volume is 1.2 liters and 1.8 moles are dissolves in it. 1.8 moles/ 1.2 liters is 1.5 moles per liter.

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If the last three animals of a species are all females, what will happen to the species?

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