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2 years ago

A student placed a crystal of a blue dye at the bottom of a beaker of water. After 5 minutes, the crystal sa

Chemistry

1 answer:

ludmilkaskok [199]2 years ago

7 0

Explanation:

It is because it is having color in it and the water is colorless

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Heisenberg’s uncertainty principle

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7.01*10^23 sulfur atoms is equal to______mol<br> Help please

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The answer is 1.164451827 moles

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2 years ago

Identify the type of igneous rock that is formed when magma cools above Earth's surface?

mylen [45]

Answer: cools down and keeps going

Explanation:

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2 years ago

Lithium arsenate (Li3AsO4) and iron nitrate (Fe(NO3)3) are dissolved in water. What precipitate would you expect to form? what i

Alla [95]

Answer:

CO₃²⁻ + H₂O <----------------> HCO₃⁻ + OH⁻

The chemical formula of the precipitate is Fe(OH)₃

Explanation:

Fe(NO₃)₃ and K₂CO₃ are strong electrolytes and completely dissociate in water. Carbonate ions is a weak base and combine with water to form hydroxide ions (OH⁻), CO₃²⁻ + H₂O <----------------> HCO₃⁻ + OH⁻
Ferric, Fe (III), combines with these hydroxide ions to form insoluble precipitates. Fe(OH)₃ is only partially soluble i.e., it does not completely dissociate in water. When the solutions of Fe(NO₃)₃ and K₂CO₃ are mixed, Fe(OH)₃ precipitates out due to the strong electrostatic attraction between Fe (III) and hydroxide ions.

8 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

2 years ago