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3 years ago

2. In a concentrated solution there is _____. (1 point)

Chemistry

2 answers:

spin [16.1K]3 years ago

6 0

Answer : The correct option is, a large amount of solute.

Explanation :

As we know that the saturated solution is made when the maximum amount of solute dissolved in the solvent.

In unsaturated solution, the undissolved substance is settle down at the bottom that means no more solute is dissolved in the solvent.

The concentrated solution means that a large amount of solute present in the solution.

Hence, in a concentrated solution there is a large amount of solute.

hram777 [196]3 years ago

4 0

In concentrated solution , there is a large amount of solute !!

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decomposition of hydrogen peroxide produce water and oxygen . calculate the volume of O2 formed from the decomposition of 150 mL

Ugo [173]

Explanation:

2H2O2 => 2H2O + O2

Moles of hydrogen peroxide = 0.150dm³ * (0.02mol/dm³) = 0.003mol .

Moles of oxygen = 0.0015mol.

Volume of oxygen = 0.0015mol * (22.4dm³/mol) = 0.0336dm³.

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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

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

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