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3 years ago

A 1.34 mole sample of LiCl dissolves in water, The volume of the final

Chemistry

1 answer:

Mamont248 [21]3 years ago

0 0

Answer:

$1.56 \,\,mol/L$

Explanation:

Molarity of the solution measures number of moles of a solute per litre of solution.

Molarity = volume of solution in litres/number of moles of solute dissolved in solution

Volume of solution in litres = 0.86 L

Also, 1.34 mole sample of LiCl dissolves in water

So,

Molarity of the Solution = $\frac{1.34}{0.86}=1.56 \,\,mol/L$

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A solution made by dissolving 9.81 g of a nonvolatile, nonelectrolyte in 90.0 g of water boiled at 100.37 °C at 760 mm Hg. What

NISA [10]

Answer:

151 g/mol

Explanation:

In order to solve this problem we need to keep in mind the formula for the boiling point elevation:

ΔT = Kb * m * i

Where:

ΔT is the temperature difference between the boiling point of the solution and that of pure water. 100.37 °C - 100.00 °C = 0.37 °C.

m is the molarity of the solution

i is the van't Hoff factor. As the solute is a nonelectrolyte, the factor is 1.

Input the data and calculate m:

0.37 °C = 0.51 °C/m * m * 1

m =0.72 m

We now can calculate the number of moles of the substance, using the definition of molarity:

molarity = moles of solute / kg of solvent

In this case kg of solvent = 90.0 g / 1000 = 0.090 kg

0.72 m = moles / 0.090 kg

moles = 0.065 mol

Finally we calculate the molar mass, using the number of moles and the mass:

9.81 g / 0.065 mol = 151 g/mol

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What conversion factor would you use to convert grams to kilograms?

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Explanation:

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The reaction we will be doing in lab is to take sand(SiO2),baking sod(NaHCO3) and reacting that with HCl. The HCl with the react

sergij07 [2.7K]

I believe that this equation is
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3 years ago

The diffusion coefficient for sodium ions crossing a biological membrane 12nm thick is 1.5 x10-18 m2/s. what flow rate of sodium

Vladimir [108]

Answer:

$-1.08\times 10^{23} mol/s$

Explanation:

We are given that

Diffusion coefficient, $D=1.5\times 10^{-18} m^2/s$

Thickness of membrane, $dx=12nm=12\times 10^{-9} m$

$1 nm=10^{-9} m$

Area, $A=12\times 12=144nm^2=144\times 10^{-18} m^2$

Concentration differences, $dc=0.60 mol/dm^3=0.60\times 1000=600mol/m^3$

We have to find the flow rate of sodium ions.

Flow rate, $\frac{dn}{dt}=-DA\frac{dc}{dx}$

Using the formula

$\frac{dn}{dt}=-\frac{1.5\times 10^{-18}\times (144\times 10^{-18})\times 600}{12\times 10^{-9}}=-1.08\times 10^{23} mol/s$

$\frac{dn}{dt}=-1.08\times 10^{23} mol/s$

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2 years ago