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2 years ago

10

A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate for

m when 10. ml of 0.0010 M Ca(NO3)2 is mixed with 10. ml of 0.00010 M NaF? Ksp for CaF2 = 3.2 x 10-11

1 answer:

laiz [17]2 years ago

5 0

Answer:

CaF2 will not precipitate

Explanation:

Given

Volume of Ca(NO3)2 $= 10$ ml

Molar concentration of Ca(NO3)2 $= 0.001$

Volume of NaF $= 10$ ml

Molar concentration of NaF $= 0.0001$

Ksp for CaF2 $= 3.2 * 10^ {-11}$

CaF2 will precipitate if Q for the reaction is greater than ksp of CAF2

Moles of calcium ion

$= 10 * 0.001\\= 0.01$

$[Ca2+] = \frac{0.01}{10 + 10} \\= \frac{0.01}{20} \\= 5 * 10^{-4}$

Moles of F- ion

$= 10 * 0.0001\\= 0.001$

$[F-] = \frac{0.001}{10 + 10} \\= \frac{0.001}{20} \\= 5 * 10^{-5}$

$Q = [Ca2+] [F-]^2\\= (5 * 10^{-4}) * (0.5* 10^-4)\\= 1.25 * 10^{-12}$

Q is lesser than Ksp value of CaF2. Hence it will not precipitate

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2 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

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In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

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2 years ago

An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is

Juli2301 [7.4K]

Answer:

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

Explanation:

The gas ideal law is

PV= nRT (equation 1)

Where:

P = pressure

R = gas constant

T = temperature

n= moles of substance

V = volume

Working with equation 1 we can get

$n =\frac{PV}{RT}$

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

$\frac{m}{mw} =\frac{PV}{RT}$ or

$m =\frac{P*V*mw}{R*T}$ (equation 2)

The cylindrical container has a constant pressure p

The volume is the volume of a cylinder this is

$V =(pi)*r^{2}*h$

Where:

r = radius

h = height

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is $V =(pi)*r^{2}*h$

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:

$T =\frac{T_{1} + T_{O}}{2}$

Replacing these values in the equation 2 we get:

$m =\frac{P*V*mw}{R*T}$ (equation 2)

$m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}$

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3 years ago

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The molarity of Sr(OH)2 solution is = 0.1159 M

calculation
write the equation for reaction
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then finds the mole of HCl used

moles = molarity x volume
=40.03 x0.1159 = 4.639 moles

by use of mole ratio between Sr(OH)2 to HCL which is 1 :2 the moles of Sr(OH)2 is therefore = 4.639 x1/2 = 2.312 moles

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Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

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$c=5760J/^oC$

$\Delta T=0.570^oC$

To calculate the enthalpy change, we use the formula:

$\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J$

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = $\frac{3283.2J}{0.1326g}=24760.181J/g$

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

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4 0

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