Explanation:
b. Bromo cyclopentane + HBr
Which led to the formation of oceans after water on Earth's surface evaporated?
Answer:
condensation
Explanation:
What led to the formation of oceans after the water on the Earth's surface evaporated is a condensation reaction of water vapour.
The water present on the surface of the earth was able to evaporate as a result of the hot condition of the primitive earth. As the earth cools down the water vapour present in the atmosphere began to condense, gradually forming puddles of water and eventually leading to the formation of various oceans as we currently have it on the earth.
Answer:
as earth cooled, water in the atmosphere condensed.
Explanation:
Two elements represents by the letter Q and R atomic number 9 and 12 respectively.
1. Write the electronic configuration of R
2. To what group does Q belongs to in the periodic table
3. Write the formula of the compound formed when Q combines with R
Answer:
The two elements with atomic number 9 and 12 are represented by letter Q and R respectively, where Q represents fluorine atom and R represents magnesium atom.
1. Electronic configuration of R that is magnesium (atomic number 12) is:
1s2 2s2 2p6 3s2
2. Q represents fluorine atom, which belongs to group 17 in periodic table that is the most reactive and lightest member of the group.
3. Q and R that is fluorine and magnesium combinely form magnesium fluoride or MgF2.
What is the mass percentage of C in CH₃CH₂OH? Provide an answer to two decimal places.
Answer:
here's your answer
Explanation:
Molar mass of CH3CH2OH = 46.06844 g/mol
This compound is also known as Ethanol.
Convert grams CH3CH2OH to moles or moles CH3CH2OH to grams
Molecular weight calculation:
12.0107 + 1.00794*3 + 12.0107 + 1.00794*2 + 15.9994 + 1.00794
Percent composition by element
Hydrogen H 1.00794 6 13.128%
Carbon C 12.0107 2 52.143%
Oxygen O 15.9994 1 34.730%
The mass percentage of C in CH₃CH₂OH is 52.14% (to two decimal places)
To calculate the mass percentage of C (Carbon) in CH₃CH₂OH (Ethanol),
First, we will determine the mass of CH₃CH₂OH
Molar mass of CH₃CH₂OH = 46.07 g/mol
Mass of C = 12.01 g/mol
Now, for the mass percentage of C in CH₃CH₂OH,
We will determine the ratio of the total mass of C to the mass of CH₃CH₂OH, and then multiply by 100%
Since we have 2C in CH₃CH₂OH
Then, total mass of C in CH₃CH₂OH = 2 × 12.01 g/mol = 24.02 g/mol
That is,
Mass percentage of C in CH₃CH₂OH = [tex]\frac{24.02}{46.07} \times 100\%[/tex]
Mass percentage of C in CH₃CH₂OH = 0.5213805 ×100%
Mass percentage of C in CH₃CH₂OH = 52.13805%
Mass percentage of C in CH₃CH₂OH ≅ 52.14%
Hence, the mass percentage of C in CH₃CH₂OH is 52.14% (to two decimal places)
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Lead ions are toxic when absorbed into the body and can interfere with the neurological development of children. Based on what you learned in this lab activity, what substance might be added to an IV solution to remove Pb2 ions from the blood of a contaminated person
Answer:
The interpretation of the particular subject is covered in the subsection below in detail.
Explanation:
Large quantities of heavy substances like Lead ions become extremely poisonous when provided by a human. The administration of the medications recognized as "chelators" will eliminate these harmful chemicals from an infected individual's blood.However, here law enforcers calcium sodium polyacrylate seems to be the safest chelator in radiation sickness. It could be administered intravenously and attaches throughout the blood system with either the lead ions and afterward, removes the metal-chelator complicated from urine.For the following set of pressure/volume data, calculate the new volume of the gas sample after the pressure change is made. Assume that the temperature and the amount of gas remain constant.
a. 125 mL at 755 mm Hg; V =2mL at 780 mm Hg
b. 223 mL at 1.08 atm; V =2mL at 0.951 atm
c. 3.02 L at 103 kPa; V= 2Lat 121 kPa
Answer:
a. 121 ml, b. 253 ml and c. 2.57 L.
Explanation:
The new volume can be calculated by using the Boyle's law equation:
P1V1 = P2V2
In the equation, P1 and P2 are the initial and final pressures and V1 and V2 are the initial and final volumes for a real gas at constant temperature.
a) Based on the given information, P1 = 755 mmHg, V1 = 125 ml, P2 = 780 mm Hg and V2 will be,
V2 = P1V1/P2
V2 = 755 mmHg × 125 ml/780 mmHg
V2 = 121 ml
b) Based on the given information, P1 = 1.08 atm, V1 = 223 ml, P2 = 0.951 atm and V2 will be,
V2 = P1V1/P2
V2 = 1.08 atm × 223 ml/0.951 atm
V2 = 253 ml
c) Based on the given information, P1 = 103 kPa, V1 = 3.02 L, P2 = 121 kPa and V2 will be,
V2 = P1V1/P2
V2 = 103 kPa × 3.02 L/121kPa
V2 = 2.57 L
The Lewis structure of N2H2 shows ________. Group of answer choices a nitrogen-nitrogen single bond each hydrogen has one nonbonding electron pair each nitrogen has one nonbonding electron pair each nitrogen has two nonbonding electron pairs a nitrogen-nitrogen triple bond
Answer:
one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.
Explanation:
H-N=N-H
Stibnite, Sb2S3, is a dark gray mineral from which antimony metal is obtained. What is the mass percent of antimony in the sulfide? If you have 1.00 kg of an ore that contains 10.6% antimony, what mass of Sb2S3 (in grams) is in the ore?
Answer:
duw8 Wert gsi
Explanation:
pues pues y 8y1rcuecisscfjfj3eoeu xv cihskdkkd HD jekifeuifkeñ elijo eh fh FC eh SSH DJ djdvheshdhs
The mass percent of antimony in the sulfide is 71.7% and the mass of Sb2S3 is 0.148 g.
What is mass percent?Mass percent is defined as a way of expressing a concentration or describing a component in a particular mixture.
To calculate the mass percent of an element in a compound, divide the mass of the element in 1 mole of the compound by the molar mass of the compound and multiply the result by 100.
Mass percent = Mass of chemical / Total mass of compound x 100
Mass of ore = 10 kg
% antimony in ore = 10.6%
= 10.6 / 100 = 0.106 g
Mass of antimony = 0.106 g
Mass percent = Mass of antimony / Mass of stibnite x 100
71.7% = 0.106 / X x 100
X = 0.106 x 1000 / 71.7 x 1000
= 106 / 717 = 0.148 g
Thus, the mass percent of antimony in the sulfide is 71.7% and the mass of Sb2S3 is 0.148 g.
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Draw Lewis structures to show how H+ is transferred when HNO₂ and NH₃ react with each other. The Lewis structure of HNO₂ is:
Attached are the Lewis structures:
Hope it helps...
[tex] \: \: \: [/tex]
PLEASE HELP ME
a)
b)
c)
or d)?
Answer:
D / 15.0 g
Explanation:
3 % volume thus shows that there are 3 g of an solute in every 100mL of solutions
.. there will be 3 × 5000÷ 100 of H2O2 in a 500 mL bottle
A chemist prepares a solution of sodium nitrate by measuring out of sodium nitrate into a volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's sodium nitrate solution. Round your answer to 3 significant digits.
Answer:
5.74M or 5.74 mol/L (to 3 sign. fig.)
Explanation:
The molar mass of NaNO3 is 85g/mol, which means that:
1 mole of NaNO3 - 85g
? moles - 122.0g
= 122/85 = 1.44 moles
Concentration in mol/L = no. of moles (moles) ÷ volume (L)
[tex]\frac{1.44}{0.250}[/tex] = 5.74M or 5.74 mol/L (to 3 sign. fig.)
I hope the steps are clear and easy to follow.
The last group of elements on the periodic table are called _____. noble gases halogens metals noble solids
Answer:
The answer is noble gases
Explanation:
Here is your explanation The vertical columns are called groups. There are eighteen groups. The last group on the far right is called the noble, or inert gases. Elements in a group have similar chemical properties. For example, elements in the noble gas group are all gases under. This is the thing from the passege bye god bless you
The last group of elements on the periodic table are called "noble gases."
The noble gases are a group of elements located in Group 18 of the periodic table. They are also known as Group 0 or the "inert gases." The noble gases include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).
The noble gases are unique because they have a full complement of valence electrons in their outermost energy level. This full electron configuration gives them exceptional stability, making them chemically unreactive or inert under normal conditions. In other words, noble gases are less likely to form chemical bonds with other elements.
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what is calcium anyone tell plz
Answer:
Calcium is a chemical element with the symbol Ca and atomic number 20.
Calcium is a mineral that is necessary for life. In addition to building bones and keeping them healthy, calcium enables our blood to clot, our muscles to contract, and our heart to beat. About 99% of the calcium in our bodies is in our bones and teeth.
Draw all four products when the following compound undergoes dehydrohalogenation and rank them in terms of stability. Which product do you expect to be the major product?
Answer:
2 Methyl pent 2 ene
Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
Answer:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Explanation:
Hello,
In this case, for the dissociation of calcium fluoride:
[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]
The equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:
[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]
Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Regards.
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
The dissociation of calcium fluoride has been given by:
[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]
The solubility constant, ksp has been given as:
[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]
From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.
The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.
The concentration of Ca in the solution has been resulted as x + 0.01 M.
The solubility product can be given as:
[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
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Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) 3 Cl2(g) + 2 Fe(s) → 6 Cl-(aq) + 2 Fe3+(aq) Cl2(g) + 2 e- → 2 Cl-(aq) E° = +1.36 V Fe3+(aq) + 3 e- → Fe(s) E° = -0.04 V
The cell potential for the electrochemical cell has been 1.40 V.
The standard reaction for the cell will be:
[tex]\rm 3\;Cl_2\;+\;2\;Fe\;\rightarrow\;6\;Cl^-\;+\;2\;Fe^3^+[/tex]
The half-reaction of the cells has been:
[tex]\rm Fe^3^+\;+\;3\;e^-\;\rightarrow\;Fe[/tex]
The potential for this reduction has been -0.04 V.
[tex]\rm Cl_2\;+\;2\;e^-\;\rightarrow\;2\;Cl^-[/tex]
The potential for the reduction has been 1.36 V.
The cell potential has been: Potential of reduction - Potential of oxidation
Cell potential = 1.36 - (-0.04) V
Cell potential = 1.40 V.
The cell potential for the electrochemical cell has been 1.40 V.
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volume of percentage =
formula??
Answer:
Volume percent is defined as: v/v % = [(volume of solute)/(volume of solution)] x 100%
Explanation:
choose isomer of hexanoic acid?
A) - penthylformiate
B) _ izopropyleacetate
C) _ methylpropanoate
D) _ a- Methylbutanoic acid
Answer:
THR ANSWER IS C) _ methylpropanoate
A compound is found to contain 27.29 % carbon and 72.71 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound
Answer:
CO2
Explanation:
Mass of carbon = 27.29
Mass of oxygen = 72.21
Step 1:
We have to first convert these masses to moles
Carbon = 2.29/2.01 = 2.274
Oxygen = 72.71/16 = 4.544
Step 2:
We have to divide the mols by the smallest mol to get simplest ratio of whole number.
The smallest mol is 2.274. we have to divide the mols by this.
2.274/2.274 = 1
4.544/2.274 = 2
Our empirical formula is therefore CO2
how can you prevent frequent landslides from heavy rains
Answer: make a drainage system or make sure that the ground does not absorb the water from the rain and cause a landslide.
Explanation: To prevent frequent landlides you have to make suee there is and area for the rain to go so it does not get stuck in the mud and destabalize the mud
B. Plant vegetation on the slops that expreience landslides.
E. Reduce the sloped where landslides occur.
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Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: specific heat (ice) = 2.1 J/g·°C; specific heat (water) = 4.18 J/g·°C; ΔH fus = 6.0 kJ/mol.
Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
Heat required to raise the temperature of ice from -20 °C to 0 °CBeing the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 [tex]\frac{J}{g*C}[/tex] and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 [tex]\frac{J}{g*C}[/tex] *20 C and solving: Q=420 J
Heat required to convert 0 °C ice to 0 °C waterThe heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*[tex]6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}[/tex]= 3.333 kJ= 3,333 J (being kJ=1,000 J)
Heat required to raise the temperature of water from 0 °C to 60 °CIn this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 [tex]\frac{J}{g*C}[/tex] and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 [tex]\frac{J}{g*C}[/tex] *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
The amount of heat to absorb is 6,261 J.
Calculation for heat:Heat required to raise the temperature of ice from -20 °C to 0 °C.
The formula for specific heat is used to calculate the amount of heat
Q = c * m * ΔT
Where,
Q =heat exchanged by a body,
m= mass of the body
c= specific heat
ΔT= change in temperature
Given:
m= 10 g,
specific heat of the ice= 2.1
ΔT=0 C - (-20 C)= 20 C
On substituting the values:
Q= 10 g*2.1 *20 C
Q=420 J
Heat required to convert 0 °C ice to 0 °C water.
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
Heat required to raise the temperature of water from 0 °C to 60 °C
m= 10 g,
Specific heat of the water= 4.18
ΔT=60 C - (0 C)= 60 C
On substituting:
Q= 10 g*4.18 *60 C
Q=2,508 J
Thus, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
The amount of heat to absorb is 6,261 J
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Which gas will have the least collisions between its particles?
A. Argon at 265 K
B. Helium at 20 C
C. Oxygen at -45 C
D. Carbon dioxide at 80 C
Answer:
C
Explanation:
This is because oxygen has the lowest temperature, thereby collision rate is least
One of the electron transitions in a hydrogen atom produces infrared light with a wavelength of
8.231 x 10°m. What amount of energy causes this transition?
A transition occurs when an electron absorbs energy and moves from one energy level to another.
Let us recall the formula;
E = hc/λ
h= planks constant = 6.6 * 10^-34 Js
c = speed of light = 3 * 10^8 ms-1
λ = wavelength =
We can see from the question that the exponent in the value for the wavelength was not properly rendered hence we can not be able to substitute values and obtain a numerical answer for the energy of the infrared radiation.
However, if all the values are properly substituted, then the energy of the infrared radiation can conveniently be obtained.
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At what temperature in K will 0.750 moles of oxygen gas occupy 10.0 L and exert 2.50 atm of pressure
Answer:
406 K.
Explanation:
The following data were obtained from the question:
Number of mole (n) = 0.750 mole
Volume (V) = 10.0 L
Pressure (P) = 2.50 atm
Temperature (T) =.?
Note: Gas constant (R) = 0.0821 atm.L/Kmol
The temperature, T can be obtained by using the ideal gas equation as follow:
PV = nRT
2.5 x 10 = 0.75 x 0.0821 x T
Divide both side by 0.75 x 0.0821
T = (2.5 x 10) /(0.75 x 0.0821 )
T = 406 K.
Therefore, the temperature is 406 K.
Answer: 406 K
Explanation:
We can rewrite the ideal gas law to solve for T:
PV = nRT
T=PV / nR
We are given the following from the problem:
n=0.750 mol P=2.50 atm V=10.0 L
Plugging in our values and using R=0.08206 L⋅atm / K⋅mol we get:
T=(2.50 atm)(10.0 L) / (0.750 mole)(0.08206L ⋅ atm ⋅ mole K) = 406 K
How many grams of H2O will be formed when 32.0 g H2 is mixed with 73.0 g of O2 and allowed to react to form water
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a)
b)
c)
d)
Answer:
B
Explanation:
the fluorine has an high tendency to gain electrons from other elements with lower electronegativities
A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONaA runner can cover 2.0 miles in 31 minutes, how long would it take for this runner to cover 6.0 Km. Hint (1 mile= 1.609 Km)
The answer to this question is approximately equal to 57.8
A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbonate that has twice the mass of the first sample
Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%
The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the standard molar Gibbs energy of formation of X(g) is 4.25 kJ·mol−1 at 2000. K and −63.12 kJ·mol−1 at 3000. K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature.
Answer:
[tex]K^{2000K}=0.774\\\\K^{3000K}=12.56[/tex]
Explanation:
Hello,
In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:
[tex]\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}[/tex]
Thus, at 2000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol[/tex]
And at 3000 K:
[tex]\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol[/tex]
Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]K=exp(-\frac{\Delta _rG}{RT} )[/tex]
Thus, at each temperature we obtain:
[tex]K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56[/tex]
In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).
Best regards.
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 sp (MgF 2) = 6.9 × 10 –9]
Answer:
No precipitate is formed.
Explanation:
Hello,
In this case, given the dissociation reaction of magnesium fluoride:
[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]
And the undergoing chemical reaction:
[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]
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:
[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]
Next, the moles of magnesium chloride consumed by the sodium fluoride:
[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]
Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:
[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]
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:
[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]
[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]
Thereby, the reaction quotient is:
[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]
In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.
Regards.