electrostatic and ionic are definitely not the answer because they have high melting point
hydrogen bonds are too weak and not permanent.
so the answer is apolar as it is soluble in polar solvents (water)
Answer:
B. Nonpolar
Explanation:
The low melting point tells you the compound is not ionic, metallic, or a network solid.
It is almost certainly a molecular solid.
It does not conduct electricity, so it is not metallic (which we have already ruled out).
It is insoluble in polar solvents (water) and soluble in nonpolar solvents (gasoline).
Since like dissolves like, the molecule is nonpolar.
The type of links must be nonpolar.
3,3-dibromo-4-methylhex-1-yne
Explanation:
see the attachment. hope it will help you...Advantages of using a resource person in handling the first aid lesson
The advantage of a resource person would be that it will provide a hands-on activity that will allow the students to experience spacing between organs and on the body of the person.
It will also allow them to identify challenges when doing this and will engage them more in the activity and lesson.
Answer:A resource person add knowledge to the course
Explanation:
An 80L capacity steel cylinder contains H2 at a pressure of 110 atm and 30 ° C, after extracting a certain amount of gas, the pressure is 80 atm at the same temperature. How many liters of hydrogen (measured under normal conditions) have been extracted?
Answer:
2200 L
Explanation:
Ideal gas law:
PV = nRT,
where P is absolute pressure,
V is volume,
n is number of moles,
R is universal gas constant,
and T is absolute temperature.
The initial number of moles is:
(110 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 353.58 mol
After some gas is removed, the number of moles remaining is:
(80 atm) (80 L) = n (0.0821 L atm / K / mol) (30 + 273.15) K
n = 257.15 mol
The amount of gas removed is therefore:
n = 353.58 mol − 257.15 mol
n = 92.43 mol
At normal conditions, the volume of this gas is:
PV = nRT
(1 atm) V = (92.43 mol) (0.0821 L atm / K / mol) (273.15 K)
V = 2162.5 L
Rounded, the volume is approximately 2200 liters.
You wish to construct a galvanic cell with the anode consisting of a Ni electrode in a 1.0 M Ni(NO3)2 solution. What would be the highest standard cell potential if used as the cathode in this galvanic cell?
Answer:
Au^3+(aq) +3e ------> Au(s). 1.50 V
Explanation:
When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.
E°cell= E°cathode - E°anode
If E°cathode= 1.50 V
E°anode= -0.25 V
E°cell= 1.50 -(-0.25)
E°cell= 1.75 V
Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential
The enthalpy change for a chemical reaction is: a. the temperature change b. the amount of heat given off or absorbed c. related to molar volume d. none of the above
Answer:
b. the amount of heat given off or absorbed
Explanation:
Hello,
In this case, we should take into account a formal definition of enthalpy change such as an energetic change that occurs in a system when matter is transformed by a given chemical reaction from reactants to products. Thus, such energetic change is macroscopically exhibited and it is related with either a temperature increase or decrease; it means that if a reaction exhibits a temperature increase, we say that heat was given off and if the temperature exhibits a decrease, we say that heat is absorbed. For that reason, answer is b. the amount of heat given off or absorbed.
Regards.
Gas is contained in a 9.00-L vessel at a temperature of 24.0°C and a pressure of 5.00 atm. (a) Determine the number of moles of gas in the vessel. mol (b) How many molecules are in the vessel? molecules
Answer:
a. Moles in the vessel = 1.85 moles of the gas
b. 1.11x10²⁴ molecules are in the vessel
Explanation:
a.It is possible to determine moles of a gas using the general law of gases:
PV = nRT
Where P is pressure: 5.00atm; V is volume = 9.00L; R is gas constant: 0.082atmL/molK; T is absolute temperature: 273.15K +24.0 = 297.15K
Computing the values:
PV / RT = n
5.00atm* 9.00L / 0.082atmL/molK*297.15K = n
Moles in the vessel = 1.85 moles of the gasb. With Avogadro's number we can convert moles of any compound to molecules thus:
Avogadro's number = 6.022x10²³ molecules / mole
1.85moles ₓ (6.022x10²³ molecules / mole) =
1.11x10²⁴ molecules are in the vesselThe direction of the functional group is called?
Explanation:
they are called hydrocarbyls
pls mark me brainliest
Answer:
The first carbon atom that attaches to the functional group is referred to as the alpha carbon.
1.40 m3 is how many mL
[tex] \LARGE{ \boxed{ \rm{ \pink{Solution:}}}}[/tex]
We know, 1 m³ of space can hold 1000 l of the substance.
⇛ 1 m³ = 1000 l----(1)
And, 1 l is 1000 times more than 1 ml
⇛ 1 l = 1000 ml------(2)
So, From (1) and (2),
⇛ 1 m³ = 1000 × 1000 ml
⇛ 1m³ = 1000000 ml
We had to find,
⇛ 1.40 m³ = 1.40 × 1000000 ml
⇛ 1.40 m³ = 140/100 × 1000000 ml
⇛ 1.40 m³ = 1400000 ml
⇛ 1.40 m³ = 14,00,000 ml / 14 × 10⁵ ml / 1.4 × 10⁶ ml
☃️ So, 1.40 m³ = 14 × 10⁵ ml / 1.4 × 10⁶ ml.
━━━━━━━━━━━━━━━━━━━━
Oxygen condenses into a liquid at approximately 90 K. What temperature, in degrees Fahrenheit, does this correspond to?
Answer:
-297.67 °F
Explanation:
Oxygen condenses into a liquid at approximately 90 K. We can convert any temperature in the Kelvin scale (absolute scale) to the Fahrenheit scale using the following expression.
°F = (K − 273.15) × 9/5 + 32
°F = (90 − 273.15) × 9/5 + 32
°F = (-183.15) × 9/5 + 32
°F = -329.67 + 32
°F = -297.67 °F
Suppose that a 100 mL sample of ideal gas is held in a piston-cylinder apparatus. Its volume could be increased to 200 mL by
Answer:
e. reducing the pressure from 608 torr to 0.40 atm at constant temperature.
Explanation:
According to Boyle's law when a gas is at the same temperature and there is a mass in a closed container so the pressure and the volume changes in the opposite direction
So here the equation is
[tex]P_1V_1=P_2V_2[/tex]
Now we choose the options
where,
[tex]V_1 = 100\ mL = 0.1\ L\\\\V_2 = 200\ mL = 0.2\ L[/tex]
[tex]P_1 = 608\ torr = 0.8\ atm \\\\P_2= 0.4\ atm[/tex]
Now applying these values to the above equation
So,
P1V1=P2V2
[tex]P_1V_1=P_2V_2[/tex]
[tex]0.8\times0.1 = 0.4\times0.2[/tex]
0.8 = 0.8
Hence, it is proved
Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation:
2ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96×10^11 at 273 K.
For an equilibrium mixture in which [Cl2O2] is 6.00 x 10^-6M, what is [ClO]?
Answer:
[ClO] = 3.48×10¯⁹ M.
Explanation:
The following data were obtained from the question:
Equilibrium constant (Kc) = 4.96×10¹¹
Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.
Concentration of ClO, [ClO] =.?
The equation for the reaction is given below:
2ClO(g) ⇌ Cl2O2(g)
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant, Kc for the reaction is given by:
Kc = [Cl2O2] / [ClO]²
Thus, we can calculate the concentration of ClO, [ClO] as follow:
Kc = [Cl2O2] / [ClO]²
4.96×10¹¹ = 6x10¯⁶ / [ClO]²
Cross multiply
4.96×10¹¹ × [ClO]² = 6x10¯⁶
Divide both side by 4.96×10¹¹
[ClO]² = 6x10¯⁶ / 4.96×10¹¹
[ClO]² = 1.21×10¯¹⁷
Take the square root of both side
[ClO] = √ (1.21×10¯¹⁷)
[ClO] = 3.48×10¯⁹ M
Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.
Q 11.20: What is the product of the reaction between t-BuCl and MeOH? A : t-BuOH B : MeOCl C : t-BuOMe D : (CH3)2CCH2
Answer:
C : t-BuOMe
Explanation:
The tert -butanol is a tertiary alcohol and when chloride ion attacks the carbocation, it forms t-BuCl.
The reaction of tert-butyl chloride or t-BuCl ((CH3)3C−Cl) with methanol and MeOH (CH3−OH) gives the product tert-Butyl methyl ether or t-BuOMe (CH3)3C−OCH3:
(CH3)3C−Cl + CH3−OH => (CH3)3C−OCH3 + HCl
Hence, the correct asnwer is C : t-BuOMe
Which of the following correctly summarizes the
relative composition of the lithosphere with
respect to inorganic and organic material?
A) inorganic >> organic
B) inorganic = organic
C) inorganic << organic
D) There is no organic matter in the lithosphere
Answer:
A
Explanation:
The lithosphere represents the layer of hardened/solid rock that makes up the hard part of the earth, including the brittle upper portion of the mantle and the crust. The lithosphere is broken into pieces that are referred to as plates. The pieces move to and away from each other in a process known as plate tectonics. The movement of plates accounts for the global locations of volcanoes, earthquakes, and mountain ranges.
The lithosphere is made of largely of inorganic materials known as silicates. The weathering of the solid rocks together with the interaction of living organisms gives rise to soil with an appreciable amount of organic materials.
The correct option is, therefore A.
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
Answer:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation
Explanation:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
What is chemical equation?To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.
Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
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candium forms the ion Sc3+. How many bromite ions could bond with Sc3+, and what would be the chemical formula?
a.3 bromite ions, Sc(Broa)2
b.2 bromite ions, Sc(BrO4)3
c.3 bromite ions, Sc(Broz)
d.2 bromite ions, Sc (BrO2)2
Answer: 3 bromite ions and [tex]Sc(BrO_2)_3[/tex]
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here scandium is having an oxidation state of +3 called as [tex]Sc^{3+}[/tex] cation and bromite is an anion with oxidation state of -1 called as [tex]BrO_2^-[/tex]. Thus 1 Scandium ion combines with three bromite ions and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Sc(BrO_2)_3[/tex]
Answer:
3 bromite ions and
Explanation:
Which of the following elements is in the same group as Sulfur (S)?
Answer:
PLEASE SHOW ME THE ELEMENTS OR I WOULD ENLIST ALL THE ELEMENTS.
Explanation:
Group 6A (or VIA) of the periodic table are the chalcogens: the nonmetals oxygen (O), sulfur (S), and selenium (Se), the metalloid tellurium (Te), and the metal polonium (Po)
Two elements represents by the letter Q and R atomic number 9 and 12 respectively. Write the electronic configuration of R
Answer:
Atomic no = 12 = Mg
Explanation:
It is given that,
The atomic number of two elements that are represented by letter Q and R are 9 and 12.
We need to write the electronic configuration of R. Atomic number shows the number of protons in atom.
For R, atomic number = 12
Its electronic configuration is : 2,8,2
It has two valance electrons in its outermost shell. The element is Magnesium (Mg).
A compound is found to contain 26.94 % nitrogen and 73.06 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 104.02 g/mol. What is the molecular formula for this compound?
Answer:
THE EMPIRICAL FORMULA OF THE COMPOUND IS NF2
THE MOLECULAR FORMULA OF THE COMPOUND IS N2F4
Explanation:
To calculate the empirical formula for the compound, we:
1. Write out the percentage weight of each elements
N = 26.94%
F = 73.06 %
2. Divide each by its atomic mass
( N= 14, F = 19)
N = 26.94 / 14 = 1.924
F = 73.06 / 19 = 3.845
3. Divide each by the smaller of the values
N = 1.924 / 1.924 = 1
F = 3.845 / 1.924 = 1,998
4. Round up to a whole number and write the empirical formula
N= 1
F = 2
So the empirical formula of the compound is N F2
To calculate the molecular formula, we:
(N F2 )n = molecular weight
( 14 + 19*2) n = 104.02
52 n = 104.02
n = 2.000
The molecular formula of the compound will be:
(N F2)2 = N2F4
In conclusion, the empirical formula of the compound is NF2 and the molecular formula of the compound is N2F4
1.78 L of gas is at a pressure of 735 torr. What is the volume in liters (L) when the pressure decreases to 700.0 torr
Explanation:
using boyles law
p1v1=p2v2
735 x 1.76 = 700 x V2
1293.6 = 700 x V2
V2 = 1293.6/700
V2 = 1.85L
Use the following equation to determine the charge on bromine when it dissociates from sodium and determine whether it is being oxidized or reduced: Cl2 + NaBr -> NaCl + Br2A. It starts with a charge of 0 and is oxidized. B. It starts with a charge of -1 and is reduced. C. It starts with a charge of 0 and is reduced. D. It starts with a charge of -1 and is oxidized.
It starts with a charge of 0 and is oxidized. Hence, option A is correct.
What is a chemical equation?A chemical reaction is a representation of symbols of the elements to indicate the number of substances and moles of reactant and product.
[tex]Cl_2 + NaBr[/tex] ->[tex]NaCl + Br_2[/tex]
The oxidation number of bromine changes from -1 (in NaBr) to 0 (in Br
2). Thus, NaBr is oxidized.
The oxidation number of chlorine changes from 0 to -1. Thus, chlorine is reduced.
Hence, option A is correct.
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f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration need to be for a precipitate to occur?
Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what will be the volume (in L) of the balloon if 0.50 moles of gas are released?
Answer:
Volume : 1.25 L
Explanation:
We are given here that the volume ( V[tex]_1[/tex] ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V[tex]_1[/tex] ) = 1.50 L,
Initial moles ( n[tex]_1[/tex] ) = 3.00 mol,
Final Volume ( n[tex]_2[/tex] ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V[tex]_2[/tex] ).
P[tex]_1[/tex]V[tex]_1[/tex] / n[tex]_1[/tex]T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / n[tex]_2[/tex]T[tex]_2[/tex] - we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V[tex]_1[/tex] / n[tex]_1[/tex] = V[tex]_2[/tex] / n[tex]_2[/tex] - isolate V[tex]_2[/tex],
V[tex]_2[/tex] = V[tex]_1[/tex] n[tex]_2[/tex] / n[tex]_1[/tex] = 1.50 L [tex]*[/tex] 2.5 mol / 3.00 mol = ( 1.5 [tex]*[/tex] 2.5 / 3 ) L = 1.25 L
The volume of the balloon will be 1.25 L.
3. Identify the reagents you would use to convert 1-bromopentane into each of the following compounds: (a) Pentanoic acid (b) Hexanoic acid (c) Pentanoyl chloride (d) Hexanamide (e) Pentanamide (f) Ethyl hexanoate
Answer:
Explanation:
a )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH
b )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH .
c )
CH₃CH₂CH₂CH₂CH₂Br + KOH ⇒ CH₃CH₂CH₂CH₂CH₂OH
CH₃CH₂CH₂CH₂CH₂OH + acidic potassium dichromate ⇒ CH₃CH₂CH₂CH₂COOH + SOCl₂ ( thionyl chloride ) ⇒ CH₃CH₂CH₂CH₂COCl
d )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + PCC ( NH₃ ) ⇒ CH₃CH₂CH₂CH₂CH₂CONH₂
e )
CH₃CH₂CH₂CH₂CH₂Br + KCN ⇒ CH₃CH₂CH₂CH₂CH₂CN Hydrolysis ⇒ CH₃CH₂CH₂CH₂CH₂COOH + C₂H₅OH ( Ethyl alcohol + H⁺ )⇒
CH₃CH₂CH₂CH₂CH₂COOC₂H₅ ( ethyl hexanoate )
If 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
The tosylate of (2R,3S)-3-phenylbutan-2-ol undergoes an E2 elimination on treatment with sodium ethoxide. Draw the structure of the alkene that is produced.
Answer:
(R)-but-3-en-2-ylbenzene
Explanation:
In this reaction, we have a very strong base (sodium ethoxide). This base, will remove a hydrogen producing a double bond. We know that the reaction occurs through an E2 mechanism, therefore, the hydrogen that is removed must have an angle of 180º with respect to the leaving group (the "OH"). This is known as the anti-periplanar configuration.
The hydrogen that has this configuration is the one that placed with the dashed bond (red hydrogen). In such a way, that the base will remove this hydrogen, the "OH" will leave the molecule and a double bond will be formed between the methyl and the carbon that was previously attached to the "OH", producing the molecule (R) -but-3- en-2-ylbenzene.
See figure 1
I hope it helps!
There are 454 grams in one pound. How many pounds are in 700 grams
Answer:
1.543 pounds = 700 grams
For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
what is a chemical that is safe to use in food but in small amounts?
Answer:
Toxins
Explanation:
Nitric oxide reacts with oxygen to give nitrogen dioxide, an important reaction in the Ostwald process for the industrial synthesis of nitric acid: 2NO(g)+O2(g)⇌2NO2(g)
Part A If Kc=6.9×105 at 227 ∘C,
what is the value of Kp at this temperature? Express your answer using two significant figures. Kp =
Part BIf Kp=1.3×10−2 at 1000 K, what is the value of Kc at 1000 K? Express your answer using two significant figures. Kc =
Answer:
Kp=1.68×10⁴∆1.7×10⁴
Kc=1.06∆1.1
Explanation:
Value of Kp at 227°C is 2.86×10² and value of Kc at 1000 K is 1.56.
How are Kp and Kc related?Kp and Kc are related by the formula Kp=Kc(RT).For part A , Kp is calculated as,
Kp=6.9×10⁵×8.314×500=28.683×10² and for part B Kc is calculated as,
Kc=1.3×10[tex]^-2[/tex]/(8.314×1000)=1.56
Kc and Kp are equilibrium constants of a mixture of ideal gases. Kp is equilibrium constant when concentrations at equilibrium are in atmospheric pressure and Kc is equilibrium constant when concentrations are in molarity. The relation is only valid for gaseous mixtures. The relation between these two parameters is obtained through ideal gas equation.
Kc and Kp of reaction change with temperature of reaction but remain unaffected by change in concentration , pressure and presence of catalyst.
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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.
