Answer:
C. 101.5°C
Explanation:
The computation of the boiling point of a solution is shown below:
As we know that
[tex]\Delta T = i K_bm[/tex]
where,
i = Factor of Van’t Hoff i.e 2KBr
K_b = Boiling point i.e 0.512° C
m = molality [tex]= \frac{Moles\ of\ solute}{Solvent\ mass}[/tex]
Now place the values to the above formula
So, the boiling point is
[tex](T - 100^\circ C) = 2\times 0.512^\circ \times \frac{3.00\ ml}{2.00\ kg}[/tex]
After solving this, the t is
= 100° C + 1.536° C
= 101.536° C
Hence, the correct option is C.
Name of th molecule
1. CH3CH2CHClCHBrCH3
2.C=C-CH3
CH3CH=CHCH2
Answer:
1: 2-bromo-3-chloropentane
Explanation:
find longest carbon chain =5
place the Br and Cl on the carbon chain
follow naming rules I guess
Of the following elements,________has the most negative electron affinity.
A. Sb
B. Sn
C. Sr
D. Te
E. I
Of the following elements,________has the most negative electron affinity.
A. Sb
B. Sn
C. Sr
D. Te
E. I
Answer:
E. Iodine
Explanation:
To answer this question, we use periodic trends of the Periodic Table of Elements to help us out. Electron Affinity increase going up and to the right. Since all of our elements are in the same period, we look to see which element is the farthest right. In that case, Iodine is our answer.
Where would you find the following symbol on a diagram of a water molecule: δ+ ?
Answer:
Near the Hydrogen Atoms.
Explanation:
δ+ would be found near the dipole end of the hydrogens. δ- would be found near the dipole end of oxygen.
How many moles of chloride ions are there in 2.5 L of 5 M magnesium chloride?
Answer:
[tex]n_{Cl^-}=25molCl^-[/tex]
Explanation:
Hello,
In this case, since the given 5-M concentration of magnesium chloride is expressed as:
[tex]5\frac{molMgCl_2}{L}[/tex]
We can notice that one mole of salt contains two moles of chloride ions as the subscript of chlorine is two, in such a way, with the volume of solution we obtain the moles of chloride ions as shown below:
[tex]n_{Cl^-}=5\frac{molMgCl_2}{L}*\frac{2molCl^-}{1molMgCl_2} *2.5L\\\\n_{Cl^-}=25molCl^-[/tex]
Best regards.
2. Using Tables 5 and 6 below, balance the following chemical reaction:
Na2CO3 (aq)+CaCl2 (aq) → CaCO3 (s)+NaCl (aq)
Answer:
Na₂CO₃ + CaCl₂ → CaCO₃ + 2 NaCl
Explanation:
Na₂CO₃ + CaCl₂ → CaCO₃ + NaCl
First, determine what is balanced.
Ca and CO₃ is balanced; there is one on each side.
Na and Cl aren't balanced. On one side, there is two of each, while on the other there is only one. To fix this add a 2 in front of NaCl. This will cause the equation to be balanced.
Na₂CO₃ + CaCl₂ → CaCO₃ + 2 NaCl
A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete precipitation of the
ions as Al(OH), and Co(OH)2. The total mass of the precipitate is 22.73 g. Find the masses of Al3+ and Co2+ in the solution.
Answer:
The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.
Explanation:
Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.
The mass of the precipitate given is 22.73 grams.
Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams
Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.
And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.
Now the equation will become,
3x/78 + 45.46-2x/92.94 = 0.647 moles
0.03846 x + 0.489 - 0.02152 x = 0.647
0.01694 x + 0.489 = 0.647
0.01694 x = 0.158
x = 0.158/0.01694
x = 9.327 grams
Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,
= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams
Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams
the mass of Co²⁺ will be,
= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams
The NaOH solution is standardized (or its true concentration) is found by reacting it with KHSO4. One of the two products from when NaOH reacts with KHSO4 is H2O. The other product is is a salt consisting of what?
a. NaK (aq)
b. (aq)
c. NaS (aq)
d. None of the above
What did Rutherford discover that Thomson did not understand?
Answer: That most of an atom's mass was packed in a central nucleus
Answer:
That most of an atom's mass was packed in the central nucleus.
Explanation:
Rutherford disproved Thomson's model in 1911 with his gold foil experiment, in which he demonstrated that the atom has a high-mass nucleus.
0.60 atm of SO3 and 0.30 atm of SO2are placed in a container and the system is allowed to reach equilibrium. Calculate the pressure of O2(g) at equilibrium.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The pressure is [tex][O_2] = 4.8 *10^{-5} \ atm[/tex]
Explanation:
From the question we are told that
The pressure of [tex]SO_3[/tex] is [tex][SO_3 ] = 0.63 \ atm[/tex]
The pressure of [tex]SO_2[/tex] is [tex][SO_ 2] = 0.30 \ atm[/tex]
The equilibrium constant is [tex]K_p = 1.2 *10^{-5}[/tex]
The reaction is
[tex]2SO_3 _{(g)}[/tex] ⇔ [tex]2SO_2_{(g)} + O_2 _{(g)}[/tex]
Generally the equilibrium constant is mathematically represented as
[tex]K_p = \frac{(SO_2)^2 * (O_2)}{(SO_3)^2}[/tex]
=> [tex][O_2] = \frac{k_p * [SO_3] ^2 }{[SO_2]^2}[/tex]
substituting values
[tex][O_2] = \frac{1.2 *10^{-5} * 0.60 ^2 }{0.30^2}[/tex]
[tex][O_2] = 4.8 *10^{-5} \ atm[/tex]
In general, the solubility of a short chain carboxylic acid is greater than an ester with the same number of carbons. For example, butyric acid (four carbon atoms) is infinitely soluble in water, whereas ethyl acetate is only partially soluble in water (8 g/100 mL).
Select the statement that correctly explains why the water solubility of the carboxylic acid is greater than that of the ester.
a. Ethyl acetate has two -CH3 groups while butyric acid has only one.
b. The carboxylic acid is much more soluble because the-COOH group is on the end of the molecule and is therefore more accessible to H2O.
c. An ester group (-COO-) cannot form hydrogen bonds with water since there are no hydrogen atoms in the ester group.
d. Carboxylic acid groups (-COOH) can form hydrogen bonds with each other, but ester groups (Coo-) cannot.
e. A carboxylic ad roup CH) can form more hydrogen bonds with water than an ester group (-COO-).
Answer:
A carboxylic acid group (-COOH) can form more hydrogen bonds with water than an ester group (-COO-).
Explanation:
The carboxylic acid group (-COOH) is found in the carboxylic acids. This group is ultimately responsible for the solubility of carboxylic acids in water. It is worthy of note that the high boiling points of low molecular weight carboxylic acids is often because they are capable of intermolecular hydrogen bonding which leads to the dimerization of carboxylic acid.
The solubility of carboxylic acids decreases as the length of the alkyl chain increases. Hence, a long chain carboxylic acid is less soluble in water than shorter chain carboxylic acids.
Ester molecules can't form hydrogen bonds with each other but they do form weak hydrogen bonds with water. This leads to the solubility of low molecular weight esters. However, if a carboxylic acid and an ester posses the same length of alky chain, the carboxylic acid will form more hydrogen bonds and thus be more soluble in water than than a corresponding ester of the same chain length.
Question 23
1 pts
When solutions of AgNO3 and NaOH react, the balanced molecular equation is:
2 AgNO3(aq) + 2NaOH(aq) O--> Ag2O(s) + 2 NaNO3(aq) + H20(1)
How much Ag2O is produced when 0.200 g of AgNO3 and 0.200 g of NaOH react?
a. 0.127 g
c. 0.273 g
b. 0.136 g
d. 0.400 g
OB
OC
OA
OD
Answer:
Option B. 0.136 g
Explanation:
The balanced equation for the reaction is given below:
2AgNO3(aq) + 2NaOH(aq) —> Ag2O(s) + 2NaNO3(aq) + H2O(l)
Next, we shall determine the masses of AgNO3 and NaOH that reacted and the mass of Ag2O produced from the balanced equation. This is illustrated below:
Molar mass of AgNO3 = 108 + 14 + (16x3) = 170g/mol
Mass of AgNO3 from the balanced equation = 2 x 170 = 340g
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the balanced equation = 2 x 40 = 80g
Molar mass of Ag2O = (108x2) + 16 = 232g/mol
Mass of Ag2O from the balanced equation = 1 x 232 = 232g
Summary:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH to produce 232g of Ag2O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted with 80g of NaOH.
Therefore, 0.2g of AgNO3 will react with = (0.2 x 80)/340 = 0.047g of NaOH.
From the calculations made above, only 0.047g out of 0.2g of NaOH given, reacted completely with 0.2g of AgNO3. Therefore, AgNO3 is the limiting reactant and NaOH is the excess reactant.
Now, we can calculate the mass of Ag2O produced from the reaction of 0.2g of AgNO3 and 0.2g of NaOH.
In this case, we shall use the limiting reactant because it will produce the maximum yield of Ag2O as all of it is used up in the reaction.
The limi reactant is AgNO3 and the mass of Ag2O produced can be obtained as follow:
From the balanced equation above,
340g of AgNO3 reacted to produce 232g of Ag2O.
Therefore, 0.2g of AgNO3 will react to produce = (0.2 x 232)/340 = 0.136g of Ag2O.
Therefore, 0.136g of Ag2O was produced from the reaction.
determine the rate of reaction that follows the rate= k[A]^m[B]^n
rate=0.2*3^1*3^2=0.2*3*9=5.4(mol/L)s so the correct answer is C.
Why does the excess of base used in these eliminations favor the E2 over the E1 mechanism for elimination
Answer:
The base is involved in the rate determining step of an E2 reaction mechanism
Explanation:
Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;
Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.
For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;
Rate = k[alkyl halide] [base]
The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.
Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.
What is the final pH of a solution with an initial concentration of 2.5mM Ascorbic acid (H2C6H6O6) which has the following Kas: 7.9x10-5 and 1.6x10-12
Answer:
pH = 3.39
Explanation:
The equilibrium in water of ascorbic acid (With its conjugate base) is:
H₂C₆H₆O₆(aq) + H₂O(l) ⇄ HC₆H₆O₆⁻(aq) + H₃O⁺(aq)
Where the acidic dissociation constant is written as:
Ka = 7.9x10⁻⁵ = [HC₆H₆O₆⁻] [H₃O⁺] / [H₂C₆H₆O₆]
H₂O is not taken in the Ka expression because is a pure liquid.
As initial concentration of H₂C₆H₆O₆ is 2.5x10⁻³M, the equilibrium concentration of each species in the equilibrium is:
[H₂C₆H₆O₆] = 2.5x10⁻³M - X
[HC₆H₆O₆⁻] = X
[H₃O⁺] = X
Replacing in the Ka expression:
7.9x10⁻⁵ = [X] [X] / [2.5x10⁻³M - X]
1.975x10⁻⁷ - 7.9x10⁻⁵X = X²
0 = X² + 7.9x10⁻⁵X - 1.975x10⁻⁷
Solving for X:
X = -0.00048566→ False solution, there is no negative concentrations
X = 0.00040666 → Right solution
As [H₃O⁺] = X, [H₃O⁺] = 0.00040666
pH is defined as -log [H₃O⁺];
pH = -log 0.00040666,
pH = 3.39The average molecular speed in a sample of Ar gas at a certain temperature is 391 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.
Answer:
550 m/s
Explanation:
The average molecular speed (v) is the speed associated with a group of molecules on average. We can calculate it using the following expression.
[tex]v = \sqrt{\frac{3 \times R \times T}{M} }[/tex]
where,
R: ideal gas constantT: absolute temperatureM: molar mass of the gasWe can use the info of argon to calculate the temperature for both samples.
[tex]T = \frac{v^{2} \times M}{3 \times R} = \frac{(391m/s)^{2} \times 39.95g/mol}{3 \times 8.314J/k.mol} = 2.45 \times 10^{5} K[/tex]
Now, we can use the same expression to find the average molecular speed in a sample of Ne gas.
[tex]v = \sqrt{\frac{3 \times R \times T}{M} } = \sqrt{\frac{3 \times (8.314J/k.mol) \times 2.45 \times 10^{5}K }{20.18g/mol} } = 550 m/s[/tex]
The next few questions will walk you through solving the following problem: At a given temperature, a 5.0M solution of hydrazine (N2H4) as a pH of 11.34. Hydrazine is base.
A. What is the concentration of hydroxide ion at equilibrium?
B. What is the pK for hydrazine reacting with water at this temperature?
Answer:
A. [OH⁻] = 2.188x10⁻³M
B. pKb = 6.02
Explanation:
When hydrazine is in equilbrium with water, its reaction is:
N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)
Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
A. From pH, you can find [OH⁻], thus:
pH = -log [H⁺]
11.34 = -log [H⁺]
4.57x10⁻¹² = [H⁺]
As 1x10⁻¹⁴ = [OH⁻] [H⁺]
1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]
[OH⁻] = 2.188x10⁻³MB. Concentrations in equilibrium of the species are:
[N₂H₄] = 5.0M - X
[HN₂H₄⁺] = X
[OH⁻] = X
Where X is reaction coordinate
As [OH⁻] = 2.188x10⁻³M
X = 2.188x10⁻³M
Replacing:
[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M
[HN₂H₄⁺] = 2.188x10⁻³M
[OH⁻] = 2.188x10⁻³M
Replacing in Kb expression:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]
Kb = 9.577x10⁻⁷
pKb is defined as -log Kb
pKb = -log 9.577x10⁻⁷
pKb = 6.02
Consider the following precipitation reaction occurring in aqueous solution:
3 SrCl2(aq)+2 Li3PO4(aq) →Sr3(PO4)2(s)+6 LiCl(aq)
Write the complete ionic equation and the net ionic equation for this reaction.
Answer:
[tex]3Sr^{+2}+6Cl^{-}+6Li^{+}+2PO_{4}^{3-}-->Sr_{3}(PO_{4})_{2}+6Li^{+}+6Cl^{-}\\\\3Sr^{+2}+2PO_{4}^{3-} --->Sr_{3}(PO_{4})_{2}[/tex]
First equation is the complete ionic equation.
Second equation is the net ionic equation.
Determine the empirical formula of Magnesium Oxide from following data. Type the calculations.
Mass of Crucible and Cover + magnesium ribbon (before heating) =27.60 g
Mass of crucible and Cover = 27.30 g
Mass of magnesium metal =??
Mass of crucible and cover + magnesium oxide (after heating) = 27.80 g
Mass of combined oxide (after heating - before heating) =??
lonic compounds can conduct electricity in
Answer:
Molten form
Explanation:
This is because they have mobile ions
How are scientific questions answered?
A. Through observing and measuring the physical world
B. Through testing a theory about the physical world
c. Through forming a hypothesis about the question
D. Through predicting a solution about the question
SUBM
Answer:
Option B
Explanation:
Scientific question are answered through experimentation, through testing the theory about the physical world.
Answer: its A
through observing and measuring the physical world
Explanation:
Which of the following alkyl halides is most likely to undergo rearrangement in an SN1 reaction? Please explain your answer.
A) 3-‐bromopentane
B) 2-‐chloro-‐3,3-‐dimethylpentane
C) 3-‐chloropentane
D) bromocyclohexane
E) 1-‐bromo-‐4-‐methylcyclohexane
If an electron has a principal quantum number (n) of 7 and an angular momentum quantum number (l) of 1, the subshell designation is ________
Answer:
7p
Explanation:
principal quantum number is 7
n=7( principle shell)
angular momentum quantum number gives sub shell
l = 1 means it is p orbital
so answer is 7p orbital
A volumetric flask contains 25.0 mL of a 14% m/V sugar solution. If 2.5 mL of this solution is added to 22.5 mL of distilled water, what is the % m/V of the new solution.
Answer:
The new solution is 1.4% m/V
Explanation:
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
We have 2.5 mL (V₁) of a concentrated solution and add it to 22.5 mL of distilled water. Assuming the volumes are additives, the volume of the new solution (V₂) is:
[tex]2.5 mL + 22.5 mL = 25.0 mL[/tex]
We want to prepare a dilute solution from a concentrated one, whose concentration is 14% m/V (C₁). We can calculate the concentration of the dilute solution (C₂) using the dilution rule.
[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{14\% m/V \times 2.5 mL}{25.0 mL} = 1.4 \% m/V[/tex]
The concentration of the new solution, obtained by adding 22.5 mL of distilled water to 2.5 mL of 14 % m/V sugar solution, is 1.4% m/V.
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The solubility of O2 in water is approximately 0.00380 g L-1 of water when the temperature is 25.0°C and the partial pressure of gaseous oxygen is 760. torr. The oxygen gas above the water is replaced by air at the same temperature and pressure, in which the mole fraction of oxygen is 0.210. What will the solubility of oxygen in water be under these new conditions?
Answer:
The correct answer is 0.00080 gram per liter.
Explanation:
Based on the given information, the solubility of water is 0.00380 gram per liter, the temperature mentioned is 25 degree C, the partial pressure of oxygen gas is 760 torr, and the mole fraction of oxygen is 0.210. There is a need to determine the solubility of oxygen in water.
Based on Henry's law,
Solubility of oxygen gas = Henry's constant × partial pressure of oxygen gas
Henry's constant, K = solubility of oxygen gas / partial pressure of oxygen gas
= 0.00380 g/L × 1 mol/32 grams / 760 torr × 1 atm/760 torr
= 0.00012 mol/L/atm
= 0.00012 M/atm
Now the partial pressure of the oxygen gas = mole fraction of oxygen × atmospheric pressure
= 0.210 × 1 atm
= 0.210 atm
Now putting the values in Henry's law equation we get,
Solubility of oxygen gas = 0.00012 mol/L/atm × 0.210 at,
= 0.000025 mol/L × 32 gram/mol
= 0.00080 gram per liter
Identify the correctly written chemical reaction
A. Reactant + Reactant = Product
B. Reactant + Reactant → Product + Product
C. Reactant + Product → Reactant + Product
D. Product + Product Reactant + Reactant
Answer:
B. Reactant + Reactant -> Product + Product
Explanation:
Reactants are substances that- as the name suggests- reacts with other substances at the beginning of a reaction
Products are substances that are produced as a result of the reaction
Typically, when writing a chemical reaction, an arrow is used to show the direction the reaction is moving. In this case, the arrows in options B and C suggest that the reaction only moves in one direction- forwards
And as mentioned above, reactants are the substances at the start of the reaction, they're what mixes together to form a new product.
To keep things simple:
Products can't be at the beginning of a reaction since they weren't formed yet.
Similarly, reactants can't be part of the products since they already existed and didn't need to be made. In a lot cases, the reactants would be completely used up to make the products
As such, only one possible chemical reaction would follow that reasoning:
Reactant + Reactant -> Product + Product
Reactant + Reactant → Product + Product is the correctly written chemical reaction. Hence, option B is correct.
What is a chemical equation?A chemical equation is a mathematical expression of the chemical reaction which represents the product formation from the reactants.
In an equation, the reactants are written on the left-hand side and the products are written on the right-hand side demonstrated by one-headed or two-headed arrows.
Hence, option B is correct.
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Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/
Answer:
10.328 m
Explanation:
normal atmospheric pressure = 101325 Pa
density of water at 25 °C = 1.0 g/cm^3 = 1000 kg/m^3
pressure = pgh
where p = density
g = acceleration due to gravity = 9.81 m/s^2
h = height of column
imputing values, we have
101325 = 1000 x 9.81 x h
height of column h = 101325/9810 = 10.328 m
Write the following isotope in nuclide notation: oxygen-14
Answer:
[tex]14\\8[/tex]O
Explanation:
The top number always represents the mass number.
The bottom number always represents the atomic number.
The element always goes after the numbers.
If charge is present, that comes after the element.
A compound is found to contain 39.99 % carbon, 6.727 % hydrogen, and 53.28 % 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:
CH₂O
Explanation:
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of C:H:O.
Assume 100 g of the compound.
1. Calculate the mass of each element.
Then we have 39.99 g C, 6.727 g H, and 53.28 g O.
2. Calculate the moles of each element
[tex]\text{Moles of C} = \text{39.99 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = \text{3.330 mol C}\\\\\text{Moles of H} = \text{6.727 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H }} = \text{6.674 mol H}\\\\\text{Moles of O} = \text{53.28 g O} \times \dfrac{\text{1 mol O}}{\text{16.00 g O }} = \text{3.330 mol O}[/tex]
3. Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
C:H:O = 3.330:6.674:3.330 = 1:2.004:1.000 ≈ 1:2:1
4. Write the empirical formula
EF = CH₂O
What happens to electrons in the photoelectric effect?
A. They are added to atoms.
B. They are knocked away from atoms.
C. They are turned into light.
D. They are made to orbit faster.
The correct answer is B. on Apex!
They have been knocked away from atoms. Hence, option B is correct.
What is the photoelectric effect?The photoelectric effect is a phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation.
When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect.
This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.
Hence, option B is correct.
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How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water
Answer:
WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.
Explanation:
Equation for the reaction:
K2SO4 + H20 ------->2 K+ + SO4^2-
When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.
1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-
At STP, 1 mole of K2SO4 will be the molar mass of the substance
Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol
Molar mass = 174 g/mol
So therefore;
1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion
When 63.7 g is used; we have:
174 g = 2 moles of K+
63.7 g = ( 63.7 * 2 / 174) moles of K+
= 0.73 moles of K+
Forr sulfate ion, we have:
174 g = 1 mole ofSO4^2-
63.7 g = (63.7 * 1 / 174) moles of SO4^2-
= 0.366 moles of SO4^2-
In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.