The first step to solve this question is to convert the given mass of copper to moles using its molecular mass (63.5g/mol):
[tex]12.8g\cdot\frac{1mol}{63.5g}=0.2mol[/tex]Now, use Avogadro's number to find the number of atoms in 0.2 moles. Remember that it states the number of atoms in 1 mole of substance is 6.022x10^23.
[tex]0.2mol\cdot\frac{6.022\times10^{23}atoms}{1mol}=1.2044\times10^{23}atoms[/tex]There are 1.2044x10^23 atoms of copper in 12.8g of copper metal.
A- Label the reactions as weak acid dissociation , strong acid, weak base, strong base dissociation, or neutralization.B- Circle the acid, base, or salt on the products side and label it as such.
To label the reactions, it is important to remember that in:
- Weak acid dissociation the acid partially dissociates.
- Strong acid dissociation the acid fully dissociates.
- Weak base dissociation the base partially dissociates.
- Strong base dissociation the base fully dissociates.
- Neutralization an acid and a base react to form a salt.
1. a: Strong acid dissociation, b: Cl- is the weak conjugate base.
2. a: Weak acid dissociation, b: CH3COO- is the strong conjugate base.
3. a: Strong base dissociation, b: Na+ is the weak conjugate acid.
4. a: Weak base dissociation, b: Na4+ is the strong conjugate acid.
5. a: Neutralization, b: NaNO3 is the salt.
Give the periodic group number and number of valence electrons for each of the following atoms:a) Ob) Bc) Nad) Mg
Answer:
a. O: Period 2. VIA family. 6 valence electrons.
b. B: Period 2. IIIA family. 3 valence electrons.
c. Na: Period 3. IA family. 1 valence electron.
d. Mg: Period 3. IIA family. 2 valence electrons.
Explanation:
There is a property of the periodic table which is: for neutral atoms, the number of valence electrons is equal to the atom's main group number. The group numbers are found in columns in the periodic table and periods are found in rows.
a. If you see the periodic table, you can see that O belongs to the second period and VIA family, VI is 6, so its valence electrons are 6.
b. You can see that B is in the second period too but is in the IIIA family, III is 3, so the valence electrons of B are 3.
c. Na is in the third period and IA family, I is 1, so Na only contains 1 valence electron.
d. Mg is in the third period too, but is in the IIA family, II is 2, so Mg has 2 valence electrons.
How many grams are in 3.00 moles of carbon? Show your work.
There are 36.03 grams in 3.00 moles of carbon.
1st) We need to look for the atomic mass of Carbon (C) in the Periodic Table of Elements:
Carbon atomic mass = 12.01 g/mol
2nd) With the atomic mass of carbon and a mathematical Rule of Three we can calculate the grams of carbon in 3.00 moles:
[tex]\begin{gathered} 1\text{ mol - 12.01 g} \\ 3.00\text{ moles - x=}\frac{3.00\text{ moles}\cdot\text{12.01 g}}{1\text{ mol}} \\ \\ x=36.03\text{ g} \end{gathered}[/tex]So, there are 36.03 grams of carbon in 3.00 moles.
hello I need help with problem number three part A and B thank you
In letter A, we have to find out the number of moles of NaOH from the given volume, which must be in Liters, and also the given concentration, which is 1.00 M of NaOH
The volume found was = 5.625 mL, which in liters will be 0.00562 Liters
The formula we must use is:
n = M * V
n = 1.00 * 0.00562
n = 0.005625 moles of NaOH
In letter B, we have to calculate the moles that are in excess of HCl, the reaction that we have is:
NaOH + HCl -> NaCl + H2O
We have:
0.005625 moles of NaOH
0.0150 moles of HCl
According to the molar ratio between NaOH and HCl, we have a ratio of 1 mol of NaOH for 1 mol of HCl, therefore:
1 NaOH = 1 HCl
0.005625 NaOH = x HCl
x = 0.005625 moles of HCl
0.0150 - 0.005625 = 0.009375 moles of HCl of excess
What is the % error if the theoretical value of the enthalpy of a reaction is 92 kJ and the experimental value is 38 kJ? Do not include % sign.
To find the % error of the enthalpy of the reaction we have to use the following formula:
[tex]\%error=\frac{|theoreticalvalue-experimentalvalue|}{theoreticalvalue}\cdot100[/tex]Replace for the given values and solve:
[tex]\begin{gathered} \%error=\frac{|92kJ-38kJ|}{92kJ}\cdot100 \\ \%error=\frac{|54kJ|}{92kJ}\cdot100 \\ \%error=0.59\cdot100 \\ \%error=59 \end{gathered}[/tex]The %error is 59.
Suggest and an oxidising agent that could be used in reaction 1.The reaction:Butan-2-ol + [O] --> Butanone
First of all, we want to oxidize secondary alcohol to ketone.
As an oxidizing agent, we can name sodium or potassium dichromate (IV), some heat is needed too and, acidified with dilute sulfuric acid.
For this reaction, Butan-2-ol to Butanone, we can use alkaline KMnO4
What is the pH of 8.63x10^-05 M KOH solution? Is the solution acid or base?
Explanation
Given
Concentration of KOH = 8.63x10^-05 M
Required: pH of the solution
Solution
pOH = -log[OH-]
pOH = -log[8.63x10^-05]
pOH = 4.07
pH = 14-pOH
pH = 14-4.07
pH = 9.93
Therefore the solution is basic since the pH is higher than 7
Answers
pH of KOH = 9.93
Solution is a base
Which is true of the thermal energy of particles some help would be greatly appreciated
The definition for thermal energy is the sum of the kinetic energy and potential energy of all the particles in a compound, this will give us the total thermal energy of this compound, therefore the best fit with this definition will be letter A
How many kJ are produced when 50.0 g of oxygen reacts with excess H2(g) in the following reaction?2H2(g) + O2(g) → 2H2011) + 572 kJYour Answer:
we are given the equation
2H₂(g) + O₂(g) → 2H₂011) + 572 kJ
and we are required to find the amount of KJ produced when 50 g of oxygen are reacted with excess hydrogen
solution
50 grams of oxygen has :
Mols = m/M
= 50/16.00
= 3.125 mols
from the balance equation we see that when 1 mol of oxygen reacts with 2 mols of Hydrogen +572KJ of energy are produces. therefore when 3.125 mols of oxygen are reacted we get :
3.125 x 572
= 1787 KJ
therefore 1787KJ of energy will be produced
how many moles of iron(lll) hydroxide contain 9.45 x10^25 atoms? Include units and name of atoms/molecule.
We are required to calculate the number of moles of iron (III) hydroxide.
We are given the number of atoms = 9.45x10^25.
We also know that 1 mol = 6.022x10^23 atoms
Iron(III) hydroxide moles = 9.45x10^25/6.022x10^23
= 156.925 mol of Iron (III) hydroxide
I need help with my homework An unknown substance contains 43.20% oxygen, 8.200% hydrogen and 48.60% carbon. Determine the empirical formula.
An unknown substance contains 43.20% oxygen, 8.200% hydrogen and 48.60% carbon. Determine the empirical formula.
When we are given the composition by mass of a substance and we have are asked to determine the empirical formula we usually have to follow some steps.
1) We can suppose that we have 100 g of that substance. So we can convert those percentages into the mass of each element.
mass of sample = 100 g
% of O = 43.20 %
mass of O = 43.2 g
% of H = 8.200%
mass of H = 8.200 g
% of C = 48.60%
mass of C = 48.60 g
2) We have to find the number of moles of each element that we have in the sample.
molar mass of O = 16.00 g/mol
molar mass of H = 1.01 g/mol
molar mass of C = 12.01 g/mol
moles of O = 43.2 g/(16.00 g/mol)
moles of O = 2.7 moles
moles of H = 8.200 g/(1.01 g/mol)
moles of H = 8.12 moles
moles of C = 48.60 g/(12.01 g/mol)
moles of C = 4.05 moles
3) The definition of empirical formula is: "it is the simplest whole number ratio of atoms present in a compound". So if we want to find the simples number ratio we have to divide the number of moles of each element by the least of them.
O = 2.7/2.7 = 1
H = 8.12/2.7 = 3
C = 4.05/2.7 = 1.5
The empirical formula seams to be:
C₁.₅H₃O
4) We can't have fractions as subscripts, we need whole numbers. So we have to multiply the formula by 2.
Empirical formula = C₃H₆O₂
Answer: The empirical of the unknown substance is C₃H₆O₂.
34 mL of nitrogen gas is measured at 21.0 Celsius and .953 ATM pressure calculate the volume at STP
To solve this question, we would use the combined gas equation which comprises of Boyle's law, Charles law and pressure law.
The equation is given as
[tex]\begin{gathered} v_1=34mL \\ T_1=21^0C=21+273.15=294.15K \\ P_1=0.953\text{atm} \end{gathered}[/tex]At STP (standard temperature and pressure), the temperature and pressure changes to
[tex]\begin{gathered} p_2=1\text{atm} \\ t_2=273.15K_{} \end{gathered}[/tex]Now, combining the data above, we can find the change in volume (v₂)
[tex]\begin{gathered} \frac{p_1v_1}{t_1}=\frac{p_2v_2}{t_2} \\ v_2=\frac{p_1v_1t_2}{p_2t_1} \\ \end{gathered}[/tex]Let's substitute the values into the equation above and solve for v₂
[tex]\begin{gathered} v_2=\frac{p_1v_1t_2}{p_2t_1} \\ v_2=\frac{0.953\times34\times273.15}{1\times294.15} \\ v_2=30.1mL \end{gathered}[/tex]From the calculations above, the volume at STP is 30.1mL
Three diagrams shown here represent solutions and one does not. Which two diagrams represent solutions in which the solutes are nonelectrolytes
Chemistry => Solution =>Solution
A solution is defined as a homogeneous mixture of two or more dissolved substances, that is, when mixing two substances there will be only one visible phase.
The first figure is not a solution since it is observed that the molecules are precipitated at the bottom, so they will be as solid particles.
Therefore, we will have to:
Electrolytes are characterized by the fact that they separate into ions in solution.
Circled images represent solutions with no electrolytes.
What is the difference between filtering and sieving?
Answer:
Explanations:
What is Filtering?Filtering is a mechanical separation method. It is a way of separating a solid from a liquid. The solids are mostly undissolved particles or suspended meechanical separa
Hi can you help me find the balanced equation for this. _CuSO4+_NaOH->_Cu(OH)2+_Na2SO4
Balance the equation:
CuSO₄ + Na(OH) -----> Cu(OH)₂ + Na₂SO₄
We will start counting the amount of each element that we have on both sides of the equation. But, to make it easier, we will consider the OH⁻ ion as a whole, and the SO₄²⁻ as a whole. So:
__ CuSO₄ + ___ Na(OH) -----> ___ Cu(OH)₂ + ___ Na₂SO₄
Na: 1 Na: 2
OH: 1 OH: 2
Cu: 1 Cu: 1
SO₄: 1 SO₄: 1
Now, we will start changing the coefficients to balance the equation. The first one is Na, and we have 2 atoms on the right side and only one on the left. That tells us that we have to change the coefficient that is in front of NaOH, and write a 2 there. Let's do that:
__ CuSO₄ + 2 Na(OH) -----> ___ Cu(OH)₂ + ___ Na₂SO₄
Na: 2 Na: 2
OH: 2 OH: 2
Cu: 1 Cu: 1
SO₄: 1 SO₄: 1
After we change that coefficient we count again and now we have two atoms of Na on the left, so it is balanced. The OH wasn't balanced but when we changed that coefficient we balanced it. And Cu and SO4 were already balanced. The equation is balanced now.
The answer to our problem is:
CuSO₄ + 2 Na(OH) -----> Cu(OH)₂ + Na₂SO₄
Convert 1.14 moles to grams .This is the mass of acetic acid in a 15.00ml sample
Answer:
68.4 g of CH3COOH.
Explanation:
Acetic acid has the formula CH3COOH. To convert from moles to grams or vice versa, we have to know the molar mass of the given compound.
If we go to see the periodic table, you can see that the molar mass of carbon (C) is 12 g/mol, of hydrogen (H) is 1 g/mol, and oxygen (O) is 16 g/mol.
To calculate the molar mass we have to do an algebraic sum with the number of atoms that we have in the formula.
In this case, you can note that from CH3COOH, we have 2 carbons, 4 hydrogens, and 2 oxygens. The calculation will look like this:
[tex]\begin{gathered} Molar\text{ mass CH}_3COOH=2\cdot12\frac{g}{mol}+4\cdot1\frac{g}{mol}+2\cdot16\frac{g}{mol}, \\ \\ Molar\text{ mass CH}_3COOH=24\frac{g}{mol}+4\frac{g}{mol}+32\frac{g}{mol}, \\ \\ Molar\text{ mass CH}_3COOH=60\frac{g}{mol}. \end{gathered}[/tex]The molar mass of CH3COOH is 60 g/mol. This is telling us that we have 60 g of CH3COOH in 1 mol of CH3COOH.
Now let's convert 1.14 moles of CH3COOH to grams using its molar mass, as follows:
[tex]1.14\text{ moles CH}_3COOH\cdot\frac{60\text{ g CH}_3COOH}{1\text{ mol CH}_3COOH}=68.4\text{ g CH}_3COOH.[/tex]The answer would be 68.4 g of CH3COOH.
Identify the products and reactants in the following chemical reaction:6 CO2 + 6 H20 > C6H12o6 + 602
Answer: considering the reaction as provided by the question, we can say that CO2 and H2O would be the reactants (in this direction) and C6H12O6 and O2 would be the products
Explanation:
The question requires us to identify the reactants and products in the provided chemical reaction:
[tex]6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2[/tex]The reactants can be defined as the compounds that take part in a reaction and suffer a change through the course of the reaction. On the other hand, the products of a reaction can be defined as the compounds that are formed when the reaction takes place.
An easy way to identify the reactants and products is observing the "arrow" that indicates the direction of a reaction. For example, in the following generic reaction
[tex]A+B\rightarrow C+D[/tex]A and B (which are before the arrow) are the reactants, while C and D (after the arrow) are the products.
The inverse reaction would be:
[tex]C+D\rightarrow A+B[/tex]where C and D are the reactants and A and B are the products.
Therefore, considering the reaction as provided by the question, we can say that CO2 and H2O would be the reactants (in this direction) and C6H12O6 and O2 would be the products.
How to answer number 3?What is the pressure of a 8.5 L cylinder filled with 23.5 g of nitrogen gas at a temperature of 298 K?
To solve this problem we will use the ideal gas law. The formula is:
P * V = n * R * T
Where P is the pressure of the gas, V is the volume, n is the number of moles of that gas, R is the ideal gas constant and T is the temperature.
We know some of those values.
V = 8.5 L
T = 298 K
R = 0.082 atm*L/(mol*K)
Then, P is our unknown and we can find the number of moles from the mass of nitogren using its molar mass.
P = ?
mass of N₂ = 23.5 g
atomic mass of N = 14.01 amu
molar mass of N₂ = 2 * 14.01
molar mass of N₂ = 28.02 g/mol
n = number of moles of N₂ = 23.5 g/(28.02 g/mol)
n = 0.839 moles
Now we can replace the values in the formula and solve it for P.
P * V = n * R * T
P = n * R * T / V
P = 0.839 moles * 0.082 atm*L/(mol*K) * 298 K/(8.5 L)
P = 2.41 atm
Answer: the pressure is 2.41 atm.
The percent composition of the compound formed when 9.17 grams iron combine with 3.94 grams oxygen is
To solve this question, we need to use the following formula:
% by mass = (mass of a component/total mass) x 100%
So first, we need to calculate the total mass of the compound:
9.17 + 3.94 = 13.11
Now we can calculate the percent composition of Iron and of Oxygen.
Iron:
% by mass = (mass of Iron/total mass) x 100%
% by mass = (9.17/13.11) x 100%
% by mass = 69.9%
Oxygen:
% by mass = (mass of Oxygen/total mass) x 100%
% by mass = (3.94/13.11) x 100%
% by mass = 30.0%
Answer: 69.9% of Iron and 30.0% of Oxygen
How Many electrons have both n=2 and ms=+ 1/2?
Answer
8 electrons for n = 2
4 electrons for ms = +1/2
Explanation
The principal quantum number, n = 2 implies the electrons are in second energy level. There can be a maximum of 8 electrons for n = 2. The 2s subshell can hold 2 electrons, and the 2p can hold 6 electrons.
Since there are maximum of 8 electrons for n = 2, then maximum of 4 electrons will spin-up (ms = +1/2), and maximum of 4 electrons will spin-down (ms = -1/2)
Is 14 rightThe mg mass are 0.33 and after is 0.49
Answer:
0.55g of MgO will be formed.
Explanation:
1st) From the balanced equation we know that with 2 moles of Mg, 2 moles of MgO are formed. It is necessary to use the molar mass of Mg and MgO to convert moles to grams:
- Mg molar mass: 24.3g/mol
- Mg conversion:
[tex]2moles\frac{24.3g}{1mole}=48.6g[/tex]- MgO molar mass: 40.3g/mol
- MgO conversion:
[tex]2moles*\frac{40.3g}{1mole}=80.6g[/tex]Now we know that with 48.6g og Mg, 80.6g of MgO are formed.
2nd) Finally, with the calculated grams from the stoichiometry of the reaction, and the starting mass of Mg (0.33g) we can calculate the grams of MgO that will be formed:
[tex]\begin{gathered} 48.6gMg-80.6gMgO \\ 0.33gMg-x=\frac{0.33gMg*80.6gMgO}{48.6gMg} \\ x=0.55gMgO \end{gathered}[/tex]So, 0.55g of MgO will be formed.
DA
THIS EQUATION IS BALANCED
O2 + 2H2 → 2H₂O
O True
O False is it true or false
Answer:
TRUE
Explanation:
If we distribute the 2 on the right we will see there are 4 moles of hydrogen and 2 moles of oxygen and on the left it’s plain to see there are 2 moles of oxygen and then we distribute the 2 to see there are 4 moles of hydrogen
Hopes this help please mark brainliest have a nice day
2Li + Cl₂ → 2 LiCl Did lithium undergo oxidation or reduction?Did chlorine undergo oxidation of reduction?
Explanation:
According to the next equation,
2Li + Cl₂ → 2 LiCl (completed and balanced)
-------
The oxidation state of each element changes as follows:
Li) goes from 0 to +1
[tex]2Li^0=>\text{ Li}^{+1}+\text{ 2e}^{-1}[/tex]Cl) goes from 0 to -1
[tex]2Cl^0+2e^{-1}=>2Cl^{-1}[/tex](these equations above are the half-reactions)
Therefore,
Answer:
Lithium undergo oxidation
Chlorine undergo reduction
metal oxide MO2 reacts with excess HCL to produce chlorine gas at STP as given by the following unbalanced equation:MO2 (s) HCl (aq) -> MCl2 (aq) + Cl2 (g) + H20(l)0.20g of MO2 was added to 25mL of 0.15M HCL solution. (Relative molecular mass for MO2 is 87 g/mol)(a) determine the limiting reactant(b) calculate the mass of MCl2 produced in the reaction(c) calculate the percentage yield if the actual mass of MCl2 produced is 0.078g.
a) Step 1 - We need to balance the equation:
MO2 (s) 4 HCl (aq) -> MCl2 (aq) + Cl2 (g) + 2 H2O(l)
Reactant side:
M - 1
O - 2
H - 4
Cl - 4
Product side
M - 1
O - 2
H - 4
Cl - 4
Step 2 - We need to transform 0.20 grams of MO2 into moles using the following formula: mole = mass/molar mass
mass = 0.20 g
molar mass = 87 g/mol
mole = 0.002299
0.002299 moles of MO2
Step 3 - let's transform mL into L and then transform 0.15 M = 0.15 mol/L into moles.
1000 mL = 1 L
25 mL = 0.025 L
0.15 moles --- 1 L
x --- 0.025 L
x = 0.00375 moles
0.00375 moles of HCl
Step 4 - Let's see the proportion of the equation and compare with the real value of moles
1 mole of MO2 reacts with 4 moles of HCl
So:
1 MO2 --- 4 HCl
0.002299 MO2 --- x
x = 0.00919 moles of HCl
1 MO2 --- 4 HCl
x MO2 --- 0.00375
x = 0.0009375 moles of MO2
We should have 0.00919 moles of HCl reacting with 0.002299 moles of MO2, but we have just 0.00375. It means that the limiting reactant is HCl.
a) HCl is the limiting reactant.
b) Step 1 - We first need to find the atomic mass of M. For this, we know that molar mass of O is 16 and the molar mass of MO2 is 87.
So:
M + (2x16) = 87
M = 87 - 32
M = 55 g/mol
molar mass of M is 55 g/mol.
Step 2 - Now we need to find the molar mass of MCl2:
(1x55) + (35.45 x 2) = 125.9 g/mol
Step 3 - Now we use the equation proportion to find the quantity in moles of MCl2 produced:
1 mole of MO2 produces 1 mole of MCl2
0.0009375 moles of MO2 produces 0.0009375 moles of MCl2
Step 4 - Transform moles of MCl2 into grams:
125.9 g --- 1 mol
x --- 0.0009375 moles of MCl2
x = 0.118 grams
b) Mass of MCl2 produced is 0.118 grams.
c) To calculate the percentage yield we use the following formula:
Percent yield = (actual yield / theoretical yield) x 100%
actual yield = 0.078
theoretical yield = 0.118
Percent yield = (0.078 / 0.118) x 100%
c) Percent yield = 66%
Answer:
a) HCl is the limiting reactant.
b) Mass of MCl2 produced is 0.118 grams.
c) Percent yield = 66%
The future of Delton’s Atomic theory that states: atoms are indivisible; they cannot be created or destroyed” is considered valid today A)TrueB) False
Dalton's atomic model was one of the firsts that tried to define what is an atom and what is its composition, one of the statements was "atoms are indivisible". Nowadays, with modern technology we can see that atoms have subparticles, and can be splitted into protons, neutrons, electrons, and even these subparticles will have smalled fundamental particles inside of it. Therefore this is statement is false
Perform the followingmathematical operation, andreport the answer to theappropriate number ofsignificant figures.5.4 – 4.21 = [?]=Enter
For addition or subtraction operations, we must first round off the values of significant figures in order to make them to the same number of decimal places:
Let's round 4.21 to 4.2.
Now we do the operation:
5.4 - 4.2 = 1.2
After the calculations, we choose as a reference the number with the fewest decimal places. For subtraction operations, we must follow the same reasoning as for addition.
Answer: 1.2
If I add water to 10mL of a 0.15M NaOH solution until the final volume is150mL, what will the molarity of the diluted solution be?
To solve this problem we need to remember the equation for dilution of solutions:
[tex]C_1V_1=C_2V_2[/tex]Where C1 is the concentration of the first solution, V1 is the volumen of the first solution, C2 is the concentration of second solution and V2 is the concentration of the second solution.
We know from the text that C1=0.15M and V2 is 150ml, and they tell us that V2 is obtained by adding V1 10ml which we can express as the following expression:
[tex]V_2=V_{1\text{ }}+10ml[/tex]Solving for V1 and using the data provided:
[tex]V_1=V_2-10ml=150ml-10ml=140ml[/tex]Now we have all de data for solving the first equation, we have to solve for C2 and substitute:
[tex]C_2=\frac{C_1V_1}{V_2}=\frac{0.15\text{ M}\cdot140ml}{150ml}=0.14M[/tex]Final concentration is 0.14M
Equilibrium Learning Check #1Sulfuric acid is an important industrial chemical that is usually produced by a series of reactions. One of these involves and equilibrium between gaseous sulfur dioxide, oxygen and sulfur trioxide. 2SO2(g) + O2(g) ↔ 2SO3(g) If 2.5 mol of sulfur dioxide gas and 2.0 mol of oxygen gas are placed in a sealed 1.0L container and allowed to reach equilibrium, 0.75 mol of sulfur dioxide remains at equilibrium. Use an ICE table to determine the concentration of the other gases at equilibrium.
answer and explanation
below we show an ICE table
now that we have the mols at equilibrium we can calculate the concentration for the gases
for SO₂
concentration = 0.75mol/1L
= 0.75M
for O₂ gms
concentration = 1.125mol/1L
= 1.125M
for SO₃
concentration = 1.75mol/1L
=1.75M
What is the pH of a saturated solution of silver hydroxide, given Ksp = 1.55×10^-8?
Answer:
[tex]pH\text{ = 10.1}[/tex]Explanation:
We start off by writing the ionization equation
We have this as follows:
[tex]AgOH\text{ }\rightarrow\text{ Ag}^+\text{ + OH}^-[/tex]Let us have the concentration of the silver and hydroxide ions as x M
Thus, we have this as:
[tex]\begin{gathered} K_{sp}\text{ = \lbrack Ag}^+]\text{ + \lbrack OH}^-] \\ 1.55\text{ }\times\text{ 10}^{-8}\text{ = x}\times x \\ x^2\text{ = 1.55 }\times\text{ 10}^{-8} \\ x\text{ = }\sqrt{(1.55\text{ }\times10^{-8})} \\ x\text{ = 0.0001245} \end{gathered}[/tex]Thus, we have the value of x as 0.0001245
Mathematically:
[tex]\begin{gathered} pOH\text{ = -log\lbrack x\rbrack} \\ pOH\text{ = -log \lparen0.0001245\rparen} \\ pOH\text{ = 3.90} \end{gathered}[/tex]However:
[tex]\begin{gathered} pH\text{ = 14 - pOH} \\ pH\text{ = 14-3.90} \\ pH\text{ = 10.1} \end{gathered}[/tex]What is the pH of a 0.050 M Ba(OH)2 aqueous solution?
First, let's see that we can extract the concentration of OH- because Ba(OH)2 is a base. Let's see the dissociation of this base:
[tex]Ba(OH)_2\to Ba^{2+}+2OH^-,[/tex]You can realize that we have 2 moles of OH-. The next step is to multiply this number of moles by the concentration (0.050 M):
[tex]\lbrack OH^-\rbrack=2\cdot0.050=0.1.[/tex]Remember that the formula of pOH is -log ( [OH-] ):
[tex]\text{pOH}=-\log (0.1)=1.[/tex]And with this result, we can find pH, using the formula:
[tex]pH+\text{pOH}=14[/tex]And we're going to obtain:
[tex]\begin{gathered} pH=14-\text{pOH}, \\ pH=14-1, \\ pH=13. \end{gathered}[/tex]The pH of the solution would be 13, so the answer is (2).