Answer:
[tex]\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}[/tex]
Explanation:
1. Concentration of SO₄²⁻
SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷
0.0150 x
[tex]K_{sp} =\text{[Sr$^{2+}$][SO$_{4}^{2-}$]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}[/tex]
2. Concentration of Pb²⁺
PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸
x 2.293 × 10⁻⁵
[tex]K_{sp} =\text{[Pb$^{2+}$][SO$_{4}^{2-}$]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb$^{2+}$ is $\large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}}$}[/tex]
Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.
Answer:
There are no forces holding the gas particles together.
Explanation:
when the temperature of an ideal gas is increased from 27C to 927C then kinetic energy increases by
Answer:
The rms speed of its molecules becomes. (T) has become four times. Therefore, v_(rms) will become two times,...
The following reaction, catalyzed by iridium, is endothermic at 700 K: CaO(s) + CH4(g) + 2H2O (g) → CaCO3 (s) + 4H2 (g) For the reaction mixture above at equilibrium at 700 K, how would the following changes affect the total quantity of CaCO3 in the reaction mixture once equilibrium is re-established?
a. Increasing the temperature
b. Adding calcium oxide (CaO)
c. Removing methane (CH4)
d. Increasing the total volume
e. Adding iridium
Answer:
A. Increasing the temperature will favor forward reaction and more CaCo3 formed.
B. More CaCo3 will be formed.
C. CaCo3 will decrease and more react ants formed.
D. Less CaCo3 will be formed.
E. Iridium is a catalyst so there is no effect
Explanation:
A. Temperature will increase because it's an endothermic reaction.
B. Adding Cao will favor forward reaction and more CaCo3 formed.
C. Removing methane, more react ants are formed and CaCo3 decreases.
D. Irridi is a catalyst so it has no effect on the CaCo3 but only speeds its rate of reaction.
When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are: potassium hydrogen sulfate (aq) potassium hydroxide (aq) potassium sulfate (aq) water (l)\
Answer:
Explanation:
Answer:
1, 1, 1, 1
Explanation:
potassium hydrogen sulfate + potassium hydroxide ⟶ potassium sulfate + water(l)
KHSO₄ + KOH ⟶ K₂SO₄ + H₂O
1. Put a 1 in front of the most complicated-looking formula (K₂SO₄?):
KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
2. Balance S:
We have fixed 1 S on the right. We need 1 S on the left. Put a 1 in front of KHSO₄ to fix it.
1KHSO₄ + KOH ⟶ 1K₂SO₄ + H₂O
3. Balance K:
We have fixed 2 K on the right and 1 K on the left. We need 1 more K on the left. Put a 1 in front of KOH.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + H₂O
4. Balance O
We have fixed 4 O on the right and 5 O on the left. We need 1 more O on the right. Put a 1 in front of H₂O.
1KHSO₄ + 1KOH ⟶ 1K₂SO₄ + 1H₂O
Every formula has a coefficient. The equation should be balanced.
5. Check that atoms balance:
[tex]\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{K} & 2 &2\\\text{H} & 2 & 2\\\text{S} & 1 & 1\\\text{O}&5&5\\\end{array}[/tex]
It checks.
The coefficients are 1, 1, 1, 1.
Find the number of ibuprofen molecules in a tablet containing 210.0 mg of ibuprofen (C13H18O2).
Answer:
the answer is 5.83x1020 molecules
Explanation:
I'd really appreciate a brainleast
"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
Answer:
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
Explanation:
The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical
First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.
pKa NH₃/NH₄⁺pKb = - log Kb
pKb = -log 1.8x10⁻⁵ = 4.74
pKa = 14 - pKb
pKa = 14 - 4.74
pKa = 9.26
Moles NH₃2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃
H-H equation:pH = pKa + log [NH₃] / [NH₄Cl]
8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]
-1.06 = log [0.400 moles] / [NH₄Cl]
0.0087 = [0.400 moles] / [NH₄Cl]
[NH₄Cl] = 0.400 moles / 0.0087
[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):
4.59 moles NH₄Cl ₓ (53.491g / mol) =
245.66g of NH₄Cl is the mass we need to add to obtain the desire pH
An element has 6 protons and 6 neutrons. What is the atomic number? What is its mass number?
Answer:
The atomic is 6, the mass number is 12
Explanation:
The element being described is Carbon.
The number of protons in the nucleus of an atom is called its atomic number, Z.
The number of protons pluss the number of neutronsin the nucleus of an atom gives the atom's mass number, A.
Which of the following is a salt that will form from the combination of a strong base with a weak acid?
Select the correct answer below:
A. NaHCO3
B. H2O
C. CH3CO2H
D. NH4Cl
Answer:
A. NaHCO₃
Explanation:
NaHCO₃ ⇒ NaOH + H₂CO₃
NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.
Hope that helps.
1. Explain the test for unsaturation.
2. Write down the balanced chemical equations for the complete and incomplete
combustion of octene
3. Explain how propanol, an alcohol, is formed from propene..
4. How is margarine formed?
Answer:
1)In organic chemistry, the bromine test is a qualitative test for the presence of unsaturation (carbon-to-carbon double or triple bonds), phenols and anilines. ... The more unsaturated an unknown is, the more bromine it reacts with, and the less coloured the solution will appear.
2)The equation for incomplete combustion of propane is: 2 C3H8 + 9 O2 → 4 CO2 + 2 CO + 8 H2O + Heat. If not enough oxygen is present for complete combustion, incomplete combustion occurs. The result of incomplete combustion is, once again, water vapour, carbon dioxide and heat. But it also produces carbon monoxide.
Explanation:
3)Propene, also known as propylene, is an unsaturated organic compound with the chemical formula {\displaystyle {\ce {CH3CH=CH2}}}. It has one double bond, and is the second simplest member of the alkene class of hydrocarbons. It is a colorless gas with a faint petroleum-like odor.
Formula: C3H6
IUPAC ID: Propene
4)Margarines are chemically created during hydrogenation which, until January 1, 2006, relied upon trans fats to solidify their vegetable oils. Food companies have been exploring options for replacing trans fat in partially hydrogenated margarine.
Which response includes all the following processes that are accompanied by an increase in entropy? 1) 2SO 2(g) + O 2(g) → SO 3(g) 2) H 2O(l) → H 2O(s) 3) Br 2(l) → Br 2(g) 4) H 2O 2(l) → H 2O(l) + 1/ 2O 2(g)
Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]
3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.
2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]
1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.
3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]
1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.
4) [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]
1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.
To find the pH of a solution of NaNO2, one would have to construct an ICE chart using:
a. the Kb of NO−2 to find the hydroxide concentration.
b. the Kb of HNO2 to find the hydronium concentration.
c. the Kb of NO-2, to find the hydronium concentration.
d. the Kb of HNO2, to find the hydroxide concentration.
Answer:
a. the Kb of NO₂⁻ to find the hydroxide concentration.
Explanation:
When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.
NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)
Na⁺ comes from NaOH (strong base) so it doesn't react with water.
NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.
NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)
This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to find the hydroxide concentration.
Three bonding pairs around a central atom results in a ________. A. octahedral B. trigonal planar compound C. tetrahedral compound D. linear compound
Answer:
B
Explanation:
because it is a trigonal planar compound
Three bonding pairs around a central atom results in a trigonal planar compound. Option B is correct.
In a trigonal planar arrangement, the central atom is surrounded by three bonding pairs of electrons, forming a flat, triangular shape. The bond angles between the bonding pairs are approximately 120 degrees.
This molecular geometry is observed when a molecule has a central atom with three bonded pairs and no lone pairs. Examples of compounds with trigonal planar geometry include boron trifluoride (BF₃) as well as formaldehyde (H₂CO).
The other options are not correct for a molecule with three bonding pairs;
Octahedral refers to a molecular geometry with six bonding pairs around a central atom.
Tetrahedral corresponds to a molecular geometry with four bonding pairs around a central atom.
Linear represents a molecular geometry with two bonding pairs around a central atom.
Hence, B. is the correct option.
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A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g, the cup floats in the seawater around his boat with the fresh water inside the cup at exactly the same level as the seawater outside the cup (see sketch at right).
Calculate the amount of salt dissolved in each liter of seawater. Be sure your answer has a unit symbol, if needed, and round it to 2 significant digits.
You'll need to know that the density of fresh water at the temperature of the sea around the sailor is 0.999/gcm3. You'll also want to remember Archimedes' Principle, that objects float when they displace a mass of water equal to their own mass.
Answer:
Amount of salt in 1 L seawater = 34 g
Explanation:
According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater
mass of freshwater = density * volume
1 cm³ = 1 mL
mass of freshwater = 0.999 g/cm³ * 735 cm³ = 734.265 g
mass of freshwater + cup = 734.265 + 25 = 759.265 g
Therefore, mass of equal volume of seawater = 759.265 g
Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)
1 liter = 1000 cm³ = 1000 mL;
Density of seawater = mass / volume
Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L
Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L
mass of 1 Liter seawater = 1033.01 g
mass of 1 Liter freshwater = 999 g
mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g
Therefore, amount of salt in 1 L seawater = 34 g
What is pH of a buffer made by combining 45.0mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate
Answer:
3.11
Explanation:
Any buffer system can be described with the reaction:
[tex]HA~->~H^+~+~A^-[/tex]
Where is the acid and is the base. Additionally, the calculation of the pH of any buffer system can be made with the Henderson-Hasselbach equation:
[tex]pH~=~pKa~+~Log(\frac{ [A^-]}{[HA]})[/tex]
With all this in mind, we can write the reaction for our buffer system:
-) Nitrous acid: [tex]HNO_2[/tex]
-) Sodium nitrate: [tex]NaNO_2[/tex]
[tex]HNO_2~->~H^+~+~NO_2^-[/tex]
In this case, the acid is [tex]HNO_2[/tex] with a concentration of 0.150 M and a volume of 45.0 mL (0.045 L). The base is [tex]NO_2^-[/tex] with a concentration of 0.175 M and a volume of 20.0 mL (0.020 L).
We can calculate the moles of each compound is we take into account the molarity equation ([tex]M=\frac{mol}{L}[/tex]). So:
-) moles of [tex]HNO_2[/tex]:
[tex]mol=0.150~M*0.045~L=0.00657[/tex]
-) moles of [tex]NO_2^-[/tex]:
[tex]mol=0.175~M*0.020~L=0.0035[/tex]
The total volume would be:
0.020 L + 0.045 L = 0.065 L
With this in mind, we can calculate the molarity of each compound:
-) Concentration of [tex]HNO_2[/tex]
[tex]M=\frac{0.00657~mol}{0.065~L}=0.101~M[/tex]
-) Concentration of [tex]NO_2^-[/tex]
[tex]M=\frac{0.0035~mol}{0.065~L}=0.0538~M[/tex]
The pKa reported is 3.39, therefore we can plug the values into the Henderson-Hasselbach equation:
[tex]pH~=~3.39~+~Log(\frac{[0.0538~M]}{[0.101~M]})~=~3.11[/tex]
The final pH value would be 3.11
I hope it helps!
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
We have a buffer made by combining 45.0mL of 0.150 M nitrous acid and 20.0mL of 0.175M sodium nitrate.
Nitrous acid is a weak acid and nitrate ion is its conjugate base.
What is a buffer?It is a solution used to resist abrupt changes in pH when acids or bases are added.
Step 1: Calculate the moles of each species.We do so by multiplying the molar concentration by the volume in liters.
HNO₂: 0.150 mol/L × 0.0450 L = 6.75 × 10⁻³ mol
NaNO₂: 0.175 mol/L × 0.0200 L = 3.50 × 10⁻³ mol
Step 2: Calculate the total volume of the mixture.The total volume will be the sum of the volumes of each solution.
V = 45.0 mL + 20.0 mL = 65.0 mL = 0.0650 L
Step 3: Calculate the molar concentration of each species in the mixture.HNO₂: 6.75 × 10⁻³ mol/0.0650 L = 0.104 M
NaNO₂: 3.50 × 10⁻³ mol/0.0650 L = 0.0538 M
Step 4: Calculate the pH of the buffer.We can calculate the pH of a buffer system using Henderson-Hasselbach's equation.
pH = pKa + log [NaNO₂]/[HNO₂]
pH = 3.16 + log 0.0538/0.104 = 2.87
The pH of a buffer made by combining 45.0 mL of 0.150M nitrous acid and 20.0mL of 0.175M sodium nitrate is 2.87.
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What electron configuration represents Nitrogen? A. 2,2 B. 2,8,4 C. 2,4 D. 2,5
Answer:
D. 2:5
Explanation:
It has 5 valency electrons
[tex].[/tex]
In the given question, the electronic configuration of nitrogen is 2,5. The correct answer is option D.
The electronic configuration of an atom describes the arrangement of its electrons in different energy levels or orbitals.
Nitrogen has an atomic number of 7, which means it has 7 electrons.
The first energy level or shell can hold up to 2 electrons, and the second can hold up to 8 electrons. Nitrogen has 2 electrons in its first energy level and 5 electrons in its second energy level.Therefore, the electronic configuration of nitrogen is 2,5, which means it has 2 electrons in its first energy level and 5 electrons in its second energy level. Option D is the correct answer.
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If 1 mol of a pure triglyceride is hydrolyzed to give 2 mol of RCOOH, 1 mol of R'COOH, and 1 mol of glycerol, which of the following compounds might be the triglyceride?
CHOC(O)R
A. CHOC(O)R
CHOC(O)R
CH,OC(O)R
B. CHOC(O)R
CH2OC(O)R
CHOC(O)R
C. CHOC(O)R
CHOC(O)R
CHOC(O)R
D. CHOC(O)R
CHOC(O)R
Answer:
The correct option is C.
Note the full question and structure of the moleculesis found in the attachment below.
Explanation:
Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.
The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.
During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.
From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.
The correct option therefore, is C
alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.
A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
Answer:
The pH of this solution = 5.06
Explanation:
Given that:
number of moles of CH3COOH = 0.100 mol
volume of the buffer solution = 1.0 L
number of moles of NaC2H3O2 = 0.100 mol
The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
we know that concentration in mole = Molarity/volume
Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex] = 0.10 M
The chemical equation for this reaction is :
[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]
The conjugate base is CH3COO⁻
The concentration of the conjugate base [CH3COO⁻] is = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]
= 0.10 M
where the pka (acid dissociation constant)for CH3COOH = 4.74
If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M
The ICE Table for the above reaction can be constructed as follows:
[tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]
Initial 0.10 0.035 0.10 -
Change -0.035 -0.035 + 0.035 -
Equilibrium 0.065 0 0.135 -
By using Henderson-Hasselbalch equation:
The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]
The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]
The pH of this solution = 4.74 + log (2.076923077 )
The pH of this solution = 4.74 + 0.3174
The pH of this solution = 5.0574
The pH of this solution = 5.06 to two decimal places
Which of the following happens to a molecule of an object when the object is heated? (1 point)
Answer:
They get more energy, so they vibrate!
Explanation:
Which of the following is an example of a physical, rather than a chemical, change?
Answer: The question is not clear
Explanation:
The gas with an initial volume of 24.0 L at a pressure of 565 torr is compressed until the volume is 16.0 L. What is the final pressure of the gas, assuming the temperature and amount of gas does not change
Answer:
848 torr
Explanation:
The only variables are the pressure and the volume, so we can use Boyle's Law.
p₁V₁ = p₂V₂
Data:
p₁ = 565 torr; V₁ = 24.0 L
p₂ = ?; V₂ = 16.0 L
Calculations:
[tex]\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{565 torr} \times \text{24.0 L} & = & p_{2} \times \text{16.0 L}\\\text{13 560 torr} & = & 16.0p_{2}\\p_{2} & = & \dfrac{\text{13 560 torr}}{16.0}\\\\& = &\textbf{848 torr}\\\end{array}\\\text{The final pressure of the gas is $\large \boxed{\textbf{848 torr}}$}[/tex]
When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+O2(g)⟶2H2O(l)
A) 11.7%
B) 17.0%
C) 38.9%
D) 65.7%
Answer:
D) 65.7%
Explanation:
Based on the reaction:
2H2(g)+O2(g)⟶2H2O(l)
2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.
To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.
Theoretical yield:
Moles of 5.58g H₂:
5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂
As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:
2.768 moles H₂O ₓ (18.015g / mol) =
49.86g H₂O is theoretical yield
Percent yield:
Percent yield = Actual yield / Theoretical yield ₓ 100
32.8g H₂O / 49.86g ₓ 100 =
65.7% is percent yield of the reaction
D) 65.7%If the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?
Answer:
14.297 g
Explanation:
From the question;
1 mo of the compound requires 1320.0 kJ
From the molar mass;
1 ml of the compound weighs 30.55g
How many grams requires 617.30kJ?
1 ml = 1320
x mol = 617.30
x = 617.30 / 1320
x = 0.468 mol
But 1 mol = 30.55
0.468 mol = x
x = 14.297 g
Plz help!!!! Solve this by using factor labeling
Answer:
the answer is 2,000 nickels.
Explanation:
we multiplied 100 by 100, because there are 100 cents in a dollar, and we divided 10,000 by 5, because there are 5 cents in a nickel.
What type of bonding is occuring in the compound below?
A. Covalent polar
B. Metallic
C. Ionic
D. Covalent nonpolar
Answer:
(B). it's metallic bonding
When the equation MnO₄⁻ + I⁻ + H₂O → MnO₂ + IO₃⁻ is balanced in basic solution, what is the smallest whole-number coefficient for OH⁻?
Answer:
The smallest whole-number coefficient for OH⁻ is 2
Explanation:
Step 1: The equation redox reaction is divided into two half equations
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
Step 5 : addition of the two half equations to yield a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole number coefficient for OH⁻ is 2
A redox reaction is divided into two half equations which are shown below:
Reduction half equation: MnO₄⁻ ----> MnO₂
Oxidation half-equation: I⁻ ---> IO₃⁻
Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;
MnO₄⁻ + 2H₂O ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O
The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation
MnO₄⁻ + 2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.
2MnO₄⁻ + 4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻
I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻
The two half equations are then added and written together to form a net ionic equation
2MnO₄⁻ + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻
The smallest whole-number coefficient for OH⁻ is therefore 2.
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a) Define typical polyfunctional acid ?
b) Show the equations of dissociation mechanism of phosphoric acid as an example.
c) Write the equation for calculating the [H3O*].
a) A polyfunctional acid is an acid that has more than one functional group.
b) The equations of dissociation of phosphoric acid are:
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺ HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺c) The equation for calculating the concentration of H₃O⁺ is [tex] [H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3} [/tex]
a) A polyfunctional acid can be defined as an acid that has more than one functional group. Phosphoric acid (H₃PO₄) is an example of polyfunctional acid since it is composed of three hydroxyl groups joined to a phosphorus atom, which is also joined to an oxygen atom by a double bound. In that structure, the three hydrogen atoms of the hydroxyl groups give the acidic behavior to this compound.
b) Phosphoric acid has three equations of dissociation:
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ (1)H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺ (2)HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺ (3)The dissociation constants for the three above equations are:
[tex] K_{1} = \frac{[H_{2}PO_{4}^{-}][H_{3}O^{+}]}{[H_{3}PO_{4}]} [/tex] (4)
[tex] K_{2} = \frac{[HPO_{4}^{2-}][H_{3}O^{+}]}{[H_{2}PO_{4}^{-}]} [/tex] (5)
[tex] K_{3} = \frac{[PO_{4}^{3-}][H_{3}O^{+}]}{[HPO_{4}^{2-}]} [/tex] (6)
c) We can calculate the concentration of H₃O⁺ for each equilibrium with the equations (4), (5), and (6).
The general reaction of dissociation of phosphoric acid is given by the sum of equations (1), (2), and (3):
H₃PO₄ + 3H₂O ⇄ PO₄³⁻ + 3H₃O⁺ (7)
The concentration of H₃O⁺ for the total dissociation reaction (eq 7) can be found as follows:
[tex] K_{t} = \frac{[PO_{4}^{-3}][H_{3}O^{+}]^{3}}{[H_{3}PO_{4}]} [/tex] (8)
Where:
[tex] K_{t} = K_{1}*K_{2}*K_{3} [/tex]
Hence, by knowing the dissociation constants K₁, K₂ and K₃, and the concentrations of PO₄³⁻ and H₃PO₄, the [H₃O⁺] is:
[tex][H_{3}O^{+}] = (\frac{K_{1}K_{2}K_{3}[H_{3}PO_{4}]}{[PO_{4}^{-3}]})^{1/3}[/tex]
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Consider the reaction: C(s) + O2(g)CO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) C(s) + 1/2 O2(g) CO(g) Ka b.) CO(g) + 1/2 O2(g) CO2(g) Kb
Answer:
A. Ka = [CO2] / [C] [O2]^1/2
B. Kb = [CO2] / [CO] [O2]^1/2
Explanation:
Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:
A. Determination of the expression for equilibrium constant Ka.
This is illustrated below:
C(s) + 1/2 O2(g) <==> CO(g)
Ka = [CO2] / [C] [O2]^1/2
B. Determination of the expression for equilibrium constant Kb.
This is illustrated below:
CO(g) + 1/2 O2(g) <==> CO2(g)
Kb = [CO2] / [CO] [O2]^1/2
Which molecule is NOT hypervalent?
Select the correct answer below:
SF
PBr3
PBr5
XeFo
Answer:
PBr3 is NOT hypervalent
Explanation:
The molecule that is not hypervalent is PBr3
A molecule can be defined as the smallest part of a substance that can exist independently.
It is formed by the chemical combination of two or more atoms.
A molecule is said to be hypervalent when more than four pairs of electrons are around the central atom.
A molecule is said to be hypovalent when less than four pairs of electrons are around the central atom.
From the question, the molecule that is hypovalent is PBr3
This is because, phosphorus can make hypervalent compounds, but in this specific example it is sharing three bonds and has one lone pair, so it has simply a full octet.
Therefore, the molecule that is not hypervalent is PBr3.
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When titrating a strong acid with a strong base, after the equivalence point is reached, the pH will be determined exclusively by: Select the correct answer below:
A) hydronium concentration
B) hydroxide concentration
C) conjugate base concentration
D) conjugate acid concentration
Answer:
B) hydroxide concentration
Explanation:
Hello,
In this case, since we are talking about strong both base and acid, since the base is the titrant and the acid the analyte, once the equivalence point has been reached, some additional base could be added before the experimenter realizes about it, therefore, since the titrant is a strong base, it completely dissociates in hydroxide ions and metallic ions which allows us to compute the pOH of the solution by known the hydroxide ions concentration.
After that, due to the fact that the pH is related with the pOH as shown below:
pH=14-pOH
We can directly compute the pH.
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What is the balanced equation for the reaction of aqueous cesium sulfate and aqueous barium perchlorate?
Answer:
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
Explanation:
When aqueous cesium sulfate and aqueous barium perchlorate are mixed together it gives white precipitate barium sulfate and aqueous solution od cesium perchlorate.
The balanced chemical reaction is given as:
[tex]Cs_2SO_4(aq)+Ba(ClO_4)_2(aq)\rightarrow BaSO_4(s)+2CsClO_4(aq)[/tex]
According to reaction, 1 mole of cesium sulfate reacts with 1 mole of barium perchlorate to give 1 mole of a white precipitate of barium sulfate and 2 moles of cesium perchlorate.