Answer:
a. NH₃(g) + HCl(g) → NH₄Cl(s)
b. HCl is the limiting reagent
c. 7.35 g of NH₄Cl
d. 0.039 moles of NH₃ remains after the reaction is complete (0.663 grams of ammonia)
Explanation:
First of all, we define the reaction where the reactants are the ammonia and the hydrogen chloride. The product is the ammonium chloride.
NH₃(g) + HCl(g) → NH₄Cl(s)
First of all, we need to determine the limiting reactant. As the ratio is 1:1, the compound that has the lowest mol, will be the limiting.
3 g / 17g/mol = 0.176 moles NH₃
5 g / 36.45 g/mol = 0.137 moles HCl (limiting reactant)
For 0.176 moles of ammonia, we need the same of HCl, but we have 0.137 moles. We convert the moles to mass, in order to know how many grams of NH₄Cl are produced → 0.137 mol . 53.45 g /1mol = 7.32 g
As ratio is 1:1, and the limiting reagent is the HCl, after the reaction goes complete, some ammonia remains. This is because, the ammonia is the excess reagent.
0.176 mol - 0.137 moles = 0.039 moles of NH₃ remains after the reaction is complete; we convert the moles to mass → 0.039 mol . 17 g /mol = 0.663 g are left over in the reaction mixture.
Final answer:
The balanced chemical equation for the reaction between ammonia and hydrogen chloride to form ammonium chloride is NH3(g) + HCl(g) → NH4Cl(s). Ammonia is the limiting reagent in the given mixture, and the theoretical yield of ammonium chloride is 9.42 g. No excess reactant will be left over after the reaction.
Explanation:
Chemical Reaction Between Ammonia and Hydrogen Chloride
When ammonia gas (NH3) combines with hydrogen chloride gas (HCl), solid ammonium chloride (NH4Cl) is formed. The balanced chemical equation for this reaction is:
NH3(g) + HCl(g) → NH4Cl(s).
Limiting Reagent and Yield
In a reaction mixture with 3.0 g of ammonia and 5.0 g of hydrogen chloride, to find the limiting reagent we must compare the mole ratio of the reactants to their stoichiometric ratio in the balanced equation. Using molar masses (NH3: 17.03 g/mol, HCl: 36.46 g/mol), we find that:
NH3: 3.0 g × (1 mol / 17.03 g) ≈ 0.176 moles
HCl: 5.0 g × (1 mol / 36.46 g) ≈ 0.137 moles
The stoichiometry of the reaction is 1:1, so ammonia is the
limiting reagent
because we have fewer moles of it relative to HCl.
The mass of ammonium chloride formed is based on the limiting reagent. Since all of the ammonia will react, the theoretical yield of NH4Cl is the moles of NH3 times the molar mass of NH4Cl (53.49 g/mol):
0.176 moles × 53.49 g/mol ≈ 9.42 g of NH4Cl.
Excess Reactant Left Over
To determine the amount of excess reactant left over, we subtract the number of moles of HCl that reacted (equivalent to the moles of NH3) from the initial moles of HCl:
0.137 moles HCl - 0.176 moles NH3 = -0.039 moles HCl (not possible, indicating no excess HCl remains).
Calculate the solubility of CuX (Ksp=[Cu2+][X2−]=1.27×10−36) in a solution that is 0.200 M in NaCN.
I have already tried to square root the Ksp value to get the answer but it was wrong.
The solubility of CuX in a 0.200 M NaCN solution cannot be calculated solely based on the Ksp of CuX, because NaCN forms a complex with Cu2+, significantly affecting its solubility. Additional information on the stability constant of the copper-cyanide complex is needed
To calculate the solubility of CuX in a solution that is 0.200 M in NaCN, we must consider the common ion effect due to the presence of the cyanide ion, CN-. The equation for the solubility product (Ksp) is given by Ksp = [Cu2+][X2−]. Since cyanide ions form a complex with copper ions, the direct precipitation of CuX is suppressed, and we cannot simply take the square root of the Ksp to find its molar solubility.
Instead, we would write the reaction of copper with cyanide: CuX(s) + 4CN−
ightleftharpoons [Cu(CN)4]3− + X2−. This complexation reaction would vastly reduce the concentration of free Cu2+ ions in solution, thereby affecting the solubility of CuX. To find the actual solubility, we would need to know the stability constant (Kf) for the copper-cyanide complex, and then set up an equilibrium calculation that includes both the Ksp of CuX and the Kf of [Cu(CN)4]3−. Since the problem doesn't provide Kf, we would not be able to calculate the solubility without additional information
The presence of NaCN significantly decreases the solubility of CuX due to the common ion effect. Therefore, the solubility of CuX in a solution that is 0.200 M in NaCN remains extremely low, approximately 1.27×10 ⁻¹⁸ M.
To calculate the solubility of CuX in a solution that is 0.200 M in NaCN, we need to consider the common ion effect. When NaCN dissolves, it produces CN⁻ ions, which can interact with Cu²⁺ ions, reducing the solubility of CuX.
Given that the solution is 0.200 M in NaCN, we can assume that the concentration of CN⁻ ions ([CN⁻]) is 0.200 M.
Now, let's denote the solubility of CuX as x M. Since CuX dissociates into Cu²⁺ and X²⁻ ions, the concentration of Cu²⁺ ions ([Cu²⁺]) and X²⁻ ions ([X²⁻]) will both be equal to x M at equilibrium.
The solubility product constant (K sp ) expression for CuX is:
K sp =[Cu²⁺ ][X²⁻ ]
Given that K sp =1.27×10⁻³⁶ , we can substitute the concentrations into the expression:
1.27×10⁻³⁶ =(x)(x)
1.27×10⁻³⁶ =x²
Now, we'll solve this equation for x to find the solubility of CuX. Let's proceed with the calculations.
To solve for x, we take the square root of both sides of the equation:
x= 1.27×10⁻³⁶
x=1.127×10⁻¹⁸
So, the solubility of CuX in a solution that is 0.200 M in NaCN is
1.127×10⁻¹⁸ M.
Calculate the value of the equilibrium constant, K c , for the reaction AgCl ( s ) + Cl − ( aq ) − ⇀ ↽ − AgCl − 2 ( aq ) K c = ? The solubility product constant, K sp , for AgCl is 1.77 × 10 − 10 and the overall formation constant, K f ( β 2 ), for AgCl − 2 is 1.8 × 10 5 .
The equilibrium constant, Kc, for the given reaction can be calculated using the solubility product constant, Ksp, and the overall formation constant, Kf. Substituting the given values, we find that Kc is equal to 3.186 × 10⁻⁵.
Explanation:The equilibrium constant, Kc, for the reaction AgCl (s) + Cl⁻ (aq) ⇌ AgCl⁻₂ (aq) can be calculated using the solubility product constant, Ksp, for AgCl and the overall formation constant, Kf, for AgCl⁻₂.
The equilibrium constant is given by the product of Ksp and Kf:
Kc = Ksp * Kf
Substituting the values given, Ksp = 1.77 × 10⁻¹⁰ and Kf = 1.8 × 10⁵, we can calculate the value of Kc:
Kc = (1.77 × 10⁻¹⁰) * (1.8 × 10⁵) = 3.186 × 10⁻⁵
For this question please enter the number of sigma (σ) and pi (π) bonds (e.g. 0,1,2,3,4, etc). How many sigma and pi bonds, respectively, are in this carboxylic acid? H2CCHCH2COOH. σ bonds and π bond(s). How many sigma and pi bonds, respectively, are in this organic molecule (an amine)? HCCCH2CHCHCH2NH2. σ bonds and π bond(s). How many sigma and pi bonds, respectively, are in this organic molecule (an alcohol)?
Answer:
Explanation:
find the solution below
The organic compounds carboxylic acid, amine, and alcohol contain different numbers of sigma and pi bonds resulting from the overlapping of atomic orbitals. In these provided examples, the carboxylic acid contains nine sigma and two pi bonds, the amine has twelve sigma and two pi bonds, and the alcohol has nine sigma and 0 pi bonds.
Explanation:Sigma
and
pi bonds
occur in various types of organic molecules, including carboxylic acids, amines, and alcohols. A sigma (σ) bond is formed by the head-on overlapping of atomic orbitals, whereas a pi (π) bond is formed by the lateral overlap of two p orbitals. In the carboxylic acid H2CCHCH2COOH, there are nine sigma and two pi bonds. As for the amine HCCCH2CHCHCH2NH2, there are twelve sigma and two pi bonds. Lastly, an alcohol molecule, such as CH3OH, consists of 9 sigma bonds and 0 pi bonds. It's essential to note that a double bond contains one sigma and one pi bond, while a triple bond contains one sigma and two pi bonds.
Learn more about Sigma and Pi Bonds here:https://brainly.com/question/31238729
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Given the equation representing a system at equilibrium:
PCl5(g) + energy ⇌ PCl3(g) + Cl2(g)
Which change will cause the equilibrium to shift to the right?
(1) adding a catalyst (2) adding more PCl3(g)
(3) increasing the pressure (4) increasing the temperature
Answer:
(4) increasing the temperature
Explanation:
PCl₅(g) + energy ⇌ PCl₃(g) + Cl₂(g)
According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.
If you increase the temperature, the position of equilibrium will move to the right to get rid of the added heat.
(1) is wrong. Adding a catalyst does not change the position of equilibrium.
(2) is wrong. If you add more PCl₃, the position of equilibrium will move to the left to get rid of the added PCl₃.
(3) is wrong. The left-hand side has fewer moles of gas. If you increase the pressure, the position of equilibrium will move to the left to relieve the pressure.
Answer: Increasing the temperature
Explanation: If you increase the temperature, the position of equilibrium will shift to the right so it can eliminate the heat.
A chemist dissolves 327.mg of pure hydrochloric acid in enough water to make up 120.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant dig
Answer:
pH → 1.13
Explanation:
Our solution is pure HCl
HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
As a strong acid, it is completely dissociated.
1 mol of HCl, can give 1 mol of H⁺ to the medium. Water does not participate. Let's find out M for the acid.
1st step: We convert the mass from mg to g → 327 mg . 1g /1000mg = 0.327 g
2nd step: We convert the mass(g) to moles: 0.327 g / 36.45 g/mol = 8.97×10⁻³ moles
3rd step: We convert the volume from mL to L → 120mL . 1L /1000 mL = 0.120L
Molarity (mol/L) = 8.97×10⁻³ mol / 0.120L = 0.075M
We propose: HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
0.075M 0.075M
pH = - log [H₃O⁺] → - log 0.075 = 1.13 → pH
What is the % by volume of 50mL of ethylene glycol dissolved in 950mL of H2O?
Answer:
5.0 %
Explanation:
Given data
Volume of ethylene glycol (solute): 50 mLVolume of water (solvent): 950 mLStep 1: Calculate the volume of solution
If we assume that the volumes are additive, the volume of the solution is equal to the sum of the volume of the solute and the solvent.
V = 50 mL + 950 mL = 1000 mL
Step 2: Calculate the percent by volume
We will use the following expression.
[tex]\% v/v = \frac{volume\ of\ solute}{volume\ of\ solution} \times 100 \% = \frac{50mL}{1000mL} \times 100 \% = 5.0\%[/tex]
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s ) + 2 O 2 ( g ) What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? Δ G ∘ rxn = kJ/mol What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? K = What is the equilibrium pressure of O2(g) over M(s) at 298 K? P O 2 = atm
Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0
[tex]deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol[/tex]
b) To calculate the constant we have the following expression:
[tex]lnK=-\frac{deltaG_{rxn} }{RT}[/tex]
Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K
[tex]lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066[/tex]
c) The equilibrium pressure of O₂ over M is:
[tex]K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm[/tex]
what is the molarity of a solution made by dissolving 25.0 grams of NaCl in 250.0 mL of solution
Answer:
Molarity= 1.7M
Explanation:
n= m/M = 25/58.5= 0.427mol
Applying
n=CV
0.427= C×0.25
C= 1.7M
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate the amount of Ga ( s ) that can be deposited from a Ga ( III ) solution using a current of 0.710 A that flows for 70.0 min .
Answer:
0.72g of gallium
Explanation:
Equation of the reduction reaction:
Ga^3+(aq) +3e --------> Ga(s)
If it takes 3F coulumbs of electricity to deposit 70g of gallium (relative atomic mass of gallium)
Then (0.710 × 70×60) coulombs of electricity will deposit (0.710 × 70×60) × 70/3F
But F = 96500C or 1Faraday
Therefore:
(0.710 × 70×60) × 70/3×96500
Mass of gallium deposited= 0.72g of gallium
A solution contains Ag and Hg2 ions. The addition of 0.100 L of 1.71 M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 39.6 g . Find the mass of AgI in the precipitate. Express your answer to two significant figures and include the appropriate units.
The mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.
To find the mass of AgI in the precipitate, we first need to determine the moles of AgI formed. We'll use the information provided and follow these steps:
1. Write the balanced chemical equation for the precipitation reaction:
[tex]\[ \text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \][/tex]
[tex]\[ \text{Hg}_2^{2+} + 2\text{I}^- \rightarrow \text{HgI}_2 \][/tex]
2. Determine the limiting reactant:
- For AgI: [tex]\( \text{Ag}^+ + \text{I}^- \)[/tex] (1 mole of Ag per mole of I)
- For HgI2: [tex]\( \frac{1}{2}\text{Hg}_2^{2+} + \text{I}^- \)[/tex] (1 mole of Hg per 2 moles of I)
The limiting reactant is the one that produces the fewer moles of I^-, as it determines the amount of AgI formed.
[tex]\[ \text{moles of I}^- = 0.100 \, \text{L} \times 1.71 \, \text{mol/L} = 0.171 \, \text{mol} \][/tex]
The limiting reactant is AgI, as it requires 0.171 moles of I^-, while HgI2 would require 0.342 moles of I^-.
3. Calculate the moles of AgI formed:
[tex]\[ \text{moles of AgI} = \text{moles of I}^- \times \frac{1 \, \text{mol AgI}}{1 \, \text{mol I}^-} = 0.171 \, \text{mol} \][/tex]
4. Calculate the mass of AgI formed:
[tex]\[ \text{mass of AgI} = \text{moles of AgI} \times \text{molar mass of AgI} \][/tex]
The molar mass of AgI is the sum of the atomic masses of Ag (107.87 g/mol) and I (126.904 g/mol).
[tex]\[ \text{mass of AgI} = 0.171 \, \text{mol} \times (107.87 + 126.904) \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of AgI} \approx 38.65 \, \text{g} \][/tex]
Therefore, the mass of AgI in the precipitate is approximately 38.65 grams (rounded to two significant figures) with the appropriate units.
For the reaction where Δn=−1Δn=−1 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants. For the reaction where Δn=0Δn=0 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants. For the reaction where Δn=+1Δn=+1 , what happens after in increase in volume? ????KQ>K so the reaction shifts toward reactants.
In chemical reactions, a change in volume affects the position of equilibrium based on stoichiometry. Increasing volume results in a shift towards the side with more gaseous molecules for reactions with Δn not equal to zero, while there is no shift if Δn equals zero. The direction of the shift is to decrease the reaction quotient (Q) in order to re-establish equilibrium with the equilibrium constant (K).
Explanation:
The effect of volume change on the position of equilibrium in chemical reactions is dependent on the stoichiometry and the reaction in question. For a reaction where Δn = -1, increasing the volume would result in a decrease in pressure and a shift towards the side with more moles of gas to re-establish equilibrium, typically the side with more molecules. However, if Δn = 0, an increase in volume has no effect on the equilibrium as there is no change in moles of gaseous substances on either side of the reaction. When Δn = +1, increasing the volume leads the equilibrium to shift towards the products, as this increases the total number of gaseous molecules which tends to lower the pressure.
When volume is increased and the reaction quotient Q becomes greater than the equilibrium constant K (Q > K), the reaction tends to shift towards the reactants to re-establish equilibrium. Conversely, when volume is decreased, and the pressure is increased, the reaction tends to shift towards the side of the reaction with fewer moles of gas.
Now, let's finish the calculation and the determination of the formula of the iron compound: Calculate the % water of hydration : Tries 0/3 Calculate the following for Fe3 : g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for K : g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for C2O42-: g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for H2O g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number)
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
A
The percentage of water of hydration is [tex]P_h= 11.01[/tex]%
Mass of [tex]Fe^{3+}[/tex] in 100mg is 10.60mg
Moles of [tex]Fe^{3+}[/tex] in 100mg is [tex]n_i= 0.19[/tex]
mol / mol Fe (3 sig figs) is [tex]= 1.00[/tex]
mol / mol Fe (whole number) is = 1
B
Mass of [tex]K^{+}[/tex] in 100mg is 27.70mg
Moles of [tex]K^{+}[/tex] in 100mg is [tex]n_i= 0.581 moles[/tex]
mol of K / mol of Fe (3 sig figs) is [tex]= 3.05[/tex]
mol of K / mol of Fe (whole number) is [tex]=3[/tex]
C
Mass of [tex]C_2O_4^{-2}[/tex] in 100mg is 55.69 mg
Moles of [tex]C_2O_4^{-2}[/tex] in 100mg is [tex]n_i= 0.633 moles[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is [tex]= 3.33[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (whole number) is [tex]=3[/tex]
D
Mass of water in 100mg is 11.01 mg
Moles of water in 100mg is [tex]n_i= 0.611 moles[/tex]
mol of water / mol of Fe (3 sig figs) is [tex]= 3.21[/tex]
mol of water / mol of Fe (whole number) is [tex]=3[/tex]
Explanation:
The percentage of water of hydration is mathematically represented as
[tex]P_h = 100 - (Pi + P_p + P_o)[/tex]
Now substituting 10.60% for [tex]P_i[/tex] (percentage of iron ) , 22.70% for [tex]P_p[/tex](Percentage of potassium) , 55.69% for [tex]P_o[/tex] (percentage of Oxlate)
[tex]P_h =100 - (10.60 + 22.70+55.69)[/tex]
[tex]P_h= 11.01[/tex]%
For IRON
Since the percentage of [tex]Fe^{3+}[/tex] is 10.60% then in a 100 mg of the sample the amount of [tex]Fe^{3+}[/tex] would be 10.60 mg
Now the no of moles is mathematically denoted as
[tex]n = \frac{mass}{molar \ mass }[/tex]
The molar mass of [tex]Fe[/tex] is 55.485 g/mol
So the number of moles of [tex]Fe^{3+}[/tex] in 100mg of he sample is
[tex]n_i = \frac{10.60}{55.485}[/tex]
[tex]n_i= 0.19[/tex]
mol / mol Fe (3 sig figs) is [tex]= \frac{0.19}{0.19} = 1.00[/tex]
FOR POTASSIUM
Since the percentage of [tex]K^{+}[/tex] is 22.70% then in a 100mg of the sample the amount of [tex]K^{+}[/tex] would be 22.70mg
The molar mass of [tex]K[/tex] is 39.1 g/mol
So the number of moles of [tex]K^{+}[/tex] in 100mg of he sample is
[tex]n_i = \frac{22.70}{39.1}[/tex]
[tex]=0.581 moles[/tex]
mol of K / mol of Fe (3 sig figs) is [tex]= \frac{0.581}{0.19} = 3.05[/tex]
FOR OXILATE [tex]C_2O_4^{-2}[/tex]
Since the percentage of [tex]C_2O_4^{-2}[/tex] is 55.69% then in a 100mg of the sample the amount of [tex]C_2O_4^{-2}[/tex] would be 55.69 mg
The molar mass of [tex]C_2O_4^{-2}[/tex] is 88.02 g/mol
So the number of moles of [tex]C_2O_4^{-2}[/tex] in 100mg of he sample is
[tex]n_i = \frac{55.69}{88.02}[/tex]
[tex]=0.633 moles[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is [tex]= \frac{0.633}{0.19} = 3.33[/tex]
FOR WATER OF HYDRATION
Since the percentage of water is 11.01% then in a 100mg of the sample the amount of water would be 11.0 mg
The molar mass of water is 18.0 g/mol
So the number of moles of water in 100mg of he sample is
[tex]n_i = \frac{11.01}{18.0}[/tex]
[tex]=0.611 moles[/tex]
mol of water / mol of Fe (3 sig figs) is [tex]= \frac{0.611}{0.19} = 3.21[/tex]
Name the following compound:
CH3
I
CH = CH2 - CH3
I
CH2
I
CH - CH3
I
CH2
I
CH
I
CH3
2-ethyl-4-methylheptane
2-ethyl-4-methylheptene
3-methyl-5-propyl-2-hexene
3,5-dimethyl-2-octene
Answer:
3,5-dimethyl-2-octene
Explanation:
Please note that there is no H at carbon 3 less carbon becomes penta hydra.
The compound is:
CH3
I
C = CH2 - CH3
I
CH2
I
CH - CH3
I
CH2
I
CH2
I
CH3
To name the above compound, do the following:
1. Locate the longest continuous chain i.e octene
2. Start counting from the side that gives the double bond the lowest low count since the double bond is the functional group. In doing this, the double bond is at carbon 2.
3. Locate the substituent groups attached and their position in the parent chain. In doing so, you will see that there are two CH3 group attached and they are at carbon 3 and carbon 5. Since the substituents attached are the same, we'll name them as 'dimethyl' indicate that they are two methyl groups
Now, we'll combine the above findings in order to obtain the name. Therefore, the name of the compound is:
3,5-dimethyl-2-octene
Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to electroplate the zinc to a thickness of 0.025 mm. It is found that a current of 20 A produces a zinc coating of sufficient quality for galvanized iron. Determine the time required to produce the desired coating, assuming 100 % efficiency.
Answer:
The time required for the coating is 105 s
Explanation:
Zinc undergoes reduction reaction and absorbs two (2) electron ions.
The expression for the mass change at electrode [tex](m_{ch})[/tex] is given as :
[tex]\frac{m_{ch}}{M} ZF = It[/tex]
where;
M = molar mass
Z = ions charge at electrodes
F = Faraday's constant
I = current
A = area
t = time
also; [tex](m_{ch})[/tex] = [tex](Ad) \rho[/tex] ; replacing that into above equation; we have:
[tex]\frac{(Ad) \rho}{M} ZF = It[/tex] ---- equation (1)
where;
A = area
d = thickness
[tex]\rho[/tex] = density
From the above equation (1); The time required for coating can be calculated as;
[tex][ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons } ] = (20 A) t[/tex]
[tex]t = \frac{2100}{20}[/tex]
= 105 s
What is the molarity of a KF solution that contains 116 grams of KF in 2 L of solution?
Answer : The molarity of KF solution is, 1 M
Explanation : Given,
Mass of [tex]KF[/tex] = 116 g
Volume of solution = 2 L
Molar mass of [tex]KF[/tex] = 58 g/mole
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Mass of }KF}{\text{Molar mass of }KF\times \text{Volume of solution (in L)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{116g}{58g/mole\times 2L}=1mole/L=1M[/tex]
Therefore, the molarity of KF solution is, 1 M
A stock solution of FeCl2 is available to prepare solutions that are more dilute. Calculate the volume, in mL, of a 2.0-M solution of FeCl2 required to prepare exactly 100 mL of a 0.630-M solution of FeCl2.
Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required
Answer:
We need 31.5 mL of the 2.0 M FeCl2 solution
Explanation:
Step 1: Data given
Molarity of a FeCl2 solution = 2.0 M
Initial volume of FeCl2 = 100 mL
Initial molarity of FeCl2 = 0.630 M
Step 2: Calculate volume of the stock solution
C1V1 = C2V2
⇒with C1 = the initial molarity FeCl2 = 0.630 M
⇒with V1 = the initial volume = 100 mL = 0.100 L
⇒with C2 = the new molarity FeCl2 = 2.0 M
⇒with V2 = the new volume = TO BE DETERMINED
0.630M * 0.100 L = 2.0 M * V2
V2 = (0.630 * 0.100) / 2.0
V2 = 0.0315 L = 31.5 mL
We need 31.5 mL of the 2.0 M FeCl2 solution
what are the smallest sub atomic structure
Answer:
Electrons
Explanation:
In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.
Zinc metal reacts with hydrochloric acid to produce hydrogen gas and an aqueous solution of zinc(II) chloride. What is the reducing agent in this reaction?
Answer:
Zinc is the reducing agent.
Explanation:
Zn is the reducing agent because its oxidation state increases from 0 to 2 and hydrochloric acid, HCl is the oxidizing agent because it loses hydrogen ions, H+ based on the following balanced equation:
Zn (s) + 2HCl (aq) -----> ZnCl2 (aq) + H2 (g)
In oxidation, there is the loss of electrons. In reduction, there is a gain of electrons. A nemonic device that might help is LEO (the lion said) GER.
Losses
Electrons
Oxidation
Gains
Electrons
Reduction
Answer:
Zinc
The reducing agent in these reaction is Zinc (Zn)
Explanation:
Zn is the reducing agent because its oxidation state increases from 0 to 2 and hydrochloric acid, HCl is the oxidizing agent because it loses hydrogen ions, H+ based on the following balanced equation:
Zn (s) + 2HCl (aq) -----> ZnCl2 (aq) + H2 (g)
In oxidation, there is the loss of electrons.
In reduction, there is a gain of electrons.
What is the pH of 9.01 x 10^-4 M Mg(OH)2
What is the pH of 2.33 x 10^-2M of NH4OH
Answer:
A. 11.26
B. 12.37
Explanation:
A. Step 1:
Dissociation of Mg(OH)2. This is illustrated below below:
Mg(OH)2 <==> Mg2+ + 2OH-
A. Step 2:
Determination of the concentration of the OH-
From the above equation,
1 mole of Mg(OH)2 produce 2 moles of OH-
Therefore, 9.01x10^-4 M Mg(OH)2 will produce = 9.01x10^-4 x 2 = 1.802x10^-3 M of OH-
A. Step 3:
Determination of the pOH. This is illustrated below:
pOH = - Log [OH-]
[OH-] = 1.802x10^-3 M
pOH = - Log [OH-]
pOH = - Log 1.802x10^-3
pOH = 2.74
A. Step 4:
Determination of the pH.
pH + pOH = 14
pOH = 2.74
pH + 2.74 = 14
Collect like terms
pH = 14 - 2.74
pH = 11.26
B. Step 1:
Dissociation of NH4OH. This is illustrated below below:
NH4OH <==> NH4+ + OH-
B. Step 2:
Determination of the concentration of the OH-
From the above equation,
1 mole of NH4OH produce 1 moles of OH-
Therefore, 2.33x10^-2M of NH4OH will also produce 2.33x10^-2M of OH-
B. Step 3:
Determination of the pOH. This is illustrated below:
pOH = - Log [OH-]
[OH-] = 2.33x10^-2M
pOH = - Log [OH-]
pOH = - Log 2.33x10^-2M
pOH = 1.63
B. Step 4:
Determination of the pH.
pH + pOH = 14
pOH = 1.63
pH + 1.63 = 14
Collect like terms
pH = 14 - 1.63
pH = 12.37
b. If 20.0 grams of Aluminum and 30.0 grams of chlorine gasſare used, and how many
grams AlCl3 can theoretically be made (3 pts)?
Answer: The mass of [tex]AlCl_3[/tex] theoretically be made can be, 37.4 grams.
Explanation : Given,
Mass of [tex]Al[/tex] = 20.0 g
Mass of [tex]Cl_2[/tex] = 30.0 g
Molar mass of [tex]Al[/tex] = 27 g/mol
Molar mass of [tex]Cl_2[/tex] = 71 g/mol
First we have to calculate the moles of [tex]Al[/tex] and [tex]Cl_2[/tex].
[tex]\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{20.0g}{27g/mol}=0.741mol[/tex]
and,
[tex]\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}=\frac{30.0g}{71g/mol}=0.422mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]
From the balanced reaction we conclude that
As, 3 moles of [tex]Cl_2[/tex] react with 2 moles of [tex]Al[/tex]
So, 0.422 moles of [tex]Cl_2[/tex] react with [tex]\frac{2}{3}\times 0.422=0.281[/tex] moles of [tex]Al[/tex]
From this we conclude that, [tex]Al[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Cl_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]AlCl_3[/tex]
From the reaction, we conclude that
As, 3 moles of [tex]Cl_2[/tex] react to give 2 moles of [tex]AlCl_3[/tex]
So, 0.422 moles of [tex]Cl_2[/tex] react to give [tex]\frac{2}{3}\times 0.422=0.281[/tex] mole of [tex]AlCl_3[/tex]
Now we have to calculate the mass of [tex]AlCl_3[/tex]
[tex]\text{ Mass of }AlCl_3=\text{ Moles of }AlCl_3\times \text{ Molar mass of }AlCl_3[/tex]
Molar mass of [tex]AlCl_3[/tex] = 133 g/mole
[tex]\text{ Mass of }AlCl_3=(0.281moles)\times (133g/mole)=37.4g[/tex]
Therefore, the mass of [tex]AlCl_3[/tex] theoretically be made can be, 37.4 grams.
Pls help ASAP, I will give brainliest and maximum points :)
Answer:
First, Second, Second-to-last, and last choice
Explanation:
Chemistry student. The other options are incorrect. If you would like a more thorough explanation, please reply to this comment.
pH according to Arrhenius definition is the measure of hydrogen ion concentration in solution:
[tex]pH=-log[H^{+}][/tex] Acids release H+ ions into solution, while bases release OH- ions into solution. This explains why second-to-last choice is correct.
Litmus paper are thin strips of paper that have been manufactured with indicators, examples being red cabbage or phenolphthalein. Indicators react to changes in pH with an according color change. This explains why the last choice is also correct.
Acid strength is better the lower the number it is, hence why pH 2 is stronger acid than 5, but in turn is a weaker base. This explains why second choice is correct.
The first option can be explained by the pH diagram. HCl is acidic, and will turn blue (basic) litmus paper more acidic, and move towards an acidic pH, hence more "red."
If you liked this solution, leave a Thanks or give a Rating!
thanks,
Answer:
I dont know if im correct number 1 is 2
Explanation:
ps im not good at this subject
Pls help ASAP I will give brainliest
Answer:
Lemon
HCI
Blood
Saliva
Bleach
NaOH
Explanation:
Blood 7.35-7.45
Bleach 12.6
Saliva 6.2-7.6
Lemon 2-3
HCI 3.01
NaOH 13
A 1.00 g sample of n-hexane (C6H14) undergoes complete combustion with excess O2 in a bomb calorimeter. The temperature of the 1502 g of water surrounding the bomb rises from 22.64◦C to 29.30◦C. The heat capacity of the hardware component of the calorimeter (everything that is not water) is 4042 J/◦C. What is ∆U for the combustion of n-C6H14? One mole of n-C6H14 is 86.1 g.
Answer:
i have an answer but i can only show you because my teacher helped my on it and wrote it down for me to remember! hope this helps!!!
Explanation:
A 1.00 g sample ofn-hexane (C6H14) under-goes complete combustion with excess O2ina bomb calorimeter. The temperature of the1502 g of water surrounding the bomb risesfrom 22.64◦C to 29.30◦C. The heat capacityof the hardware component of the calorimeter(everything that is not water) is 4042 J/◦C.What is ΔUfor the combustion ofn-C6H14?One mole ofn-C6H14is 86.1 g.The specificheat of water is 4.184 J/g·◦C.1.-9.96×103kJ/mol2.-7.40×104kJ/mol3.-1.15×104kJ/mol4.-4.52×103kJ/mol5.-5.92×103kJ/molcorrectExplanation:mC6H8= 1.00 gmwater= 1502 gSH = 4.184 J/g·◦CHC = 4042 J/◦CΔT= 29.30◦C-22.64◦C = 6.66◦CThe increase in the water temperature is29.30◦C-22.64◦C = 6.66◦C. The amount ofheat responsible for this increase in tempera-ture for 1502 g of water isq= (6.66◦C)parenleftbigg4.184Jg·◦Cparenrightbigg(1502 g)= 41854 J = 41.85 kJThe amount of heat responsible for the warm-ing of the calorimeter isq= (6.66◦C)(4042 J/◦C)= 26920 J = 26.92 kJ
Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation of PbCl2(s) occurs. The PbCl2(s) precipitate is collected by filtration, dried, and weighed. A total of 12.79 grams of PbCl2(s) is ob- tained from 200.0 milliliters of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.
Answer:
0.23 mol/L
Explanation:
Step 1:
The balanced equation for the reaction. This is illustrated below:
Pb(NO3)2(aq) + 2NaCl(aq) —> PbCl2(s) + 2NaNO3(aq)
Step 2:
Determination of the number of in 12.79g of PbCl2. This is illustrated below:
Mass of PbCl2 = 12.79g
Molar Mass of PbCl2 = 207 + (2x35.5) = 207 + 71 = 278g/mol
Number of mole of PbCl2 =?
Number of mole = Mass/Molar Mass
Number of mole of PbCl2 = 12.79/278
Number of mole of PbCl2 = 0.046 mole
Step 3:
Determination of the number of mole of Pb(NO3)2 that reacted.
This is illustrated below:
From the balanced equation above,
1 mole of Pb(NO3)2 reacted to produce 1 mole of PbCl2.
Therefore, it will also take 0.046 mole of Pb(NO3)2 to react to produce 0.046 mole of PbCl2.
Step 4:
Determination of the molarity of Pb(NO3)2. This is illustrated:
Mole of Pb(NO3)2 = 0.046 mole
Volume of the solution = 200 mL = 200/1000 = 0.2 L
Molarity =?
Molarity is defined as the mole of solute per unit litre of solution. It is given by:
Molarity = mole of solute /Volume
Molarity of Pb(NO3)2 = 0.046/0.2
Molarity of Pb(NO3)2 = 0.23 mol/L
A solution was prepared by dissolving 125.0 g of KCl in 275 g of water. Calculate the mole fraction of KCl. (The formula weight of KCl is 74.6 g/mol. The formula weight of water is 18.0 g/mol.)
Answer:
The mole fraction of [tex]KCl[/tex] is [tex]N_{KCl}=0.099[/tex]
Explanation:
Generally number of mole is mathematically represented as
[tex]n = \frac{mass}{Molar mass }[/tex]
The number of mole of [tex]KCl[/tex] is
[tex]n_{KCl} = \frac{mass \ of \ KCl}{Molar\ mass \ of \ KCl }[/tex]
[tex]n_{KCl} = \frac{125}{74.6}[/tex]
[tex]=1.676 \ moles[/tex]
The number of mole of [tex]H_2O[/tex] is
[tex]n_{H_2O} = \frac{mass \ of \ H_2O}{Molar\ mass \ of \ H_2O }[/tex]
[tex]n_{H_2O} = \frac{275}{18}[/tex]
[tex]= 15.28 \ moles[/tex]
Mole fraction of [tex]N_{KCl}= \frac{n_{KCl}}{n_{KCl} + n_{H_2O}}[/tex]
[tex]= \frac{1.676}{15.28 +1.676}[/tex]
[tex]N_{KCl}=0.099[/tex]
Answer:
The mole fraction of KCl is 0.13
Explanation:
no of moles of KCl (n KCl) = W/G.F.Wt
= 135/74.6
= 1.81moles
no of moles of H2O (nH2O) = W/G.F.Wt
= 225/18 = 12.5 moles
mole fraction of KCl ( XKCl) = nKCl/nKCl + nH2O
= 1.81/(1.81+12.5)
= 1.81/14.31
The mole fraction of KCl is = 0.13
Which best defines nitrogen fixation?
A.) the process of creating free nitrogen for plants to absorb
B.) the breakdown of nitrogen in the soil
C.) the conversion of nitrogen gas into a usable form
D.) the destruction of the bonds between nitrogen and other elements
Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 8.83 g of sulfuric acid is mixed with 9.1 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
13 grams Na₂SO₄ (2 sig.figs.)
Explanation:
1st convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.
Determine limiting reactant and
Complete problem by converting yield into grams.
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄
moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH
Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.
H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)
NaOH => (0.2275/2) = 0.1138
NOTE: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
moles 0.0901 mole 0.2275 mol 0.0901 mol 2(0.0901 mol)
mass (g) Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 12.798 grams ≅ 13 grams Na₂SO₄ (2 sig.figs.)
Answer:
13 grams Na₂SO₄ (2 sig.figs.)
Explanation:
Firstly
We convert mass values to moles and solve yield using limiting reactant principles and reaction ratio of balance equation.
To determine limiting reactant and
Complete problem by converting yield into grams. We write the equations for the reaction
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
To give
moles H₂SO₄ = 8.83g/98g·mol⁻¹ = 0.0901 mole H₂SO₄
Then,
moles NaOH = 9.1g/40g·mol⁻¹ = 0.2275 mole NaOH
Determine Limiting Reactant by dividing each mole value by the respective coefficient in the balanced equation. The smaller value is always the limiting reactant.
H₂SO₄ => (0.0901/1) = 0.0901 <= Limiting Reactant (smaller value)
NaOH => (0.2275/2) = 0.1138
N/B: when working the problem use the calculated moles of reactant NOT the LR test number. In this problem, use 0.0901 mole H₂SO₄. (yeah, it is the same but this does not occur for the LR in many other problems). Anyways...
H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O
moles 0.0901 mole 0.2275 mol 0.0901 mol 2(0.0901 mol)mass (g)
Na₂SO₄ = 0.0901 mole x 142.04 g/mol = 1
2.798 grams ≅ 13 grams Na₂SO₄ (2 sig.figs.)
Zinc can be removed from bronze by placing bronze in hydrochloric acid. The zinc reacts with the hydrochloric acid producing zinc chloride and hydrogen gas, leaving copper behind. If the reaction yields 0.680 g H2, what is the percent yield?
Answer:The Zinc Reacts With The Hydrochloric Acid Producing Zinc Chloride And Hydrogen Gas, And Leaving The Copper Behind. A. If 25.0 G Of Zinc ... Zn+ 2 HCI --> ZnCl2 + H2 (answer .771 G H2) B. If The Reaction Yields . ... If 25.0 g of zinc are in a sample of bronze, determine the theoretical yield of hydrogen gas. Zn+ 2 HCI
What do all acids produce when dissolved in water
Answer:
hydrogen ions
Explanation:
Acids are substances that when dissolved in water release hydrogen ions, H+(aq). Bases are substances that react with and neutralise acids, producing water. When dissolved, bases release hydroxide ions, OH-(aq) into solution. Water is the product of an acid and base reacting.
When acids dissolve in water, they produce hydrogen ions (H+). This is known as the process of ionization.
The hydrogen ions are responsible for the acidic properties of the solution.When HCl is added to water, it dissociates into hydrogen ions (H+) and chloride ions (Cl-). The presence of hydrogen ions gives the solution acidic properties.
Similarly, other acids such as sulfuric acid, nitric acid, and acetic acid also produce hydrogen ions when dissolved in water.
It's important to note that not all substances that dissolve in water are acids. Acids have a pH value less than 7, and their properties can vary depending on their strength. Strong acids, like hydrochloric acid, completely dissociate in water, producing a large number of hydrogen ions. Weak acids, like acetic acid, only partially ionize, producing fewer hydrogen ions.
Learn more about acids,here:
https://brainly.com/question/29796621
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Your value for the stoichiometric ration (slope) is most likely slight different than the predicted value of 1. List a reasonable error that could have caused this discrepancy and briefly explain.
Answer:
The stopper is not fitted on the flask quickly.
Explanation:
The reaction has to do with a gas. The predicted value of the slope is 1. The value of slope obtained from the experiment varies slightly from the predicted value of 1. This discrepancy may be caused by not fitting the stopper on the flask quickly enough. This means that some gas may still be left in the flask leading a discrepancy in the slope obtained.