The reaction will consume 3 moles of iron(III) oxide and will produce 3 moles of aluminum oxide and 12 moles of iron, following the mole ratio from the balanced chemical equation.
Explanation:Starting from the balanced chemical equation, we can see that the ratio of "Fe2O3" (iron(III) oxide) to "Al" (aluminum) is 1:2. This ratio allows us to calculate the amount of moles that will react or produce. We can apply these ratios in the following way:
i. Since the ratio of Fe2O3 to Al is 1:2, and we have 6 moles of Al, then it will consume 6/2 = 3 moles of Fe2O3.ii. Looking at the product side, the ratio of Al2O3 to Al is also 1:2, so the reaction will produce 6/2 = 3 moles of Al2O3.The ratio of Fe to Al is 1:1, so the reaction will produce 6 x 2 = 12 moles of Fe.Learn more about Stoichiometry here:https://brainly.com/question/34828728
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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
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
2K + 2HBr → 2 KBr + H2
When 5.5moles of K reacts with 4.04moles of HBr, to produce Hydrogen gas(H₂)
●a). What is the limiting reactant?
●b.)What is the excess reactant?
●C.)How much product is produced?
Answer:
[tex]\large \boxed{\text{0.0503 g}}[/tex]
Explanation:
The limiting reactant is the reactant that gives the smaller amount of product.
Assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 39.10 80.41 2.016
2K + 2HBr ⟶ 2KBr + H₂
m/g: 5.5 4.04
a) Limiting reactant
(i) Calculate the moles of each reactant
[tex]\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}[/tex]
(ii) Calculate the moles of H₂ we can obtain from each reactant.
From K:
The molar ratio of H₂:K is 1:2.
[tex]\text{Moles of H}_{2} = \text{0.141 mol K} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol K}} = \text{0.0703 mol H}_{2}[/tex]
From HBr:
The molar ratio of H₂:HBr is 3:2.
[tex]\text{Moles of H}_{2} = \text{0.049.93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol HBr}} = \text{0.024 97 mol H}_{2}[/tex]
(iii) Identify the limiting reactant
HBr is the limiting reactant because it gives the smaller amount of NH₃.
b) Excess reactant
The excess reactant is K.
c) Mass of H₂
[tex]\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\ \text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$ }[/tex]
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
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
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
Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochemical series for ligands and metals? (Select one answer. There is only one correct answer)
a. Strong o-donor ligands raise the energy of the eg orbitals of an octahedral complex.
b. Higher oxidation state metals form stronger bonds with ligands.
c. Ligands such as CO, PPhy, and bipyridine can act as T-acids.
d. The 4d and 5d transition metals are stronger Lewis acids than the 3d transition metals.
e. The electronegativity of the donor atom is the most important factor for ligands
Answer:
Higher oxidation state metals form stronger bong with ligands
Explanation:
Ligand strength are based on oxidation number, group and its properties
Statement e is not an accurate description because the ordering of the spectrochemical series depends on multiple factors, including ligand size, pi-donor or pi-acceptor abilities, and the metal's identity and oxidation state, in addition to electronegativity of the donor atom.
The correct answer to the question "Which of the following statements is not an accurate description of a factor that contributes to the ordering of the spectrochemical series for ligands and metals?" is e. The electronegativity of the donor atom is the most important factor for ligands.
The spectrochemical series is an ordering of ligands based on the magnitude of the ligand field splitting parameter (Δoct) they produce in coordination complexes. Contrary to option e, it is not just the electronegativity of the donor atom but also a series of other factors such as the size of the donor atom, the ligand's field strength (pi-donor or pi-acceptor abilities), and the identity of the metal ion and its oxidation state that contribute to the ordering of the spectrochemical series. For example, small neutral ligands with highly localized lone pairs, such as NH₃, produce larger Δoct values which illustrates more than just electronegativity playing a role.
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.
Warning signs are sometimes placed on aerosol cans to prevent people from throwing them into a fire. What would be true about the contents of the gas in an aerosol can just after it is placed in a fire?
Answer:
Volume stays.
Explanation:
If aerosol can throwing into the fire than the temperature of the aerosol can & the temperature of the contents of the gas in an aerosol will also increase, and it induces the pressure to rise.
Against the sides of the can, the gas molecules will smash rapidly with each other which are present inside an aerosol can, and the proportion of gas will stay the same as earlier. According to the law of conservation of matter, as a result of the heat, the gas particles can not be eradicated or expanded.
What temperature must be maintained to ensure that a 1.00 L flask containing 0.0400 mol of oxygen will show a continuous pressure of 0.981 atm?
Answer:
The temperature is 298.9K = 25.75 °C
Explanation:
Step 1: Data given
Volume of the flask = 1.00 L
Number of moles oxygen = 0.0400 moles
Pressure = 0.981 atm
Step 2: Calculate the temperature
p*V = n*R*T
⇒with p = the pressure of oxygen gas = 0.981 atm
⇒with V = the volume of the flask = 1.00 L
⇒with n = the number of moles of oxygen = 0.0400 moles
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temeprature = TO BE DETERMINED
T = (p*V) / (n*R)
T = (0.981 atm * 1.00 L) / (0.0400 moles * 0.08206 L*atm/mol*K)
T = 298.9 K
The temperature is 298.9K = 25.75 °C
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
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.)
How many grams of carbon dioxide are created from the complete combustion of 21.3 L of butane at STP?
Answer:
165.5 g of CO2
Explanation:
We must first put down the balanced reaction equation:
C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H2O(g)
From the reaction equation, one mole of butane occupies 22.4 L hence we can establish the stoichiometry of the reaction thus:
22.4 L of butane created 174 g of CO2
Therefore 21.3 L of butane will create 21.3 × 174/22.4 = 165.5 g of CO2
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
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
Calculate the amount of heat needed to melt 148. g of solid octane (C8H18 ) and bring it to a temperature of 117.8 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer:
The amount of heat is 84.4894kJ
Explanation:
Given data:
mass of octane=148g
The moles of octane is:
[tex]n_{octane} =148g*\frac{1mol}{114.23g} =1.2956moles[/tex]
The melting point is=-56.82°C
ΔHfus=20.73kJ/mol
The heat of fusion is:
[tex]Q_{fus} =n_{octane} *delta-H_{fus} =1.2956*20.73=26.8578kJ[/tex]
The heat to bring the octane to a temperature of 117.8°C is:
[tex]Q_{2} =mCp(117.8-(56.82))=148*2.23*(117.8+56.82)=57631.5848J=57.6316kJ[/tex]
The total heat in the process is:
Qtotal=Qfus+Q₂=26.8578+57.6316=84.4894kJ
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.
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]
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:
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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]
The acids and bases shown right cover a range of pH values. Use what you know about acids, bases, and concentration to label the test tubes, in order, from most acidic to most basic.
Answer:
.1m HCl 0.001HCl .00001m HCl Distilled water .00001NaOH .001NaOH .01NaOH
Explanation:
Thanks to the person who posted the answers, I just wrote it out so its easier to see. :D
In the given compounds, the most acidic compound is 0.1m HCl and most basic compound is 0.01NaOH.
What are acids and bases?
Acids are those compounds whose has a pH value in the range from 0 to 7 and bases are those compounds which has a pH range from 7 to 14.
pH of any solution will be calculated as:
pH = -log[H⁺], where
[H⁺] = concentration of H⁺ ion and this concentration is present in the form of molarity (molar concentration).
So, pH value is directly proportional to the concentration value of H⁺ ion as concentration of H⁺ ion decreases so acidity also decreases and sequence of given compounds from most acidic to most basic is represented as:
0.1m HCl > 0.001HCl > 0.00001m HCl > Distilled water > 0.00001NaOH > 0.001NaOH > 0.01NaOH.
Hence 0.1m HCl is most acidic and 0.01 NaOH is most basic.
To know more about acidity & basicity, visit the below link:
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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 proposed mechanism is: Step 1: H2(g) + ICl(g) → HI(g) + HCl(g) (slow) Step 2: HI(g) + ICl(g) → I2(g) + HCl(g) (fast) Which of the following species is a catalyst? ICl HCl This mechanism has no catalyst. HI H2
Answer: This mechanism has no catalyst
Explanation:
The proposed mechanism is :
Step 1 : [tex]H_2(g)+ICl(g)\rightarrow HI(g)+HCl(g)[/tex] (slow)
Step 2: [tex]HI(g)+ICl(g)\rightarrow I_2(g)+HCl(g)[/tex] (fast)
The combined chemical equation will be :
[tex]H_2(g)+2ICl(g)\rightarrow I_2(g)+2HCl(g)[/tex]
A catalyst is a substance which enhances the rate of chemical reaction without being consumed in the chemical reaction. Thus catalyst gets used up in first step and gets regenerated in second.
HI is formed as an intermediate as it gets formed in first step and gets used up in the second step.
Here there is no substance which is used as a catalyst.
Can 1750 mL of water dissolve 4.6 moles of copper sulfate CuSO4
Answer:
No
Explanation:
We first find the solubility of the copper sulphate salt
Number of moles= 4.6 moles
Volume =1.75 L
Molarity= number of moles/ volume = 4.6/1.75= 2.6 molL-1
Hence, only 2.6 moles of Copper II sulphate dissolves in 1.75L of water. Hence 4.6 moles of copper II Sulphate dies not dissolve in 1.75 L of water.
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 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.
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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
Identify each of these substances as acidic, basic, or neutral.
rainwater, pOH = 8.5
cola, pOH = 11
tomato juice, POH = 10
liquid drain cleaner, poH = 0
Answer:
Rain Water: Acidic
Cola: Acidic
Tomato Juice: Acidic
Liquid Drain Cleaner: Basic
Rain Water: Acidic
Cola: Acidic
Tomato Juice: Acidic
Liquid Drain Cleaner: Basic
pHIf the value of pH is 7, then the solution is neutral, if greater than 7 then basic, and if less than 7 then acidic.
The relation between pH and pOH is as follows:-
[tex]pH+pOH=14[/tex]
The value of pH for rain water is 14-8.5=5.5.
So, the rain water is acidic.
The value of pH for cola is 14-11=3.
So, the cola is acidic.
The value of pH for tomato juice is 14-10=4.
So, the tomato juice is acidic.
The value of pH for liquid drain cleaner is 14-0=14
So, the liquid drain cleaner is basic.
Find more information about the pH here,
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Determine the amount of heat energy in joules required to raise the temperature of 7.40g of water from 29.0OC to 46.0 OC.
Answer: The amount of heat energy in joules required to raise the temperature is 526 Joules
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed = ?
m= mass of substance = 7.40 g
c = specific heat capacity = [tex]4.184J/g^0C[/tex]
Initial temperature of the water = [tex]T_i[/tex] = 29.0°C
Final temperature of the water = [tex]T_f[/tex] = 46.0°C
Change in temperature ,[tex]\Delta T=T_f-T_i=(46.0-29.0)^0C=17.0^0C[/tex]
Putting in the values, we get:
[tex]Q=7.40g\times 4.184J/g^0C\times 17.0^0C[/tex]
[tex]Q=526J[/tex]
The amount of heat energy in joules required to raise the temperature is 526 Joules
The amount of heat energy will be "526 J".
Given:
Mass,
m = 7.40 gInitial temperature,
[tex]T_i[/tex] = 29.0°CFinal temperature,
[tex]T_f[/tex] = 46.0°CChange in temperature,
[tex]\Delta T = T_f -T_i[/tex][tex]= 46.0-29.0[/tex]
[tex]= 17.0^{\circ} C[/tex]
The amount of heat energy,
→ [tex]Q = mc \Delta T[/tex]
By substituting the values, we get
[tex]= 7.40\times 4.184\times 17.0[/tex]
[tex]= 526 \ J[/tex]
Thus the above answer is right.
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From this combustion equation, 2CH22 + 3102 - 22H,0 + 2000, calculate the liters of
carbon dioxide produced when 16.9 grams of CH are combusted
Answer : The volume of [tex]CO_2[/tex] produced are, 26.7 liters.
Explanation :
First we have to calculate the moles of [tex]C_{10}H_{22}[/tex]
[tex]\text{Moles of }C_{10}H_{22}=\frac{\text{Given mass }C_{10}H_{22}}{\text{Molar mass }C_{10}H_{22}}=\frac{16.9g}{142g/mol}=0.119mol[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The given combustion reaction is:
[tex]2C_{10}H_{22}+31O_2\rightarrow 22H_2O+20CO_2[/tex]
From the balanced chemical reaction we conclude that,
As, 2 moles of [tex]C_{10}H_{22}[/tex] react to give 20 moles of [tex]CO_2[/tex]
So, 0.119 moles of [tex]C_{10}H_{22}[/tex] react to give [tex]\frac{20}{2}\times 0.119=1.19[/tex] moles of [tex]CO_2[/tex]
Now we have to calculate the volume of [tex]CO_2[/tex] produced.
As we know that, 1 mole of gas occupies 22.4 L volume of gas.
As, 1 mole of [tex]CO_2[/tex] gas occupies 22.4 L volume of [tex]CO_2[/tex] gas.
So, 1.19 mole of [tex]CO_2[/tex] gas occupies 1.19 × 22.4 L = 26.7 L volume of [tex]CO_2[/tex] gas.
Therefore, the volume of [tex]CO_2[/tex] produced are, 26.7 liters.