Answers
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + Al(OH)₃
Ca are balance on both sides,
There are 2 Al at left side and one at right so, multiply Al(OH)₃ by 2 to balance,
So,
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + 2 Al(OH)₃
Now, Ca and Al are balanced, now balance SO₄, which is 3 at left hand side and one at right hand side, so multiply CaSO₄ on right side by 3, so,
Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
Again Ca got imbalance, so multiply Ca(OH)₂ by 3 to balance Ca, So,
3 Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
The Equation is balance now with respect to every element.
Result:
The Ratio is 3 : 1 : 3 : 2, so, Option-G is correct.
Related Questions
Determine the overall charge on each complex tetraquadichlorochromium (iii)
Answers
+1
Explanation:
Molecular Formula of Tetraquadichlorochromium (III) is as follow,
[Cr(H₂O)₄Cl₂]ˣ
The oxidation states of elements present in complex are,
H₂O = 0 (Neutral Ligand)
Cl = -1
Cr = +3 (Given)
x = Overall charge
So,
[+3 + (0)₄ + (-1)₂] = x
[+3 - 2] = x
x = +1
Result:
Overall charge on Tetraquadichlorochromium (III) is +1.
Write the balanced nuclear equation for β− decay of sodium−26. include both the mass numbers and the atomic numbers with each nuclear symbol. use the sup-subscript button in the answer palette to enter these numbers correctly. greek letters can be accessed in the drop-down menu that says -select−.
Answers
In beta minus decay, neutron is converted to a proton and an electron and an electron antineutrino and in beta plus decay, a proton is converted to a neutron and positron and an electron neutrino, so mass number does not change.
How can erosion form new land? (site 2)
Answers
Answer:
The procedure by which soil, land, or rock get slowly worn away by the natural elements like wind or water is known as erosion. The landforms refer to the natural characteristics found on the surface of the Earth that exhibit different shape and origin. The landforms can be destroyed and created by erosion.
The landforms created by the process of erosion are known as fluvial erosion landforms. With the passing of water across the land, the sediments and other kinds of natural debris also get carried with it. With time, the gathering of the debris and sediments generate deposits that ultimately turn into a landform. Some of the examples of fluvial erosion landforms comprise flood plains, sandbars, and levees.
Which of the following phase changes would release energy as it occurs? (4 points) Melting
Boiling
Evaporating
Freezing
Answers
What is the value of δg°' (or, to put it another way, the cost) when 2nadp+ and 2h2o are converted to 2nadph plus 2h+ plus o2?
Answers
The chemical reaction is given as:
[tex]2NaDP^{+} +2H_{2}O\rightarrow 2NaDPH+2H^{+}+O_{2}[/tex]
Here, oxygen is oxidised and [tex]NaDP^{+}[/tex] is reduced. Thus, redox reaction occurs.
For cell reaction, [tex]\Delta G^{o} = -nFE^{o}_{cell}[/tex] (2)
where, [tex]\Delta G^{o} [/tex] = standard state free energy
n= number of electrons
F= Faraday constant ([tex]96485.33 C/mol[/tex])
[tex]E^{o}_{cell}[/tex] = cell potential
Substitute the value of number of electrons i.e. 2, Faraday constant and cell potential in the formula to determine the value of [tex]\Delta G^{o} [/tex].
Now, calculate the value of cell potential
[tex]E^{o}_{cell} = E^{o}_{cathode}- E^{o}_{anode}[/tex] (1)
[tex]E^{o}_{cathode}[/tex] = [tex]-0.324 V[/tex] (standard reduction potential of [tex]NaDP^{+}[/tex])
[tex]E^{o}_{anode}[/tex] = [tex]1.23 V[/tex] (standard reduction potential of [tex]O_{2}[/tex])
Put the above values in formula (1), we get:
[tex]E^{o}_{cell} = -0.324 V-1.23 V[/tex]
= [tex]-1.554 V[/tex]
Now, substitute above value in formula (2)
[tex]\Delta G^{o} = -2\times 96485.33 C/mol \times(-1.554 V) [/tex]
= [tex]299876.40564 CV/mol[/tex]
Since, one coulomb volt is equal to one joule.
Thus, value of [tex]\Delta G^{o}[/tex] is equal to [tex]299876.40564 J/mol[/tex] or [tex]299.87640564 kJ/mol[/tex]
A sample of he gas (2.35 mol) occupies 57.9 l at 300.0 k and 1.00 atm. the volume of this sample is _____ l at 423 k and 1.00 atm.
Answers
V1=75.9 L
T1= 300 K
V2=?
T2=423 k
by making the V2 the subject of the formula
V2=T2V1/T1
V2= 57.9 L x 423 K/300 K= 81.64 L is the volume of the sample
The ideal gas relates to the pressure, temperature, pressure, and mole of the gas. The volume of the sample at 423 K and 1 atm is 81.64L.
What is Charles's law?Charles law gave the relation between the temperature and the volume of gas. According to him, the volume of the gas is directly proportional to the temperature of the gas.
Given,
Initial volume = 75.9 L
Initial temperature = 300 K
Final volume = ?
Final temperature = 423 K
The relation between the temperature and the volume of the two gases is shown as:
[tex]\rm \dfrac{V_{1}}{T_{1}} = \rm \dfrac{V_{2}}{T_{2}}[/tex]
Substituting values in the above equation:
[tex]\begin{aligned}\rm V_{2}& = \rm \dfrac{T_{2}V_{1}}{T_{1}}\\\\&= \dfrac{57.9\times 423}{300}\\\\&= 81.64 \;\rm L \end{aligned}[/tex]
Therefore, 81.64 L is the final volume at 423 K.
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Determine the molecular mass ratio of two gases whose rates of effusion have a ratio of 16 : 1.
Answers
rate of effusion of gas1 : rate of effusion of gas = 16 : 1.
rate of effusion of gas1 = 1/√M(gas1).
rate of effusion of gas2 = 1/√M(gas2).
rate of effusion of gas1 = rate of effusion of gas2 · 16.
1/√M(gas1) = 1/√M(gas2) · 16 /².
1/M(gas1) = 1/M(gas2) · 256.
M(gas1) · 256 = M(gas2).
When hydrochloric acid is added to sodium hydrogen carbonate?
Answers
Hope this helps!
The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] = 6.3×10−4m , and [hno2] = 0.21m ? be sure to express your answer in units of kj in the box below. answers without units will not be given credit.
Answers
By applying the modified Gibbs Free Energy formula with given values for equilibrium constants, atom concentrations, and other parameters, we find ΔG for nitrous acid at 25°C to be around 27.94 KJ/mol.
Explanation:In chemistry, the Gibbs free energy (ΔG) is calculated using the equation ΔG = -RTlnK, where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. However, since we're given Ka, the equation must be adapted. Therefore, we use ΔG = -RTlnKa + RTlnQ, where Q is the reaction quotient given by [NO₂⁻][H⁺] / [HNO₂].
Inserting the given values, such as Ka = 4.5×10⁻⁴, R (in appropriate units) as 0.0083145 KJ/(mol.K), T as 298.15K (25°C in Kelvin), and Q = ([NO₂⁻][H⁺]) / [HNO₂] = (6.3×10⁻⁴ × 5.9×10−2) / 0.21, we can now solve for ΔG. Doing the math, we find that ΔG ≈ 27.94 KJ/mol.
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A strontium-90 atom that has lost 2 electrons has ________ protons, ________ neutrons, and ________ electrons.
Answers
For an atom, the atomic number is equal to number of protons and number of electrons. Mass number is sum of number of protons and neutrons in nucleus of an atom. Atomic number is denoted by symbol Z and mass number is denoted by symbol A.
The atomic number of strontium is Z=38 and mass number is A=90.
Now, [tex]Z=n_{p}=n_{e}=38[/tex]
Also, [tex]A=n_{p}+n_{n}=90[/tex]
Putting the value of [tex]n_{p}[/tex] in above equation,
[tex]A=38+n_{n}=90[/tex]
Or,
[tex]n_{n}=90-38=52[/tex]
Thus, number of neutrons are 52.
Now, after losing two electrons, number of protons and neutrons remains the same but number of electrons becomes 38-2=36
Therefore, a strontium-90 atom that has lost 2 electrons has 38 protons, 52 neutrons and 36 electrons.
A strontium-90 atom that has lost two electrons has 38 protons, 52 neutrons, and 36 electrons.
Explanation:A strontium-90 atom originally has 38 protons, 52 neutrons, and 38 electrons. If it has lost 2 electrons, then it has 36 electrons left. So, a strontium-90 atom that has lost 2 electrons has 38 protons, 52 neutrons, and 36 electrons. The atomic number of strontium is 38, which tells us its number of protons, and by subtracting this number from the atomic mass, we get the number of neutrons. The number of electrons in a neutral atom should be equal to the number of protons, but if the atom becomes an ion and loses electrons in the process, the number of electrons decreases.
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Consider the electrolysis of molten barium chloride (bacl2). write the half-reactions. include the states of each species.
Answers
BaCl₂(l) → Ba(l) + Cl₂(g),
but first ionic bonds in this salt are separeted because of heat:
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).
Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).
Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.
The anode is positive and the cathode is negative.
What is the percent of MgSO4 in magnesium sulfate heptahydrate?
48.83 %
51.17 %
95.43 %
86.98 %
Answers
one mole of MgSO4.7H2O weighs 246.4696 g
in that one mole are seven moles of H2O, weighing 126.1036 g
The percentage of water is this calculation:
(126.1036 / 246.4696) * 100 = 51.17%
elaborate on the difference in natural occurrences between nuclear fission and nuclear fusion reactions. A) Neither nuclear fission nor nuclear fusion reactions naturally occur on earth or outer space. B) Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars. C) Both nuclear fission and nuclear fusion reactions readily naturally occur in the pressure of the earth's core. D) Nuclear fusion reactions do not naturally occur while nuclear fission reactions occur in the volcanoes.
Answers
Option B: Nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.
Nuclear fission reaction is defined as formation of two or more atoms by splitting of large atoms. On the other hand, nuclear fusion reaction is formation of large atom from small atoms.
The splitting of large atoms into small atoms (nuclear fission) generally does not occur naturally because it requires high speed neutrons and critical mass of the substance undergoing fission.
Nuclear fusion reaction occurs in stars such as sun because it requires extremely high energy to bring two more proton closer to each other by overcoming electronic repulsion.
Therefore, nuclear fission reactions rarely occur naturally while nuclear fusion reactions occur in the stars.
Use appropriate data to calculate δg∘ for the reaction. 2mno−4(aq)+cd(s)→2mno2−4(aq)+cd2+(aq) express your answer to three significant figures and include the appropriate units.
Answers
Reaction at cathode: 2MnO4- + 2e- → MnO4^2-, Eo = 1.5 v
Net reaction: Cd + 2MnO4- ⇆ Cd2+ + MnO4^2-
Net cell representation: Cd/Cd2+// MnO4-/MnO4^2-
Now, standard EMF of cell = E(o)cell = Er - El
= 1.5 - (-0.403)
= 1.903 v
Now, ΔGo = -nFE(o)cell
where n = number of electrons = 2
F = faraday's constant = 96500
ΔGo = - 2 X 96500 X 1.903
= 367.27 kJ
Answer:
∆G° = -185 kJ
Explanation:
First, use the half-reaction method:
OX: Cd ⟶ Cd2+ + 2e- where E° = -0.40 V (anode)
RED: 2MnO-4 + e- ⟶ 2MnO2-4 where E° = 0.56 V (cathode)
Balance the chemical equation with the correct stoichiometric coefficients:
2 × (2MnO-4 + e- ⟶ 2MnO2-4) = 4MnO-4 + 2e- ⟶ 4MnO2-4
1 × (Cd ⟶ Cd2+ + 2e-) = Cd ⟶ Cd2+ + 2e-
Cancel out e- on both sides to get:
4MnO-4 + Cd ⟶ 4MnO2-4 + Cd2+
Using this balanced equation, we can determine:
number of moles of electrons exchanged in the cell reaction, n = 2
E°cell = E°cathode - E°anode = 0.56 - (-0.40) = 0.96 V
F, Faraday's constant: 96485 / mol e-
∆G° = -nFE°cell = -(2)(96485)(0.96)
∆G° = -185251.2 J = -185 kJ
A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate the freezing point of this solution. [kfp(benzene) = 5.12°c/m, kbp(benzene) = 2.53°c/m] (the boiling point and freezing point of benzene are 80.1°c and 5.5°c, respectively).
Answers
PV = nRT
when P is the pressure = 748 mmHg/760 = 0.984 atm
V is the volume = 4 L
R is ideal gas constant = 0.0821
T is the temperature in Kelvin = 300 K
∴ n = 0.984atm*4L/0.0821*300
= 0.1598 moles
when the concentration = moles * (1000g / mass)
= 0.1598 * (1000g / 58 g )
= 2.755 M
when the freezing point = 5.5 °C
and Kf = - 5.12 °C/m
∴ the freezing point for the solution = 5.5 °C + (Kf*m)
= 5.5 °C - (5.12°C/m * 2.755m)
= -8.6 °C
The new freezing point of the solution is -8.63 °C.
Using the ideal gas law, PV = nRT, we need to convert 4.00 L of gas at 27°C (which is 300 K) and 748 mmHg to moles of gas.
P = 748 mmHg / 760 mmHg atm = 0.984 atm
[tex]n = \frac{PV}{RT} = \frac{0.984 \, \text{atm} \times 4.00 \, \text{L}}{0.0821 \, \text{L atm / K mol} \times 300 \, \text{K}} = 0.160 \, \text{mol}[/tex]
Calculate molality:
Molality (m) = moles of solute / kg of solvent = 0.160 mol / 0.0580 kg = 2.759 m
Calculate freezing point depression (ΔTf):
ΔTf = i * Kf * m
For non-electrolytes, i = 1.
ΔTf = 1 * 5.12 °C/m * 2.759 m = 14.13 °C
Determine the new freezing point:
The freezing point of pure benzene is 5.5 °C.
New freezing point = 5.5 °C - 14.13 °C = -8.63 °C
Thus, the freezing point of the solution is -8.63 °C.
Wolff-kishner reduction (hydrazine, koh, ethylene glycol, 130°c) of the compound shown gave compound
a. treatment of compound a with m−chloroperoxybenzoic acid (mcpba) gave compound b, which on reduction with lithium aluminum hydride gave compound
c. oxidation of compound c with chromic acid gave compound d (c9h14o). identify compounds a through d in this sequence, and do not indicate stereochemistry.
Answers
Step one: Reduction:
The carbonyl group of given compound on reduction using Wolf Kishner reagent converts the carbonyl group into -CH₂- group.
Step two: Epoxidation:
The double bond present in starting compound when treated with m-CPBA (meta-Chloroperoxybenzoic acid) gives corrsponding epoxide.
Step three: Reduction:
The epoxide is reduced to alcohol on treatment with Lithium Aluminium Hydride (LiAlH₄) followed by hydrolysis.
Step four: Oxidation:
The hydroxyl group (alcohol) is oxidized to carbonyl (ketonic group) using oxidizing agent Chromic acid (H₂CrO₄).
What are the boiling point and freezing point of a 3.47 m solution of naphthalene in benzene? (the boiling point and freezing point of benzene are 80.1°c and 5.5°c respectively. the boiling point elevation constant for benzene is 2.53°c/m, and the freezing point depression constant for benzene is 5.12°c/m.)?
Answers
ΔTb = Kb X m
where Kb = boiling point elevation constant = 2.53°c/m (for benzene)
m = molality of solution = 3.47 m (given)
∴ ΔTb = 2.53 X 3.47
= 8.779 oC
But, boiling point of benzene = 80.1 oC
∴ Boiling point of solution = 88.879 oC
Now, Depression in freezing point = ΔTf = Kf X m
where, Kf = freezing point depression constant = 5.12°c/m (for benzene)
∴ ΔTf = 5.12 X 3.47
= 17.766 oC
But freezing point of benzene = 5.5 oC
∴ Freezing point of solution = -12.266 oC
The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.
The correct answer is -12.266.
What is the boiling point?The boiling point of a liquid varies according to the applied pressure; the normal boiling point is the temperature at which the vapor pressure is equal to the standard sea-level atmospheric pressure (760 mm [29.92 inches] of mercury). At sea level, water boils at 100° C.Elevation in boiling point is mathematically expressed as
[tex]Tb = Kb * m[/tex]
Where
Kb = boiling point elevation constant = 2.53°c/m (for benzene)m = molality of solution = 3.47 m (given)ΔTb[tex]= 2.53 * 3.47 = 8.779 oC[/tex]
But, the boiling point of benzene = 80.1 oC
Boiling point of solution = 88.879 oC
Now, Depression in freezing point = ΔTf[tex]= Kf * m[/tex]
where,
Kf = freezing point depression constant = 5.12°c/m (for benzene)ΔTf =[tex]5.12 X*3.47 = 17.766 oC[/tex]
But the freezing point of benzene = 5.5 oC.
Freezing point of solution = -12.266 oC.
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Draw any one of the skeletal structures of a 2° alkyl bromide having the molecular formula of c6h13br and two stereogenic centers. indicate chirality by using wedge and hashed wedge notation. lone pairs do not need to be shown.
Answers
The skeletal structures of a 2° alkyl bromide with two stereogenic centers is shown below.
What are the stereogenic centers in 2° alkyl bromide ?
A carbon atom in a molecule that is linked to four separate substituents is referred to as a stereogenic center, also referred to as a chiral center. Because secondary carbon atoms normally have two identical alkyl groups and two hydrogen atoms linked to them.
The two of the substituents (two alkyl groups) are the same. Tetrahedral carbon compounds with four distinct substituents are frequently where chirality occurs.
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N aqueous solution of sodium hydroxide is standardized by titration with a 0.154 m solution of hydrochloric acid. if 17.5 ml of base are required to neutralize 17.6 ml of the acid, what is the molarity of the sodium hydroxide solution?
Answers
NaOH + HCl --> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
Number of HCl moles reacted - 0.154 mol/L x 0.0176 L = 0.00271 mol
the number of NaOH moles reacted = number of HCl moles reacted
number of NaOH moles reacted - 0.00271 mol
number of NaOH moles in 17.5 mL - 0.00271 mol
therefore NaOH moles in 1000 mL - 0.00271 mol / 17.5 mL x 1000 mL
molarity of NaOH - 0.155 M
When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction caco3(s)→cao(s)+co2(g) what is the mass of calcium carbonate needed to produce 61.0 l of carbon dioxide at stp? express your answer with the appropriate units?
Answers
Chemical reaction: CaCO₃ → CaO + CO₂.
V(CO₂) = 61.0 L.
n(CO₂) = V(CO₂) ÷ Vm.
n(CO₂) = 61 L ÷ 22.4 L/mol.
n(CO₂) = 2.723 mol.
From chemical reaction: n(CO₂) : n(CaCO₃) = 1 : 1.
n(CaCO₃) = 2.273 mol.
m(CaCO₃) = n(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 2.273 mol · 100 g/mol.
m(CaCO₃) = 227.3 g.
For producing 61 L of [tex]\rm CO_2[/tex], 227.23 grams of [tex]\rm CaCO_3[/tex]. is required.
Any gas occupies 22.4 L/mol space at STP.
So, 61.0 L of gas will be;
Moles of [tex]\rm CO_2[/tex] = [tex]\rm \frac{61.0}{22.4}[/tex]
Moles of [tex]\rm CO_2[/tex] = 2.723 moles
From the reaction,
1 mole of [tex]\rm CO_2[/tex] has been produced by 1 mole of [tex]\rm CaCO_3[/tex].
2.723 moles of [tex]\rm CO_2[/tex] has been produced by 2.723 mole of [tex]\rm CaCO_3[/tex].
Mass = [tex]\rm moles\;\times\;molecular\;weight[/tex]
Mass of [tex]\rm CaCO_3[/tex]. = 2.723 [tex]\times[/tex] 100
Mass of [tex]\rm CaCO_3[/tex]. = 227.23 grams.
For producing 61 L of [tex]\rm CO_2[/tex], 227.23 grams of [tex]\rm CaCO_3[/tex]. is required.
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a student reacts 0.600 g of lead (ii) nitrate with 0.850 g of potassium iodide
write and balance the chemical equation.
how many miles of Pb(No3)2 were used?
how many miles of KI were used?
how many miles of PbI2 would form based on the moles of Pb(No3)2 used?
how many miles of PbI2 would form based on the moles of KI used?
which is the limiting reactant?
what is the theoretical yeild of PbI2 in grams?
if the student obtained 0.475 grams of PbI2 product after conducting it by filtration, what is the percent yeild of PbI2 obtained ?
Answers
the reaction that takes place is
lead nitrate reacting with potassium iodide to form lead iodide and potassium nitrate
balanced chemical equation for the reaction is as follows
Pb(NO₃)₂ + 2KI ----> PbI₂ + 2KNO₃
Q2)
mass of lead nitrate present - 0.600 g
number of moles = mass present / molar mass
number of moles - 0.600 g / 331.2 g/mol = 0.00181 mol
Q3)
mass of potassium iodide present - 0.850 g
number of moles = mass present / molar mass
number of moles of potassium iodide = 0.850 g / 166 g/mol = 0.00512 mol
Q4)
we have to calculate the number of moles of PbI₂ formed based on the number of moles of Pb(NO₃)₂ present assuming the whole amount of Pb(NO₃)₂ was used up
stoichiometry of Pb(NO₃)₂ to PbI₂ is 1:1
number of Pb(NO₃)₂ moles reacted - 0.00181 mol
therefore number of PbI₂ moles formed - 0.00181 mol
Q5)
next we have to calculate the number of moles of PbI₂ formed based on the amount of KI moles present , assuming all the moles of KI were used up in the reaction
stoichiometry of KI to PbI₂ is 2:1
number of moles of KI reacted - 0.00512 mol
then number of moles of PbI₂ formed - 0.00512 x 2 = 0.0102 mol
0.0102 mol of PbI₂ is formed
Q6)
limting reactant is the reactant that is fully consumed during the reaction. the amount of product formed depends on the amount of limiting reactant present
if lead nitrate is the limiting reactant
if 1 mol of Pb(NO₃)₂ reacts with 2 mol of KI
then 0.00181 mol of Pb(NO₃)₂ reacts with - 2 x 0.00181 mol of KI = 0.00362 mol
but 0.00512 mol of KI is present and only 0.00362 mol are required
therefore KI is in excess and Pb(NO₃)₂ is the limiting reactant
Pb(NO₃)₂ is the limiting reactant
Q7)
then the amount of PbI₂ formed depends on amount of Pb(NO₃)₂ present
therefore number of moles of PbI₂ formed is based on number of Pb(NO₃)₂ moles present
as calculated in Question number 4 - Q4
number of PbI₂ moles formed - 0.00181 mol
mass of PbI₂ formed - 461 g/mol x 0.00181 mol = 0.834 g
mass of PbI₂ formed - 0.834 g
Q8)
actual yield obtained is not always equal to the theoretical yield . therefore we have to find the percent yield. This tells us the percentage of the theoretical yield that is actually obtained after the experiment
percent yield = actual yield / theoretical yield x 100 %
percent yield = 0.475 g / 0.834 g x 100 % = 57.0 %
percent yield of lead iodide is 57.0 %
How much 0.100 m hcl is required to completely neutralize 20.0 ml of 0.250 m naoh?
Answers
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of NaOH moles reacted - 0.250 mol/L x 0.0200 L = 0.005 mol
therefore number of HCl moles reacted - 0.005 mol
molarity of given HCl solution - 0.100 M
number of HCl moles in 1 L - 0.100 mol
then 0.005 mol of HCl in - 0.005 mol / 0.100 mol/L = 50.0 mL
volume of HCl required is 50.0 mL
To neutralize 20.0 mL of 0.250 M NaOH, you will need 50.0 mL of 0.100 M HCl.
Explanation:To determine how much 0.100 M HCl is required to neutralize 20.0 mL of 0.250 M NaOH, we can use the balanced equation for the reaction between HCl and NaOH: HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq). The mole ratio between HCl and NaOH is 1:1, meaning that 1 mole of HCl reacts with 1 mole of NaOH. We can use this ratio to calculate the amount of HCl needed.
First, find the number of moles of NaOH:
0.250 M NaOH x 0.0200 L = 0.005 moles NaOH
Since the mole ratio between HCl and NaOH is 1:1, we need 0.005 moles of HCl to neutralize the NaOH.
Now, calculate the volume of 0.100 M HCl needed to contain 0.005 moles:
0.005 mol HCl / 0.100 mol/L = 0.050 L = 50.0 mL
Therefore, 50.0 mL of 0.100 M HCl is required to completely neutralize 20.0 mL of 0.250 M NaOH.
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What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?
Answers
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required
Which of the two substances would have the higher boiling point ch4 or c?
Answers
c boiling point: 8,721 F (4,827 C)
therefore: carbon has a higher boiling point
ch4 is also a gas at room temp which is why it’s boiling point is so low.
c is a solid at room temp
Carbon has a higher boiling point than [tex]CH_4[/tex].
What is Boiling Point?The boiling point of a substance is defined as the temperature at which the vapor pressure of the liquid becomes equal to the pressure surrounding the liquid and the liquid changes into vapor. This point of liquid varies depending on the surrounding environmental pressure.
The boiling point is the temperature at which the pressure exerted by the surroundings on a liquid is equal to the pressure exerted by the liquid's vapor, in which case the liquid changes to vapor without additional heat raising the temperature.
For above given example,
[tex]CH_4[/tex] is a gas at room temperature having a boiling point of -258.7 F (-161.5 C) which is why it’s boiling point is so low while Carbon is a solid at room temperature having a boiling point of 8,721 F (4,827 C)
Thus, Carbon has a higher boiling point than [tex]CH_4[/tex].
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What is the average kinetic energy kavg of the hydrogen molecules at a temperature of 100k? assume that the molecules have only three degrees of freedom at this temperature?
Answers
The average kinetic energy of hydrogen molecules at 100 K with three degrees of freedom can be calculated using the formula kavg = 3/2 * k * T, where k is Boltzmann's constant and T is the temperature in Kelvin.
The average kinetic energy kavg of hydrogen molecules at a temperature of 100 K with three degrees of freedom can be calculated using the formula:
kavg = 3/2 * k * T
where k is Boltzmann's constant (1.38 x 10^-23 J/K) and T is the temperature in Kelvin. Plugging in the values:
kavg = 3/2 * 1.38 x [tex]10^{-23}[/tex] J/K * 100 Kkavg = 2.07 x [tex]10^{-21}[/tex] JA solution has [oh−] = 2.3×10−2 m . use the ion product constant of water kw=[h3o+][oh−] to find the [h3o+] of the solution.
Answers
Answer:
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
Explanation:
Concentration of hydroxide ions = [tex][OH^-]=2.3\times 10^{-2}[/tex]
Concentration of hydroxide ions = [tex][H_3O^+]=?[/tex]
[tex]H_2O+H_2O\rightleftharpoons H_3O^++OH^-[/tex]
The ionic product of water is given as:
[tex]K_w=[H_3O^+][OH^-][/tex]
The value of ionic product of water, [tex]K_w=1\times 10^{-14}[/tex]
[tex]K_w=1\times 10^{-14}=[H_3O^+][OH^-][/tex]
[tex]1\times 10^{-14}=[H_3O^+]\times 2.3\times 10^{-2}M[/tex]
[tex][H_3O^+]=\frac{1\times 10^{-14}}{2.3\times 10^{-2}}=4.35\times 10^{-13} M[/tex]
The concentration of hydronium ions in the solution is [tex]4.35\times 10^{-13} M[/tex].
The concentration of hydronium, ion [H₃O⁺] in the solution containing 2.3×10⁻² M concentration of hydroxide ion, [OH⁻] is 4.35×10⁻¹³ M
How to determine the concentration of hydronium, ion [H₃O⁺]?The following data were obtained from the question given above:
Concentration of hydroxide ion, [OH⁻] = 2.3×10⁻² MConcentration of hydronium, ion [H₃O⁺] = ?[H₃O⁺] × [OH⁻] = 10¯¹⁴
[H₃O⁺] × 2.3×10⁻² = 10¯¹⁴
Divide both side by 2.3×10⁻²
[H₃O⁺] = 10¯¹⁴ / 2.3×10⁻²
= 4.35×10⁻¹³ M
Thus, we can conclude that the concentration of hydronium ion, [H₃O⁺] in the solution is 4.35×10⁻¹³ M
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A 3.00-l flask is filled with gaseous ammonia, nh3. the gas pressure measured at 15.0 ∘c is 2.15 atm . assuming ideal gas behavior, how many grams of ammonia are in the flask? express your answer to three significant figures and include the appropriate units.
Answers
PV = nRT
where
P - pressure - 2.15 atm x 101 325 = 2.18 x 10⁵
V - volume - 3.00 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 15.0 °C + 273 = 288 K
substituting these values in the equation
2.18 x 10⁵ Pa x 3.00 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K
n = 0.273 mol
number of moles of NH₃ is 0.273 mol
molar mass of NH₃ - 17.0 g/mol
mass pf ammonia present - 0.273 mol x 17.0 g/mol = 4.64 g
mass of NH₃ present is 4.64 g
How many atoms are in 1.00 gram of Ag? Express your answer in scientific notation. _____× 10^___ atoms
Answers
calculate the moles of Ag = mass/molar mass
= 1g/ 107.87 g/mol= 9.27 x10 ^-3 moles
by use of Avogadro law constant 1 mole = 6.02 x10^23 atoms
what about 9.27 x10 ^-3 moles
= (9.27 x10^-3 moles/ 1mole) x6.02 x10 ^23 atoms = 5.581 x10^ 21 atoms
Equilibrium is established in a reversible reaction when:
a) the [product] = [reactants]
b) rate of reaction of products = rate of reaction of reactants
c) all the reactants dissolve or dissociate
d) when product are no longer produced
Answers
Equilibrium in a reversible reaction is established when the rate of the forward reaction (reactants transforming into products) equals the rate of the backward reaction (products reconverting into reactants). This indicates that the amounts of products and reactants are no longer changing over time.
Explanation:In a reversible reaction, equilibrium is established when option (b) is correct: the rate of the forward reaction (reactants transforming into products) becomes equal to the rate of the backward (or reverse) reaction (products converting back into reactants). This does not necessarily mean the concentrations of the products and reactants are equal. Rather, it means the amounts of products and reactants are no longer changing over time, demonstrating a state of dynamic equilibrium. The equilibrium can shift depending on external factors and conditions such as temperature, pressure, or concentration changes, which is described by Le Chatelier's Principle.
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Which would be most useful if you were lost in the Sahara desert? A. Landsat satellite B. OSTM/Jason-2 satellite C. Global Positioning System D. topographic map of Africa