Answer:
884.8 mL
Explanation:
If we have STP condition (Standard temperature and pressure), we will have the following relationship:
[tex]1~mol=22.4~L[/tex] or [tex]1~mol=22400~mL[/tex]. With this in mind we can do the conversion:
[tex]0.0395~mol\frac{22400~mL}{1~mol}=884.8~mL[/tex]
In the 0.0395 mol of fluorine gas, we will have 884.8 mL of gas at STP conditions.
Just as carbon dating is used to measure the age of organic material, Argon-40 can be used to measure the age of rocks. A volcanic eruption melts a large chunk of rock, and all gasses are expelled. After cooling, Argon-40 accumulates from the ongoing decay of potassium-40 in the rock (t_1/2 = 1.25E9 years). When a piece of rock is analyzed, it is found to contain 1.38 mmol of potassium-40 and 1.14 mmol of Argon-40. How long did the rock cool?
Answer:
3.77 mg of K-40 decayed into Ar-40.
Data:
1) K-40, Ca-40, Ar-40: all three have the same atomic mass
2) 90% of the potassium-40 will decay into calcium-40
3) 10% of the potassium-40 will decay into argon-40.
4) K-40 inside the rock = 0.81 mg
5) Ar-40 trapped = 0.377 mg
Soltuion:
1) 0.377 mg of Ar-40 is the 10% of the mass of the K-40 that decayed
=> x * 10% = 0.377 mg => x * 0.1 = 0.377mg
=> x = 0.377 mg / 0.1 = 3.77 mg
That means that 3.77 mg of K-40 decayed into Ar-40. And this is the answer to the question.
Additionaly, you can analyze the content of all K-40 and Ca-40, to understand better the case.
2) The mass of the K-40 that decayed into Ca-40 is 9 times (ratio 9:1) the amount that decayed into Ar-40 =>
mass of K-40 that decayed into Ca-40 = 9 * 0.377 = 3.393 mg
3) Total amount of K-40 that decayed = amount that decayed into Ar-40 + amount that decayed into Ca-40 = 0.377mg + 3.393mg = 3.77 mg
4) Original amount of K-40 = amount of K-40 that decayed + amount of K-40 present in the rock = 3.77mg + 0.81 mg = 4.58 mg
5) amount of K-40 that decayed into Ar-40 as percent
% = [3.77 mg / 4.58mg] * 100 = 82.31 %.
Br2(g) + 3 F2(g) ⇄ 2 BrF3(g) Kp = 5.4 × 108 0.30 atm of Br2 and 0.60 atm of F2 are placed in a 3.0 L container and the system is allowed to reach equilibrium. Calculate the pressureof Br2 at equilibrium.
Answer:
The pressure at equilibrium of Br₂ is 0.10048 atm
Explanation:
Based on the reaction:
Br₂(g) + 3 F₂(g) ⇄ 2 BrF₃(g)
Kp is defined as:
[tex]Kp = \frac{P_{BrF_3}^2}{P_{Br_2}P_{F_2}^3}[/tex] = 5.4x10⁸
If initial pressures of Br₂ and F₂ are 0.30atm and 0.60atm respectively, the pressures in equilibrium are:
Br₂ = 0.30atm - X
F₂ = 0.60atm - 3X
BrF₃ = 2X
Replacing in Kp formula:
5.4x10⁸ = [2X]² / [0.30atm - X] [0.60atm - 3X]³
5.4x10⁸ = 4X² / [0.30 - X] [0.216 - 3.24 X + 16.2 X² - 27 X³]
5.4x10⁸ = 4X² / 0.0648 - 1.188 X + 8.1 X² - 24.3 X³ + 27 X⁴
Solving for X:
X = 0.3000 → False answer because produce negative concentrations.
X = 0.19952. Replacing in equation of Br₂:
Br₂ = 0.30atm - 0.19952atm = 0.10048 atm
A galvanic (voltaic) cell consists of an electrode composed of magnesium in a 1.0 M magnesium ion solution and another electrode composed of silver in a 1.0 M silver ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C.
Answer:
3.17 V
Explanation:
The cell is operating under standard conditions. These standard conditions include; that the reaction takes place at 298 Kelvin (room temperature), the pressure of the system is 1 atmosphere (standard pressure), and the solutions have a Molarity of 1.0 M for both the anode and cathode solutions. All these conditions are satisfied in the cell under review in the question.
Hence;
E°anode (magnesium)= -2.37 V
E°cathode (silver) = 0.80 V
E°cell= E°cathode -E°anode
E°cell= 0.80-(-2.37)
E°cell= 0.80 + 2.37
E°cell= 3.17 V
Hence the standard cell potential of this cell at 25°C is 3.17 V
Answer:
The standard potential at 25ºC is 3.17 V.
Explanation:
The anode in a galvanic cell is the electrode at which oxidation occurs and the cathode is the electrode at which reduction occurs.
The overall cell reaction will be the sum of two half-cell reactions. The standard reduction potentials are:
Mg²⁺ (1.0 M) + 2e⁻ → Mg (s) Eº = -2.37
Ag⁺ (1.0 M) + e⁻ → Ag (s) Eº= +0.80
Since the reactants are in their standard states (1.0 M) and at 25ºC we can write the half-cell reactions as follows:
Anode (oxidation): Mg (s) → Mg²⁺ (1.0 M) + 2e⁻
Cathode (reduction): 2Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s)
Overall: Mg (s) +2 Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s) + Mg²⁺ (1.0 M) + 2e⁻
In order to balance the overall equation we multiply the reduction of Ag⁺ by 2. We can do so because, as an intensive property, E° is not affected by this procedure.
The standard emf of the cell, E°cell , which is composed of a contribution from the anode and a contribution from the cathode, is given by:
[tex] Eº cell = Eº cathode - Eº anode [/tex]
[tex] Eº cell = EºAg⁺/Ag - Eº Mg²⁺/Mg [/tex]
[tex] Eº cell =0.80 V - (-2.37 V) [/tex]
Eº cell = 3.17 V
Aqueous solutions of ammonium oxalate and lithium acetate react.
Answer:
The reaction is a double displacement reaction
Explanation:
Let us consider the reaction equation of the reaction between ammonium oxalate and lithium acetate.
(NH4)2C2O4(aq) +2 CH3COOLi(aq) -------> 2NH4CH3CO2(aq) + Li2C2O4(s)
This is a displacement reaction. A double displacement reaction is a type of reaction in which two reactants exchange their ions to form two new compounds. Double displacement reactions usually lead to the formation of a solid product which is also called a precipitate.
The general form of a Double displacement reaction is of the format:
AB + CD → AD + CB
Where A,B,C and D represents different ions respectively.
A double displacement reaction can also be referred to as salt metathesis reaction, double replacement reaction, exchange reaction, or a double decomposition reaction, although the latter term is more strictly used when one or more of the reactants does not dissolve in the solvent.
There is a container on the lab bench containing some unknown chemical. You should:
a. Dump it, wash the glassware. Some people are slobs.
b. Move it out of the way in case it is dangerous. Otherwise, not your problem.
c. Leave it. The rightful owner will claim it eventually.
d. Tell your lab supervisor.
Answer:
D
Explanation:
Your supervisor is in charge of you so therefore you must always tell your supervisor first.
Answer:
C.
Explanation:
Because you might get hurt, could be dangerous.
In the laboratory hydrogen gas is usually made by the following reaction:Zn(s)+2HCl(aq)=H2(g)+ZnCl2(aq)How many liters of H2 gas , collected over water at an atmospheric pressure of 752 mmHg and a temperture of 21 Co, can made from 3.566 g of Zn and excess HCl?the partial pressure of water vapor is 18.65mmHg at 21C0.A) 0.68LB) 2.72LC) 1.36LD) 1.33LE) 0.0975L
Answer:
The correct answer is option C, that is, 1.36 L.
Explanation:
The reaction mentioned in the question is:
Zn (s) + 2HCl (aq) ⇒ H2 (g) + ZnCl2 (aq)
It is clear that one mole of zinc is generating one mole of hydrogen gas as seen in the reaction. The mass of zinc mentioned in the question is 3.566 grams, the no of moles can be determined by using the formula,
n = mass / molecular mass
The molecular mass of zinc is 65.39 g/mol, now putting the values in the formula we get,
n = 3.566 g/ 65.39 g/mol
= 0.0545 mol
Based on the question, the partial pressure of water vapor is 18.65 mmHg and the atmospheric pressure is 752 mmHg. Therefore, the pressure of hydrogen gas will be,
Pressure of hydrogen gas (H2) = 752 mmHg - 18.65 mmHg
= 733.35 mmHg
The liters of hydrogen gas produced can be calculated by using the equation, PV =nRT
R is the gas constant, having the value 62.36 L mmHg/K/mol and T is the temperature (273 + 21 = 294 K).
Now putting the values in the equation we get,
733.35 mmHg * V = 0.0545 mol * 62.36 mmHg/K/mol * 294 K
= 999.19 L / 733.35
= 1.36 L
Hence, the volume of hydrogen gas is 1.36 L.
How much of a 0.250 M lithium hydroxide is required to neutralize 20.0 mL of 0.345M chlorous acid?
Answer:
27.6mL of LiOH 0.250M
Explanation:
The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:
LiOH + HClO₂ → LiClO₂ + H₂O
That means, 1 mole of hydroxide reacts per mole of acid
Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:
0.0200L ₓ (0.345mol / L) = 6.90x10⁻³ moles of HClO₂
To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:
6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =
27.6mL of LiOH 0.250MThe pressure gauge on a tank registers the gauge pressure, which is the difference between the interior pressure and exterior pressure. When the tank is full of oxygen (O2), it contains 15.0 kg of the gas at a gauge pressure of 36.0 atm. Determine the mass of oxygen that has been withdrawn from the tank when the pressure reading is 21.6 atm. Assume the temperature of the tank remains constant.
Answer:
GOOD DAY ON YOUR QUIZ
Explanation:
Draw a chair conformation for cis-1-bromo-2,4-dimethylcyclohexane showing equatorial and axial positions at the carbon atoms bearing the substituents (all substituents are at cis configuration) g
Answer:
See attachment.
Explanation:
Mono-substituted cyclohexanes are more stable with their substituents in an equatorial position. However, with poly-substituted cyclohexanes, the situation is more complex since the steric effects of all substituents have to be taken into account. In this case, you can see that the interconversion is shifted towards the conformation in the bottom because there is less tension between the substituents.
1. The chemical formula for sodium bromide is NaB.*
True
False
Answer:
NaBr
Explanation:
Bromine's symbol is Br, not B, for B is for Boron
Structurwl formula for (CH3)2CHCH2OH
Answer:
Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol)
Explanation:
Isobutanol (IUPAC nomenclature: 2-methylpropan-1-ol) is an organic compound with the formula (CH3)2CHCH2OH (sometimes represented as i-BuOH). Hope this helped!
Water is intentionally present in a state where its molecules are far apart during a change of state is molecule slow down which change of state has most likely taking place
Answer:
melting
Explanation:
Each of these geometric shapes has a different number of sides. Arrange the shapes in order from the shape with the greatest number of sides to the shape with the fewest number of sides. Rank these shapes from greatest to fewest number of sides. To rank items as equivalent, overlap them.
triangle , square, rectangle, octagon, hexagon, pentagon
1. Octagon (8 sides)
2. Hexagon (6 sides)
3. Pentagon (5 sides)
4. Square/Rectangle (4 sides)
5. Triangle (3 sides)
Hope this helps :)
If the specific heat of gold is 0.13 J/gC, what is the amount of energy (heat) required to raise 30.0g of gold from 15 degrees Celsius to 41 degrees Celsius?
Answer: The amount of energy required is 101 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 gold = 30.0 g
c = specific heat capacity of gold = [tex]0.13J/g^0C[/tex]
Initial temperature of the water = [tex]T_i[/tex] = 15°C
Final temperature of the water = [tex]T_f[/tex] = 41°C
Change in temperature ,[tex]\Delta T=T_f-T_i=(41-15)^0C=26^0C[/tex]
Putting in the values, we get:
[tex]Q=30.0g\times 0.13J/g^0C\times 26^0C[/tex]
[tex]Q=101J[/tex]
The amount of energy required is 101 Joules
When a 20.0 g sample of an unknown compound is dissolve in 500. G of benzene, the freezing point of the resulting solution is 3.77 o C. The freezing point 9f pure benzene is 5.48 o C and Kf for benzene is 5.12 o C/m. Calculate the molar mass of the unknown
Answer:
The correct answer is 116.3 g/mol
Explanation:
When a solute is added to a solvent, the freezing point decreases. That is called freezeing point depression and is given by the following expression:
ΔTf= Kf x m
Where ΔTf is the decrease in freezing point (final temperature - initial temperature), Kf is the freezeing point constant, m is the molality of the solution.
We have the following data:
Kf= 5.12ºC/m
ΔTf= 5.48ºC - 3.77ºC = 1.71ºC
We can calculate the molality of the solution from the mathematical expression for freezeing point as follows:
m = ΔTf/Kf = 1.71ºC/(5.12º C/m)= 0.334 m
The molality is the number of moles of solute in 1 kg of solvent. The solute was added to 500 g of solvent (benzene), so we can multiply the molality by the grams of solvent (500 g = 0.5 kg) to obtain the number of moles:
0.334 mol/kg x 0.5 kg = 0.172 moles
We know that the molar mass is equal to the mass of compound per mole, so we divide the mass into the moles to obtain the molar mass as follows:
MM = mass/moles= 20.0 g/0.172 moles = 116.3 g/mol
Convert 3.4 x 10^23 molecules of NaCl to grams
Answer:
[tex]m_{NaCl}=33g[/tex]
Explanation:
Hello,
In this case, in order to compute the grams of sodium chloride starting by the molecules, the first step is to compute the moles contained in the given amount of molecules by using the Avogadro's number:
[tex]n_{NaCl}=3.4x10^{23}molecules*\frac{1mol}{6.022x10^{23}molecules} =0.56mol[/tex]
Then, by using the molar mass of sodium chloride (58.45 g/mol) we can directly compute the grams:
[tex]m_{NaCl}=0.56mol*\frac{58.45g}{1mol} \\\\m_{NaCl}=33g[/tex]
Regards.
Determine the mass of CaBr2 needed to create a 500. ml solution with a concentration of 1.15 M.
Answer:
[tex]m_{CaBr_2}=115gCaBr_2[/tex]
Explanation:
Hello,
In this case, we remember molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, therefore, for the given concentration and volume, we should solve the moles of calcium bromide to subsequently compute the mass, as shown below:
[tex]M=\frac{n}{V}\\ \\n=M*V=1.15\frac{mol}{L} *500mL*\frac{1L}{1000mL}=0.575mol[/tex]
Next, since molar mass of calcium bromide is 200 g/mol, we can compute the mass:
[tex]m_{CaBr_2}=0.575mol*\frac{1mol}{200g} \\\\m_{CaBr_2}=115gCaBr_2[/tex]
Regards.
1. An isotope of cesium-137 has a half-life of 30 years. If 5.0 g of cesium-
137 decays over 60 years, how many grams will remain?
Answer:
1.25 g
Explanation:
Now we have to use the formula;
N/No = (1/2)^t/t1/2
N= mass of cesium-137 left after a time t (the unknown)
No= mass of cesium-137 present at the beginning = 5.0 g
t= time taken for 5.0 g of cesium-137 to decay =60 years
t1/2= half life of cesium-137= 30 years
Substituting values;
N/5= (1/2)^60/30
N/5= (1/2)^2
N/5= 1/4
4N= 5
N= 5/4
N= 1.25 g
Therefore, 1.25 g of cesium-137 will remain after 60 years.
Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 374 mL. Express your answer to three significant figures.
Answer:
Explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thnk me...
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanisms for this reaction are: Mechanism A (1) H2(g)+I2(g)−→k12HI(g)(one-step reaction) Mechanism B (1) I2(g)⥫⥬=k−1k12I(g)(fast, equilibrium) (2) H2(g)+2I(g)−→k22HI(g) (slow) Mechanism C (1) I2(g)⥫⥬=k−1k12I(g)(fast, equilibrium) (2) I(g)+H2(g)−→k2HI(g)+H(g) (slow) (3) H(g)+I(g)−→k3HI(g) (fast) Which of these mechanisms are consistent with the observed rate law? mechanism A mechanism C mechanism B In 1967, J. H. Sullivan showed that this reaction was dramatically catalyzed by light when the energy of the light was sufficient to break the I−I bond in an I2 molecule. Which mechanism or mechanisms are consistent with both the rate law and this additional observation? mechanism A mechanism C mechanism B
Answer:
Mechanism A and B are consistent with observed rate law
Mechanism A is consistent with the observation of J. H. Sullivan
Explanation:
In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.
In the proposed mechanisms:
Mechanism A
(1) H2(g)+I2(g)→2HI(g)(one-step reaction)
Mechanism B
(1) I2(g)⇄2I(g)(fast, equilibrium)
(2) H2(g)+2I(g)→2HI(g) (slow)
Mechanism C
(1) I2(g) ⇄ 2I(g)(fast, equilibrium)
(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)
(3) H(g)+I(g)→HI(g) (fast)
The rate laws are:
A: rate = k₁ [H2] [I2]
B: rate = k₂ [H2] [I]²
As:
K-1 [I]² = K1 [I2]:
rate = k' [H2] [I2]
Where K' = K1 * K2
C: rate = k₁ [H2] [I]
As:
K-1 [I]² = K1 [I2]:
rate = k' [H2] [I2]^1/2
Thus, just mechanism A and B are consistent with observed rate law
In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just mechanism A is consistent with the observation of J. H. Sullivan
PLEASE HELPPPP!!!
diameter & volume question
Answer:
65450 in^3
Explanation:
Someone deleted my old answer
1. Legătura covalentă este legătura chimică ce se stabileşte: a) prin punere în comun de electroni; b) prin transfer de electroni; c) prin punere în comun sau prin transfer de electroni. 2. Legătura covalentă se stabileşte între: a) două metale ; b) două nemetale; c) un metal şi un nemetal. 3. O legătură covalentă se stabileşte: a) între hidrogen şi sodiu ; b) între azot şi hidrogen; c) între aluminiu şi oxigen.. 4. Legăturile covalente sunt de două feluri: a) legături covalente simple si legaturi covalente duble; b) legături covalente polare şi legături covalente nepolare; c) nici un raspuns nu este corect. 5. Selectează substanțele în moleculele cărora atomii sunt uniți prin legătură ionică: a) oxid de fosfor P2O5; b)clorură de potasiu KCl; c) apă H2O
Answer:
1) Option A is correct.
The covalent bond is the chemical bond that is established by pooling of electrons.
2) Option B is correct.
The covalent bond is established between two non-metals.
3) Option B is correct.
A covalent bond is established between nitrogen and hydrogen.
4) Option B is correct.
Covalent bonds are of two types; polar covalent bonds and non-polar covalent bonds.
5) Option B is correct.
Select the substances in the molecules of which the atoms are united by ionic bond.
potassium chloride KCl
1) Opțiunea A este corectă.
Legătura covalentă este legătura chimică care se stabilește prin punerea în comun a electronilor.
2) Opțiunea B este corectă.
Legătura covalentă este stabilită între două nemetale.
3) Opțiunea B este corectă.
Se stabilește o legătură covalentă între azot și hidrogen.
4) Opțiunea B este corectă.
Legăturile covalente sunt de două tipuri; legături covalente polare și legături covalente nepolare.
5) Opțiunea B este corectă.
Selectați substanțele din moleculele cărora atomii sunt uniți prin legătură ionică. clorură de potasiu KCl
Explanation:
English Translation
1) The covalent bond is the chemical bond that is established: a) by pooling electrons; b) by electron transfer; c) by pooling or electron transfer.
2) The covalent bond is established between: a) two metals; b) two non-metals; c) a metal and a non-metal.
3) A covalent bond is established: a) between hydrogen and sodium; b) between nitrogen and hydrogen; c) between aluminum and oxygen.
4) Covalent bonds are of two types: a) single covalent bonds and double covalent bonds; b) polar covalent bonds and non-polar covalent bonds; c) no answer is correct.
5) Select the substances in the molecules of which the atoms are united by ionic bond: a) phosphorus oxide P2O5; b) potassium chloride KCl; c) H2O water
Solution
Covalent bonds are bonds that result from the sharing/pooling of electrons by two non-metallic atoms. When the electronegativity difference between the two atoms in a cobalent bond is substantial, a polar covalent bond results (in which the shared electrons are closer to the more electronegative atom of the pair). If the electronegativity difference isn't significant, a non-polar covalent bond ensues, in which the shared electrons are approximately equidistant from both atoms.
1) The covalent bond is the chemical bond that is established by pooling of electrons.
Already explained above.
2) The covalent bond is established between two non-metals.
Already explained too.
3) A covalent bond is established between nitrogen and hydrogen.
Of the three options, only option B contains two elements, nitrogen and hydrogen, which are both non-metals amd this satisfy the condition for the formation of a covalent bond.
4) Covalent bonds are of two types; polar covalent bonds and non-polar covalent bonds.
Already explained above.
5) Select the substances in the molecules of which the atoms are united by ionic bond.
potassium chloride KCl
Ionic bonds result from the transfer of electrons from metals (electropositive elements) to non-metals (electronegative elements). Of the 3 options provided, only potassium chloride is a coming together of a metal and a non metal.
In Romanian/In limba romana
Legăturile covalente sunt legături care rezultă din împărțirea / îmbinarea electronilor de către doi atomi nemetalici. Când diferența de electronegativitate între cei doi atomi dintr-o legătură cobalentă este substanțială, rezultă o legătură covalentă polară (în care electronii partajați sunt mai apropiați de atomul mai electronegativ al perechii). Dacă diferența de electronegativitate nu este semnificativă, apare o legătură covalentă non-polară, în care electronii partajați sunt aproximativ echidistanți de la ambii atomi.
Legăturile ionice rezultă din transferul electronilor de la metale (elemente electropozitive) la nemetalele (elemente electronegative).
Hope this Helps!!!
Sper că acest lucru vă ajută!!!!
Acetone is a solvent used in fingernail polish remover. Acetone is polar. It will form
solutions, or be soluble, in all the following except
O A. ethanol, which has molecules that have
one end with a slight charge
B. ammonia, which is made of molecules with
a slightly negative nitrogen atom and three
slightly positive hydrogen atoms
O C. water, which has oxygen atoms with a
slight negative charge
D. vegetable oil, which has atoms that share
electrons equally, so that there is no charge
difference
Read Question
Correct:
Answer:
d
Explanation:
he Punnett square shows the possible genotype combinations for the offspring of two parents. What are the genotypes of the parents? Gg and Gg GG and GG gg and gg Gg and gg
Answer:
Its A! Gg and Gg
Explanation:
0.76 g of lead(2) nitrate was dissolvedin 50.00ml of water a d treated with 25.00 ml of 0.2010M sodium sulfate inoder to determine a content of lead 2 ion. A white precipitate was formed and it was collected and dried. Calculate the amount of precipitate formed in this reaction
Answer:
Mass of precipitate (PbSO₄) formed = 0.696 g
Explanation:
The reaction of Lead II Nitrate and Sodium Sulfate is given as
Pb(NO₃)₂ (aq) + Na₂SO₄ (aq) → 2NaNO₃ (aq) + PbSO₄ (s)
The precipitate from this reaction is the PbSO₄
0.76 g of Pb(NO₃)₂ is available for reaction, we convert to number of moles
Number of moles = (Mass)/(Molar Mass)
Mass = 0.76 g
Molar mass of Pb(NO₃)₂ = 331.2 g/mol
Number of moles of Pb(NO₃)₂ = (0.76/331.2) = 0.002294686 moles = 0.002295 moles
25.00 ml of 0.2010M sodium sulfate is available for the reaction, we also conert to number of moles.
Number of moles = (Concentration in mol/L) × (Volume in L)
Concentration of Na₂SO₄ in mol/L = 0.201 M
Volume of Na₂SO₄ in L = (25/1000) = 0.025 L
Number of moles of Na₂SO₄ = 0.201 × 0.025 = 0.005025 moles
From the stoichiometric balance of the reaction, 1 mole of Pb(NO₃)₂ reacts with 1 mole of Na₂SO₄
Hence, this indicates that Pb(NO₃)₂ is the limiting reagent as it is in short supply and it will therefore dictate the amount of precipitate (PbSO₄) formed.
1 mole of Pb(NO₃)₂ gives 1 mole of the precipitate (PbSO₄)
0.002295 mole of Pb(NO₃)₂ will give 0.002295 mole of the precipitate (PbSO₄)
Mass = (Number of moles) × (Molar mass)
Number of moles of precipitate (PbSO₄) formed = 0.002295 mole
Molar mass of PbSO₄ = 303.26 g/mol
Mass of precipitate (PbSO₄) formed = 0.002295 × 303.26 = 0.6958864734 = 0.696 g
Hope this Helps!!!
convert 4.56 cm to mm
Answer:
45.6 mm
Explanation:
1 centimeter = 10 millimeter
4.56 cm = x mm
x = 4.56 × 10
= 45.6 mm
After doing a correct and careful recrystallization, you isolate your product by filtration. Which of the following will be present in the mother liquor? (select all that apply)
a. Impurities
b. Large amount of product
c. Recrystallizing solvent
d. Small amount of produc
Answer:
a. Impurities
c. Recrystallizing solvent
Explanation:
In this type of reaction the products are never considered totally pure, that is why as a final product it must always be taken into account that it is proportional, it will be a recrystallization solvent and other impurities with which the products were mixed.
What volume of a 3.5 M LiOH solution is needed to titrate 253 ml of a 2.75 M HF solution?
Answer:
198.8mL
Explanation:
Step 1:
Data obtained from the question:
Molarity of base (Mb) = 3.5M
Volume of base (Vb) =..?
Molarity of acid (Va) = 2.75M
Volume of acid (Va) = 253mL
Step 2:
The balanced equation for the reaction.
HF + LiOH —> LiF + H2O
From the balanced equation above,
The mole ratio of the acid (nA) = 1
The mole ratio of the base (nB) = 1
Step 3:
Determination of the volume of the base, LiOH needed for the reaction.
The volume of the base needed for the reaction can be obtained as follow:
MaVa /MbVb = nA/nB
2.75 x 253 / 3.5 x Vb = 1
Cross multiply
3.5 x Vb = 2.75 x 253
Divide both side by 3.5
Vb = 2.75 x 253 / 3.5
Vb = 198.8mL
Therefore, the volume of the base needed for the reaction is 198.8mL
Which evidence supports the idea that Cosmic Microwave Background radiation is a remnant of the Big Bang?
A. Its temperature is uniform.
B. Its temperature fluctuates greatly.
C. Its mass fluctuates greatly.
D. Its mass is uniform.
Answer:
A. Its temperature is uniform.
Explanation:
Cosmic Microwave Background radiation is a remnant of the Big Bang talks about the universe being very hot thereby causing its expansion with a corresponding cooling of the gases within it. The Cosmic Microwave Background radiation being what is left of the radiation after the Big Bang occurrence.
The temperature was uniform with very little fluctuations which makes the A the right choice.
Answer:
a
Explanation:
A sample of sulfuryl fluoride, SO2F2, has a mass of 19.2 g. What amount, in moles, does this mass represent?
Answer:
0.188 moles
Explanation:
moles=mass/molar mass
moles= 19.2/ (32+32+38)
moles= 0.188
Answer:
0.188 moles.
Explanation:
Molar mass of SO2F2
= 32 + 2*16 + 2 * 19
= 102 g.
So 19.2g = 19.2 / 102
= 0.188 moles.