Answer:
The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.
Explanation:
As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,
Moles of KCl = 0.275 L × 1.62 M = 0.445 moles
Now the reaction will be,
Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺
Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.
Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams
Now the mass of PbCl2 formed will be,
y × molecular mass = y × 278.1 grams = 278.1 y grams
Now the total precipitate will be,
62.37 grams = 143.32 x + 278.1 y -----------(i)
Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,
x + 2y = total mass of KCl
x + 2y = 0.445 moles ------ (ii)
On solving equation (i) and (ii) we get,
x as 0.115 and y as 0.165
Now the mass of AgCl will be,
143.32 × 0.115 = 16.48 grams
The mass of PbCl₂ will be,
278.1 × 0.165 = 45.88 grams.
Consider the following reaction where Kc = 6.50×10-3 at 298 K: 2NOBr(g) 2NO(g) + Br2(g) A reaction mixture was found to contain 9.83×10-2 moles of NOBr(g), 5.44×10-2 moles of NO(g), and 4.13×10-2 moles of Br2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc
Answer:
This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely [tex]\rm NOBr\; (g)[/tex]) in order to reach an equilibrium.
For this mixture, the reaction quotient is [tex]Q_c = 0.0126[/tex].
Explanation:
A reversible reaction is at equilibrium if and only if its reaction quotient [tex]Q_c[/tex] is equal to the equilibrium constant [tex]K_c[/tex].
Start by calculating the equilibrium quotient [tex]Q_c[/tex] of this reaction. Given the reaction:
[tex]\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g)[/tex].
Let [tex][\mathrm{NOBr\; (g)}][/tex], [tex][\mathrm{NO\; (g)}][/tex], and [tex][\mathrm{Br_2\; (g)}][/tex] denote the concentration of the three species. The formula for the reaction quotient of this system will be:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}[/tex].
(Note, that in this formula, both [tex][\mathrm{NO\; (g)}][/tex] and [tex][\mathrm{NOBr\; (g)}][/tex] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)
Calculate the reaction quotient given the concentration of each species:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126[/tex].
(Note that the unit is ignored.)
Apparently, [tex]Q_c > K_c[/tex]. Since [tex]Q_c[/tex] and [tex]K_c[/tex] are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,
Keep in mind that [tex]Q_c[/tex] denotes a quotient. To reduce the value of a quotient, one may:
reduce the value of the numerator, increase the value of the denominator, orboth.In [tex]Q_c[/tex], that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.
what is the polarity of black pepper
Answer:
Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.
Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.
It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.
3A 2B --> 5C If compound A has a molar mass of 159.7 g/mol and compound C has a molar mass of 57.6 g/mole, how many grams of compound C will be produced from 18.24 grams of compound A and excess compound B
Answer:
10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.
Explanation:
3A + 2B ⇒ 5C
By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) the following amounts of reagent and products participate in the reaction:
A: 3 molesB: 2 molesC: 5 molesThe excess reagent will be that which is not completely depleted during the reaction.
The amount of product obtained from the reaction will always depend on the amount of limiting reagent in the reaction. Then, being B the excess reagent and therefore A the limiting reagent and knowing that compound A has a molar mass of 159.7 g/mole and compound C has a molar mass of 57.6 g/mole, by stoichiometry the following mass amounts of A and C participate in the reaction:
A: 3 moles* 159.7 g/mole= 479.1 gC: 5 moles* 57.6 g/mole= 288 gThen it is possible to apply the following rule of three: if by stoichiometry of the reaction 479.1 grams of A produce 288 grams of C, 18.24 grams of A, how much mass of C does it produce?
[tex]mass of C=\frac{18.24 grams of A*288 grams of C}{479.1 grams of A}[/tex]
mass of C= 10.96 grams
10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.
Which of the following pairings usually forms molecular compounds?
Select the correct answer below:
metal, nonmetal
nonmetal, nonmetal
metal, metal
none of the above
A nonmetal and a nonmetal will make molecular compounds like H2O and CO2
What is the systematic name for the given compound? (R)-2-bromobutane (S)-2-bromobutane (R)-2-butyl bromide (S)-2-butyl bromide
Answer:
See explanation below
Explanation:
IUPAC came up with the idea of an unambiguous system of nomenclature for organic compounds. This unambiguous system relates the structure of a compound with its name. Thus, IUPAC has established a worldwide standard for the unambiguous naming of organic compounds. Scientists all over the world can now have a uniform system of nomenclature for compounds in order to facilitate easy communication of scientific information.
The systematic names of the following compounds listed in the question are shown below;
(R)-2- butyl bromide has the systematic name (R)-2-bromobutane
(S)-2-butyl bromide has the systematic name (S)-2-bromobutane
This unified system of nomenclature avoids the confusion created by the use of different trivial names in deferent localities and by various scientific academies. This is a major advantage of the systematic nomenclature.
What is the specific heat of a metal with a mass of 14.0 g, heat of 3.45 kJ and a change in temperature of 3.2 ℃?
Multiply each half reaction by the correct number, in order to balance charges for the two half reactions. Mg--> mg2++2c Au++e--> Au Please i need the answer quick i couldnt find it anywhere on the web
Answer:
Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)
Explanation:
The key to balancing redox reaction equations is this; ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation. The both half reaction equations can now be combined to give the overall reaction equation.
For the redox reaction under consideration;
Oxidation half equation;
Mg(s) ------> Mg^2+(aq) + 2e
Reduction half equation;
2Ag^+(aq) +2e ----> 2Ag(s)
Overall balanced redox reaction equation;
Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)
Recall that when a reaction is at equilibrium, the forward and reverse reactions occur at the same rate. To illustrate this idea, consider the reaction of A (small, red spheres) and B (large, blue spheres) to form AB.
A+B ⇌ AB
Notice that the reaction never stops. Even after several minutes, there is A and B left unreacted, and the forward and reverse reactions continue to occur. Also note that amounts of each species (i.e., their concentrations) stay the same.
Required:
What is the value of the equilibrium constant for this reaction?
Assume each atom or molecule represents a 1 M concentration of that substance.
Answer:
Equilibrium constant Kc = [x]² / [A - x] [B - x]
Explanation:
The equilibrium constant is defined as the ratio of the concentration of the products to that of the reactants at equilibrium
ie Kc = [products] / [reactants].
The balanced equation of the reaction is given as : A + B ⇄ AB
At the beginning of the reaction,
Initial concentration I = A = 1M
B = 1M
AB = 0M
After a period of time and assuming 'x' to be the concentration of product AB formed, the concentrations become
C = reactant A = [A - x] M
rectant B = [B - x] M
Product AB = [x] [x] M
At equilibrium, the concentrations are,
E = rectant A = [A - x] M
reactant B = [B - x] M
product AB = [x]² M
therefore , the equilibrium constant, Kc = [products]/[reactants]
= [x]² / [A - x] [B - x]
8. Exactly 169.5 J will raise the temperature of 10.0 g of a metal from 25.0°C to 60.0°C. What is the specific heat capacity of the metal?
Answer:
THE SPECIFIC HEAT OF THE METAL IS 0.4843 J/G °C
Explanation:
Heat involved in the reaction is 169.5 J
mass of the metal = 10 g
Change in temperature = 60 °C - 25 °C = 35 °C
Specific heat of the metal = unknown
Specific heat of a metal or substance is the amount of heat per unit mass required to raise the temperature by one degree Celsius.
Using the formula for heat of a reaction, which is:
Heat = mass * specific heat * change in temperature
Heat = m C ΔT
Re-arranging the formula by making C the subject of the formula, we get:
C = Heat / m ΔT
C = 169.5 J / 10 g * 35 C
C = 169.5 J / 350 g °C
C = 0.4843 J/g°C
So therefore, the specific heat of the metal is 0.4843 J/ g°C
Calcium carbide, CaC2, reacts with water to form calcium hydroxide and the flammable gas ethyne (acetylene) in the reaction: What mass of ethyne can be produced
Answer:
1 mole of CaC₂ will produce 26g of C₂H₂ or 64.1g of CaC₂ will produce 26g of C₂H₂
Explanation:
Hello,
To solve this question, we'll require a balanced chemical equation of reaction between calcium carbide and water.
Equation of reaction
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
Molar mass of calcium carbide (CaC₂) = 64.1g/mol
Molar mass of water (H₂O) = 18g/mol
Molar mass of calcium hydroxide (Ca(OH)₂) = 74g/mol
Molar mass of ethyne (C₂H₂) = 26g/mol
From the equation of reaction, 1 mole of CaC₂ will produce 1 mole of C₂H₂
1 mole of CaC₂ = mass / molar mass
Mass = 1 × 64.1
Mass = 64.1g
1 mole of C₂H₂ = mass / molar mass
Mass = 1 × 26
Mass = 26g
Therefore, 1 mole of CaC₂ will produce 26g of C₂H₂
Note: this is a hypothetical calculation since we were not given the initial mass of CaC₂ that starts the reaction
What volume of water is required to dilute 120 cm3 of 10 mol dm–3 sulphuric acid to a concentration of 2 mol dm–3?
Answer:
0.48 dm3 (or 480 cm3)
Explanation:
First find the original no. of moles existing in the sulphuric acid:
no. of moles = volume (in dm3) x concentration
= 120/1000 x 10
= 1.2 mol
Then let the total volume of the diluted acid be v dm3.
Since
Concentration = no. of moles / volume,
so by substituting the given information,
2 = 1.2 / v
v = 0.6 dm3
Hence, the volume of water required
= 0.6 - 120/1000
= 0.48 dm3 (or 480 cm3)
Considering the definition of dilution, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].
First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.
Dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one and consists of reducing the amount of solute per unit volume of solution. This is accomplished simply by adding more solvent to the solution in the same amount of solute.
In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.
A dilution is mathematically expressed as:
Ci×Vi = Cf×Vf
where
Ci: initial concentration Vi: initial volume Cf: final concentration Vf: final volume
In this case, you know:
Ci= 10 [tex]\frac{mol}{dm^{3} }[/tex] Vi= 120 cm³ Cf= 2 [tex]\frac{mol}{dm^{3} }[/tex] Vf= ?Replacing in the definition of dilution:
10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³= 2 [tex]\frac{mol}{dm^{3} }[/tex]× Vf
Solving:
Vf= (10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³) ÷2 [tex]\frac{mol}{dm^{3} }[/tex]
Vf= 600 cm³
In summary, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].
Learn more about dilution:
brainly.com/question/20113402?referrer=searchResults brainly.com/question/22762236?referrer=searchResultsIf 0.98 g of an unknown was dissolved in 10.30 g of solvent and the resulting solution has a molality of 0.45 m, what is the molar mass of the unknown
Answer:
211.4g/mol.
Explanation:
Data obtained from the question includes:
Mass of unknown compound = 0.98g
Mass of solvent = 10.30g
Molality = 0.45 M
Next, we shall determine the number of mole of the unknown compound present in the solution.
This can be obtained as follow:
Molality = mole /kg of solvent
Mole of the unknown compound =.?
Mass of solvent = 10.30g = 10.30/1000 = 0.0103Kg
Molality = 0.45 M
Molality = mole /kg of solvent
0.45 = mole /0.0103
Cross multiply
Mole = 0.45 x 0.0103
Mole = 4.635×10¯³ mole
Therefore the mole of the unknown compound that dissolve in solution is 4.635×10¯³ mole
Now, we can obtain the molar mass of the unknown compound as follow:
Mole of the unknown compound = 4.635×10¯³ mole
Mass of unknown compound = 0.98g
Molar mass of the unknown compound =?
Mole = mass /Molar mass
4.635×10¯³ = 0.98 /Molar mass
Cross multiply
4.635×10¯³ x molar mass = 0.98
Divide both side by 4.635×10¯³
Molar mass = 0.98 / 4.635×10¯³
Molar mass = 211.4g/mol.
Therefore, the molar mass of the unknown compound is 211.4g/mol.
The molecular mass of the unknown has been 211.66 g/mol.
Molality can be defined as the moles of the solute per kg of solvent.
Molality can be expressed as:
Molality = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{Mass\;of\;solvent\;(g)}[/tex] ......(i)
The given unknown has been the solute.
The mass of solute = 0.98 g.
The mass of solvent = 10.30 g.
The molality of the solution formed has been = 0.45 m.
Substituting the values in equation (i):
0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{10.30\;g}[/tex]
0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;97.087[/tex]
[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex] = [tex]\rm \dfrac{0.45}{97.087}[/tex]
[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex] = 0.00463
Molecular mass of solute = [tex]\rm \dfrac{0.98}{0.00463}[/tex]
Molecular mass of solute = 211.66 g/mol.
The molecular mass of the unknown has been 211.66 g/mol.
For more information about the molality of the solution, refer to the link:
https://brainly.com/question/8103026
Place ~0.8 g Fe(NO3)3·9 H2O into a 10 mL volumetric flask and dissolve to the volumetric line with 0.1 M HNO3(aq). (Make sure you pre-rinse the volumetric flask with the 0.1 M HNO3(aq) solution before making the solution. Also, 9 waters of hydration (·9H2O) are present in this chemical, and must be included in the formula weight calculation.) Record the exact mass of iron nitrate used and show the calculation for the solution concentration in your observations.
Before lab, create a plan for the dilution of the Fe(NO3)3(aq) solution created above with DI water. This plan should be clearly outlined in your ELN. Create enough 0.0020 M Fe(NO3)3
How do I go about solving for the dilution plan?
Answer:
You can take 1mL of your stock solution in a 100mL volumetric flask and complete to volume.
Explanation:
You need to create a 0.00200M solution of Fe(NO₃)₃. First, you have to obtain the concentration of the first solution you made. That is:
0.8g Fe(NO₃)₃.9H₂O × (1mol / 403.9972g) =
0.0020 moles of Fe(NO₃)₃.9H₂O = Moles of Fe(NO₃)₃
In 10mL = 0.010L:
0.0020 moles of Fe(NO₃)₃ / 0.010L = 0.20M Fe(NO₃)₃
This is the concentration of your stock solution, as you want to obtain a 0.0020M solution, you dilution factor must be:
0.20M / 0.0020M = 100
That means you need to dilute your stock solution 100 times.
You can make this dilution, for example,
taking 1mL of your stock solution in a 100mL volumetric flask completing to volume with the solvent, 0.1M HNO₃(aq).A gas sample is collected in a 0.279 L container at 22.7 °C and 0.764 atm. If the sample has a mass of 0.320 g, what is the identity of the gas? Group of answer choices g
Answer:
HCl
Explanation:
Choices:
CO: 28.01g/mol
NO₂: 46g/mol
CH₄: 16.04g/mol
HCl: 36.4g/mol
CO₂: 44.01g/mol
It is possible to identify a substance finding its molar mass (That is, the ratio between its mass in grams and its moles). It is possible to find the moles of the gas using general ideal gas law:
PV = nRT
Where P is pressure of gas 0.764atm; V its volume, 0.279L; n moles; R gas constant: 0.082atmL/molK and T its absolute temperature, 295.85K (22.7°C + 273.15).
Replacing:
PV = nRT
PV / RT = n
0.764atm*0.279L / 0.082atmL/molKₓ295.85K = n
8.786x10⁻³ = moles of the gas
As the mass of the gas is 0.320g; its molar mass is:
0.320g / 8.786x10⁻³moles = 36.4 g/mol
Based in the group of answer choices, the identity of the gas is:
HCl
Molar mass of substance is the mass of one mole of substance. The gas is identified as HCl as it has 36.4 g/mol of molar mass.
The substance can be identified from its molar mass. The formula to calculate the molar mass is
[tex]\bold {m = \dfrac w{n}}[/tex]
Where,
n - number of moles
w - given mass of the substance
The number of moles can be calculated by the using ideal gas law
PV = nRT
[tex]\bold {n = \dfrac {PV }{ RT }}[/tex]
Where
P - pressure of gas 0.764 atm;
V its volume, 0.279 L;
n moles;
R gas constant: 0.082atmL/molK
T its absolute temperature, 295.85 K (22.7°C + 273.15).
Put the values in the formula,
[tex]\bold {n = \dfrac {0.764\ atm \times , 0.279 }{ 0.082 \times 295.85 }}\\\\\bold {n =8.786x10^-^3}[/tex]
Now, put the values in the molar mass formula,
[tex]\bold {m = \dfrac {0.320g}{ 8.786x10^-^3} = 36.4\ g/mol}[/tex]
Therefore, the gas is identified as HCl as it has 36.4 g/mol of molar mass.
To know more about molar mass,
https://brainly.com/question/12127540
A piece of silver with a mass 368 g has a heat capacity of 87.2 J/°C. What is the specific heat of
silver?
A. 0.385 J/g.°C
B. 0.237 J/g.°C
C. 0.184 J/g.°C
D. 1.322 J/g.°C
E. 4.184 J/g.°C
Answer:
B
Explanation:
Heat capacity = mass x specific heat capacity.
(C = mc)
87.2 = 368 x c
= 0.237 J/g.°C
Calculate the mass of sodium nitrate( NaNO3), in grams, required to produce 128 g of oxygen, using the following equation: 2NaNO3 --> 2NaNO2 O2
Answer:
680g of NaNO3.
Explanation:
The balanced equation for the reaction is given below:
2NaNO3 —> 2NaNO2 + O2
Next, we shall determine the mass of NaNO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:
Molar mass of NaNO3 = 23 + 14 + (16x3) = 85g/mol
Mass of NaNO3 from the balanced equation = 2 x 85 = 170g
Molar mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32g
From the balanced equation above,
170g of NaNO3 decomposed to produce 32g of O2.
Now, we can obtain the mass of NaNO3 needed to produce 128g of O2 as shown below:
From the balanced equation above,
170g of NaNO3 decomposed to produce 32g of O2.
Therefore, Xg of NaNO3 will decompose to produce 128g of O2 i.e
Xg of NaNO3 = (170 x 128)/32
Xg of NaNO3 = 680g
Therefore, 680g of NaNO3 are needed to produce 128g of O2.
Based on the Valence Shell Electron Pair Repulsion Theory (or VSEPR), molecules will arrange to keep the following as far apart as physically possible
a) mobile electrons
b) valence electron pairs
c) inner shell electrons
d) the electrons closest to the nuclei
Answer:
B. Valence Electron Pairs
Explanation:
Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.
The Ka1 value for oxalic acid is 5.9 x10-2 , and the Ka2 value is 4.6 x 10-5 . What are the values of Kb1 and Kb2 of the oxalate ion
Answer:
2.17x10⁻¹⁰ = Kb1
1.69x10⁻¹³ = Kb2
Explanation:
Oxalic acid, C₂O₄H₂, has two intercambiable protons, its equilibriums are:
C₂O₄H₂ ⇄ C₂O₄H⁻ + H⁺ Ka1 = 5.9x10⁻²
C₂O₄H⁻ ⇄ C₂O₄²⁻ + H⁺ Ka2 = 4.6x10⁻⁵
Oxalate ion, C₂O₄²⁻, has as equilibriums:
C₂O₄²⁻ + H₂O ⇄ C₂O₄H⁻ + OH⁻ Kb1
C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2
Also, you can know: KaₓKb = Kw
Where Kw is 1x10⁻¹⁴
Thus:
Kw = Kb2ₓKa1
1x10⁻¹⁴ =Kb2ₓ4.6x10⁻⁵
2.17x10⁻¹⁰ = Kb1And:
Kw = Kb1ₓKa2
1x10⁻¹⁴ =Kb1ₓ5.9x10⁻²
1.69x10⁻¹³ = Kb1
That is because the inverse reaction of, for example, Ka1:
C₂O₄H⁻ + H⁺ ⇄ C₂O₄H₂ K = 1 / Ka1
+ H₂O ⇄ H⁺ + OH⁻ K = Kw = 1x10⁻¹⁴
=
C₂O₄H⁻ + H₂O ⇄ C₂O₄H₂ + OH⁻ Kb2 = Kw × 1/Ka1
What is the new mass/volume percent (m/v) of a KOH solution that is prepared by diluting 110 mL of a 6.0% (m/v) KOH solution to 330 mL
Answer:
The new mass/volume percent is 2.0% (m/v)
Explanation:
Dilution is a procedure by which the concentration of a solution is decreased, generally with the addition of a diluent. In other words, dilution is a process in which a concentrated solution is always started, to which a greater volume of solvent is added, causing the concentration and volume of the resulting solution to change. But the amount of solute used to prepare the initial solution remains the same.
The calculation of a dilution is made by:
Cinitial. Vinitial = Cfinal. Vfinal
where C indicates concentration and V indicates volume.
In this case:
Cinitial: 6.0% (m/v)Vinitial: 110 mLCfinal: ?Vfinal: 330 mLReplacing:
6.0% (m/v) * 110 mL= Cfinal* 330 mL
Solving:
[tex]Cfinal=\frac{ 6.0 (m/v)*110 mL}{330 mL}[/tex]
Cfinal= 2.0% (m/v)
The new mass/volume percent is 2.0% (m/v)
What volume of a 2.25 M sodium chloride solution will contain 4.58 moles of sodium chloride
(NaCl)?
A. 0.252 L
B. 119 L
C. 5.62 L
D. 0.491 L
E. 2.04 L
Answer:
Option E. 2.04 L
Explanation:
Data obtained from the question include:
Molarity of NaCl = 2.25 M
Mole of NaCl = 4.58 moles
Volume =..?
Molarity is simply defined as the mole of solute per unit litre of the solution. It is represented mathematically as:
Molarity = mole /Volume
With the above formula, we can obtain the volume of the solution as follow:
Molarity = mole /Volume
2.25 = 4.58/volume
Cross multiply
2.25 x volume = 4.58
Divide both side by 2.25
Volume = 4.58/2.25
Volume = 2.04 L
Therefore, the volume of the solution is 2.04 L
The monomer of poly(vinyl chloride) has the formula C2H3Cl. If there are 1,565 repeat units in a single chain of the polymer, what is the molecular mass (in amu) of that chain? Enter your answer in scientific notation.
Answer:
[tex]\large \boxed{9.780 \times 10^{4}\text{ u}}[/tex]
Explanation:
The molecular mass of a monomer unit is:
C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u
For 1565 units,
[tex]\text{Molecular mass} = \text{1565 units} \times \dfrac{\text{62.494 u}}{\text{1 unit }} = \mathbf{9.780 \times 10^{4}}\textbf{ u}\\\\\text{The molecular mass of the chain is $\large \boxed{\mathbf{9.780 \times 10^{4}}\textbf{ u}}$}[/tex]
Strontium crystallizes in a face-centered cubic unit cell having an edge length of 77.43 pm. What is the atomic radius of strontium (in picometers) based on this structure
Answer:
Atomic radius of Strontium is 27.38pm
Explanation:
In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:
a² + a² = b² = (4r)²
2a² = 16r²
a = √8 r
As edge length of Strontium is 77.43pm:
77.43pm / √8 = r
27.38pm = r
Atomic radius of Strontium is 27.38pmhelp please !!!!!!!!
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
The IE1, for iodine, is 1009 kJ/mol. Calculate the wavelength (in nm) of electromagnetic energy need to ionize an iodine atom. 118.6 743.2 488.3 1042
Answer:
118.6nm
Explanation:
It is possible to calculate wavelength of any energetic process (As an ionization) using:
E = hc / λ (1)
Where E is Energy, h is Planck constant (6.626x10⁻³⁴Js), c speed of light (3x10⁸ms⁻¹) and λ is wavelength In meters.
As the energy to ionize 1 mole of iodine is 1009kJ, one atom requires:
(1009kJ / mol) ₓ (1mol / 6.022x10²³ atoms) = 1.6755x10⁻²¹kJ / atom. = 1.6755x10⁻¹⁸J
Replacing in (1):
λ = hc / E
λ = 6.626x10⁻³⁴Js*3x10⁸ms⁻¹ / 1.6755x10⁻¹⁸J
λ = 1.186x10⁻⁷m
As 1m = 1x10⁹nm:
1.186x10⁻⁷m ₓ (1x10⁹nm / 1m) =
118.6nmg A 25.00 mL sample of 0.0988 M benzoic acid (HC7H5O2, Ka = 6.5 · 10-5) is a monoprotic acid that is titrated with 0.115 M sodium hydroxide. Calculate the pH at the equivalence point.
Answer:
pH at equivalence point is 8.47
Explanation:
Benzoic acid react with NaOH, thus:
HC₇H₅O₂ + NaOH → C₇H₅O₂⁻ + H₂O + Na⁺
You reach equivalence point when moles of the acid = moles of NaOH.
Moles of benzoic acid are:
0.025L ₓ (0.0988mol / L) = 0.00247 moles
To have 0.00247 moles of NaOH in solution and reach equivalence point you need to add:
0.00247 moles NaOH ₓ (1L / 0.115mol) = 0.0215L of NaOH solution.
Total volume is 0.0465L.
There are produced 0.00247 moles of C₇H₅O₂⁻ and its molarity will be:
0.00247 mol C₇H₅O₂⁻ / 0.0465L = 0.0531M C₇H₅O₂⁻
C₇H₅O₂⁻ is in equilibrium with water, thus:
C₇H₅O₂⁻(aq) + H₂O ⇄ HC₇H₅O₂(aq) + OH⁻(aq)
Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.5x10⁻⁵ = 1.54x10⁻¹⁰ is:
Kb = 1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]
The concentrations in equilibrium of the species are:
[HC₇H₅O₂] = X
[OH⁻] = X
[C₇H₅O₂⁻] = 0.0531M - X
Where X represents how much C₇H₅O₂⁻ react, X is reaction coordinate
Replacing in Kb expression:
1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]
1.54x10⁻¹⁰ = [X] [X] / [0.0531 - X]
8.169x10⁻¹² - 1.54x10⁻¹⁰X = X²
8.169x10⁻¹² - 1.54x10⁻¹⁰X - X² = 0
Solving for X:
X = -2.858x10⁻⁶M → False solution, there is no negative concentrations
X = 2.858x10⁻⁶M → Right solution
As [OH⁻] = X
[OH⁻] = 2.858x10⁻⁶M
pOH is -log [OH⁻]
pOH = 5.54
pH = 14 - pOH
pH = 8.46
pH at equivalence point is 8.47Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light?
1. Gamma rays
2. Infrared radiation
3. Ultraviolet liht
4. X-rays
5. Microwaves
6. Radio waves
Answer:
3,4,1 and 6,5,2
Explanation:
In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;
Radio waves
Microwaves
Infrared waves
Visible light rays
Ultraviolet rays
X-rays
Gamma rays
(a simple mnemonic is RMIVUXG)
7. An element's most stable ion forms an ionic compound with chlorine having the formula XCl2. If the ion of element X has a mass of 89 and 36 electrons, what is the identity of the element, and how many neutrons does it have
Answer:
The element is strontium and the number of neutrons it have is 51.
Explanation:
Based on the given information, the ionic compound is,
XCl₂ ⇔ X₂⁺ + 2Cl⁻
X2+ is the ion of the mentioned element
As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,
No. of electrons = atomic number - charge
36 = atomic number - 2
Atomic number = 38
Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.
Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,
No. of neutrons = mass number - atomic number
= 89 - 38
= 51 neutrons.
A certain radioactive element has a half life of 8694 years. How much of a 8.30 g sample is left after 8323 years
Answer: The amount of sample left after 8323 years is 4.32g
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{8694years}=7.97\times 10^{-5}years^{-1}[/tex]
b) amount left after 8323 years
[tex]t=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]
[tex]8323=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]
[tex]0.285=\log\frac{8.30}{a-x}[/tex]
[tex]\frac{8.30}{a-x}=1.92[/tex]
[tex](a-x)=4.32g[/tex]
The amount of sample left after 8323 years is 4.32g
The elements will lose of gain electrons as needed to have an electron configuration that matches a noble gas.
a) true
b) false
Answer:
true
Explanation:
This is true because elements aim to have a full octet of electrons in their outermost (also called valence) shell. Noble gases already have a full valence shell which is why the elements that are not noble gases aim to be like them.
Answer:
a) true
Explanation:
this is the answer coz elements aim to have a full octet of electrons in their outermost (also called valence) shell.
How much heat is liberated at constant pressure when 1.41 g of potassium metal reacts with 6.52 mL of liquid iodine monochloride (d = 3.24 g/mL)? 2K(s) + ICl(l) → KCl(s) + KI(s)
Answer:
The correct answer is - 13.33 kJ of heat
Explanation:
To know which one is the limiting reagent, determine the number of moles of each reagent in order .
n(K) = mass/atomic weight = 1.41/39 = 0.036 moles
Density of ICl = Mass/Volume
3.24 = Mass/6.52
Mass of ICl = 21.12 g
n(ICl) = mass/molar mass = 21.12/162.35 = 0.130 moles
2 moles of K reacts with 1 mole of ICl
0.036 moles of K will react with = 0.036/2 = 0.018 moles of ICl
since the amount of moles of ICl is more than 0.018, it is in excess and hence K is the limiting reagent. Now, use the balance equation to determine the amount of heat liberated:
2 moles of K gives out -740.71 kJ of heat
1 mole of K will give out = -740.71/2 = 370.36 kJ of heat
0.036 moles of K will give out = 0.036 × 370.36 = 13.33 kJ of heat
Thus, the correct answer is - 13.33 kJ of heat
The amount of heat liberated at constant pressure is -13.33 kJ
The given parameters are:
Mass of potassium metal = 1.41gAmount of liquid iodine monochloride = 6.52 mLStart by calculating the number (n) of moles of each reagent using:
[tex]n = \frac{Mass}{Atomic\ weight }[/tex]
For the potassium metal, we have:
[tex]n_k = \frac{1.41g}{39g/mole}[/tex] ---where 39 is the atomic weight of potassium
[tex]n_k = 0.036\ moles[/tex]
For the liquid iodine monochloride, we start by calculating its mass using:
[tex]Mass = Density \times Volume[/tex]
So, we have:
[tex]Mass = 3.24 \times 6.52[/tex]
[tex]Mass = 21.12g[/tex]
The number of moles is then calculated as:
[tex]n_I=\frac{21.12g}{162.35}[/tex]
[tex]n_I = 0.130\ moles[/tex]
The reaction equation 2K(s) + ICl(l) → KCl(s) + KI(s) means that:
2 moles of potassium reacts with 1 mole of liquid iodine monochloride.
So, 0.036 moles of potassium will react with the following moles of liquid iodine monochloride.
[tex]A = \frac{0.036}{2}[/tex]
[tex]A = 0.018\ moles[/tex]
i.e. 0.036 moles of potassium will react with of liquid iodine monochloride
By comparison: 0.018 moles is less than 0.036 moles
So, the amount of heat liberated at constant pressure is:
[tex]Amount = 0.036 \times -\frac{740.71}{2} kJ[/tex]
[tex]Amount = -13.33 kJ[/tex]
Hence, the amount of heat liberated at constant pressure is -13.33 kJ
Read more about heat at:
https://brainly.com/question/13439286