Answer: (a) The solubility of CuCl in pure water is [tex]1.1 \times 10^{-3} M[/tex].
(b) The solubility of CuCl in 0.1 M NaCl is [tex]9.5 \times 10^{-3} M[/tex].
Explanation:
(a) Chemical equation for the given reaction in pure water is as follows.
[tex]CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)[/tex]
Initial: 0 0
Change: +x +x
Equilibm: x x
[tex]K_{sp} = 1.2 \times 10^{-6}[/tex]
And, equilibrium expression is as follows.
[tex]K_{sp} = [Cu^{+}][Cl^{-}][/tex]
[tex]1.2 \times 10^{-6} = x \times x[/tex]
x = [tex]1.1 \times 10^{-3} M[/tex]
Hence, the solubility of CuCl in pure water is [tex]1.1 \times 10^{-3} M[/tex].
(b) When NaCl is 0.1 M,
[tex]CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)[/tex], [tex]K_{sp} = 1.2 \times 10^{-6}[/tex]
[tex]Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)[/tex], [tex]K = 8.7 \times 10^{4}[/tex]
Net equation: [tex]CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)[/tex]
[tex]K' = K_{sp} \times K[/tex]
= 0.1044
So for, [tex]CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)[/tex]
Initial: 0.1 0
Change: -x +x
Equilibm: 0.1 - x x
Now, the equilibrium expression is as follows.
K' = [tex]\frac{CuCl_{2}}{Cl^{-}}[/tex]
0.1044 = [tex]\frac{x}{0.1 - x}[/tex]
x = [tex]9.5 \times 10^{-3} M[/tex]
Therefore, the solubility of CuCl in 0.1 M NaCl is [tex]9.5 \times 10^{-3} M[/tex].
Final answer:
The solubility of CuCl in pure water and in 0.100 M NaCl solution both turn out to be 1.1 x 10^-3 M when formation of CuCl2- is ignored, determined using the provided Ksp.
Explanation:
To calculate the solubility of CuCl in pure water, we can use the solubility product constant (Ksp). The dissociation of CuCl can be represented as:
CuCl(s) ⇌ Cu+(aq) + Cl-(aq)
Let the solubility of CuCl be 's' moles per liter. At equilibrium, the concentration of Cu+ and Cl- ions will both be 's' M. So, the Ksp expression for CuCl is:
Ksp = [Cu+][Cl-] = (s)(s) = s²
Giving us:
s = √(1.2 x 10^-6) = 1.1 x 10^-3 M
This is the solubility of CuCl in pure water.
For the solubility of CuCl in 0.100 M NaCl solution, we have to consider the common ion effect. In this case, Cl- is the common ion. With the presence of 0.100 M NaCl, the concentration of Cl- ions at equilibrium will be higher. However, since we ignore the formation of CuCl2-, the solubility will still be governed by the Ksp of CuCl. The calculation will remain the same as in pure water, due to the assumption of ignoring CuCl2- formation:
s = 1.1 x 10^-3 M
Therefore, the presence of NaCl does not affect the solubility under the given conditions for this specific case.
Suppose the formation of iodine proceeds by the following mechanism:
step elementary reaction rate constant
1 H2 (g) + ICl (g) → HI (g) + HCl (g) k1
2 HI (g) + ICl (g) → I2 (g) + HCl (g) k2
Suppose also ki « k2. That is, the first step is much slower than the second. Write the balanced chemical equation for the overall chemical reaction. Write the experimentally- observable rate law for the overall chemical reaction.
Answer:
Overall reaction
H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)
Overall Rate = k1[H2] [ICl]
Explanation:
Overall reaction
H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)
The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.
Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence
Overall Rate = k1[H2] [ICl]
The balanced chemical equation for the overall reactions is H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g). The rate of the reaction is determined by the slowest step, in this case, the first one. So, the experimentally-observable rate law for the overall reaction would be Rate = k1[H2][ICl].
Explanation:The overall reaction is obtained by adding up the given elementary reactions. Adding step 1 and 2, we have:
H2 (g) + 2ICl (g) → 2HI (g) + HCl (g) → H2 (g) + I2 (g) + HCl (g)
After cancelling like terms the balanced chemical equation is:
H2 (g) + 2ICl (g) → I2 (g) + 2HCl (g)
Since the first step is much slower than the second, it's the rate-determining step. The experimentally-observable rate law for the overall reaction will depend on the rate-determining step, hence, Rate = k1[H2][ICl]
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A chemistry student needs 10.0g of ethanolamine for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of ethanolamine is ·1.02gcm−3. Calculate the volume of ethanolamine the student should pour out.
To determine the volume of ethanolamine needed, the student should divide the desired mass (10.0g) by the density of ethanolamine (1.02g/cm³), which gives approximately 9.8 cm³. A trusted source, such as the CRC Handbook of Physics and Chemistry was referenced.
Explanation:To calculate the volume of ethanolamine needed for the experiment, we need to use the formula for density, which is mass/volume. By rearranging the formula to solve for volume gives us volume = mass/density. So, the volume of ethanolamine will be 10.0g / 1.02g/cm3. This calculation gives us approximately 9.80 cm3. Therefore, the student needs to pour out about 9.8 cm3 of ethanolamine for the experiment. The student was able to easily determine this by consulting a trusted resource, the CRC Handbook of Physics and Chemistry.
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Final answer:
To find the volume of ethanolamine needed, divide the mass needed (10.0 g) by the density (1.02 g/cm³) to obtain approximately 9.80 cm³.
Explanation:
To calculate the volume of ethanolamine the student should pour out, we can use the density formula, which is density (d) equals mass (m) divided by volume (V), rearranged to solve for volume. Given that the density of ethanolamine is 1.02 g/cm³ and the student needs 10.0 g of ethanolamine, the equation would be:
V = m / d
Substituting the given values, we get:
V = 10.0 g / 1.02 g/cm³
The calculation results in a volume of approximately 9.80 cm³ of ethanolamine the student should pour out.
Which of the following are correct statements about reactions?
Check all that apply.
A. If a reaction happens, it will be very fast.
B. During a reaction, a catalyst is always used up along with the
reactants.
C. A catalyst could decrease the activation energy of a reaction.
D. Some reactions can occur more slowly than others.
Answer:
I think it is D.
Explanation:
Answer: i know it d but i think it is c to
Explanation:
It is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The nuclide captures a neutron to form , which in turn undergoes β decay. The daughter nuclide produces the characteristic γ rays used for the analysis. What is the daughter nuclide?
Answer:
The daughter nuclide is selenium; 76:34 Se
Explanation:
The complete question is as follows;
is desired to determine the concentration of arsenic in a lake sediment sample by means of neutron activation analysis. The nuclide 75:33 As captures a neutron to form 76:33 As, which in turn undergoes beta decay. The daughter nuclide produces the characteristic gamma rays used for the analysis. What is the daughter nuclide?
solution:
Please check attachment for decay equations and explanations
An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the ratio of the ionic radius of the cation to the ionic radius of the anion is 0.840, what is the ionic radius of the anion
Answer:
the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
Explanation:
From the diagram shown below :
The anion [tex]Cl^-[/tex] is located at the corners
The cation [tex]Cs^+[/tex] is located at the body center
The Body diagonal length = [tex]\sqrt{3 \ a }[/tex]
∴ [tex]2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a[/tex]
Given that :
[tex]\frac{r^+}{r^-} =0.84[/tex] (i.e the ratio of the ionic radius of the cation to the ionic radius of
the anion )
[tex]0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a[/tex]
Also ; a = 664 pm
Then :
[tex]r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm[/tex]
Therefore, the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
The ionic radius of the anion [tex]r^-=312.52pm[/tex]
Primitive cubic structure:The anion is placed on the corners and the cation is placed on the frame center.
The Body diagonal length = [tex]\sqrt{3a}[/tex]
[tex]2r^++2r^-=\sqrt{3a} \\\\r^++r^-=\sqrt{3}/2a }[/tex]
Given:
Ratio= 0.840
[tex]\frac{r^+}{r^-}=0.840[/tex]
[tex]0.84r^-+r^-=\sqrt{3}/2a \\\\1.84r^-=3/2a\\\\r^-=\sqrt{3}/2*1.84a[/tex]
Also ; a = 664 pm
Then : [tex]r^-[/tex] =312.52 pm
Therefore, the ionic radius of the anion = 312.52 pm
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5. You are investigating an arson scene and you find a corpse in the rubble, but you suspect that the victim did not die as a result of the fire. Instead, you suspect that the victim was murdered earlier, and that the blaze was started to cover up the murder. How would you go about determining whether the victim died before the fire
Answer:
See the answer below.
Explanation:
Fire has three major components:
HeatSmokeGases ( in form of CO, CO2 etc)If the victim had died as a result of the fire, he/he would have inhaled smoke and hot gases from the fire. These components would have resulted in traces of burns and soot deposition in the trachea and lungs as well as traces of CO in the blood of the victim.
If the analysis of the victim's corpse does not reflect some of the results above, it can be effectively concluded that the victim has been dead before the fire.
The single most important indicator of death by the fire would be the presence of CO in the blood of the victim's corpse. All others might be to a less significant degrees.
Final answer:
To determine if the victim died before the fire, forensic anthropologists analyze perimortem injuries, signs of decomposition, and toxicology for smoke inhalation indicators.
Explanation:
To determine whether a victim died before a fire, a forensic anthropologist would analyze the remains for signs of perimortem trauma. This includes examining cut marks or injuries sustained around the time of death which could indicate homicide. Disarticulation of joints and the degree of skeletal preservation can also help establish the time since death, negating the possibility that the fire caused the death if the individual was already deceased and decomposing. Moreover, the presence of smoke inhalation in the lungs can suggest if the person was alive during the fire. Toxicology reports may provide evidence of substances such as carbon monoxide or soot. The presence or absence of such substances can indicate whether death occurred before or during the fire.
Why would you use a solution, such as a cabbage pH indicator, to measure the pH of household items?
a) to see if food went bad
b) to test the safety of water
c) to make sure conditions are safe
d) to use up old cabbage
Answer:
-to see if food went bad
-to test the safety of water
-to make sure conditions are safe
Explanation:
Answer:
A, B, C
Explanation:
A mixture of propane and butane is fed into a furnace where it is mixed with air. The furnace exhaust leaves the furnace at 337�C, 786.0 mmHg and contains only N2, O2, CO2, and H2O. The partial pressure of O2 in the exhaust is 10.38 mmHg and the partial pressure of CO2 in the exhaust is 88.03 mmHg.
1.What is the mole fraction of propane in the fuel stream?
2.What is the mole fraction of water in the exhaust stream?
3.What is the dew point temperature of the exhaust gas?
Answer:
1. 3.29mol
2. 0.125molH2o/mol
3. 52.5'C
Explanation: The step by step explanation are attached to the answer.
Answer:
Mole fraction of propane = 0.74 mol C₃H₈/mol
Mole fraction of water = 0.29 mol H₂0 /mol
Dew point temperature = 52.5°C
Explanation:
See the attached file for the calculation.
When 63.4 g of glycine (C2HNO2 are dissolved in 700. g of a certain mystery liquid X, the freezing point of the solution is 7.9 °C lower than the freezing point of pure X. On the other hand, when 63.4 g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 13.3 °C lower than the freezing point of pure X Calculate the van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits. x 10
Answer:
3.8 is the van't Hoff factor for iron(III) chloride in X.
Explanation:
[tex]\Delta T_f=i\times K_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] =depression in freezing point =
[tex]K_f[/tex] = freezing point constant
m = molality =[tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}[/tex]
i = van't Hoff factor
we have :
Mass of glycine = 63.4 g
Molar mass of glycine = 71 g/mol
Mass of solvent X = 700. g = 0.7 kg
1 g = 0.001 kg
[tex]K_f[/tex] of solvent X= ?
i = 1 (non electrolyte)
Depression in freezing point= [tex]7.9^oC[/tex]
[tex]7.9^oC=1\times K_f \times \frac{63.4 g}{71 g/mol\times 0.7 kg}[/tex]
[tex]K_f=6.19 ^oC/m[/tex]
When iron(III) chloride is dissolved in 0.7 kg of solvent X
Mass of iron(III) chloride = 63.4 g
Molar mass of iron(III) chloride= 162.5 g/mol
Mass of solvent X = 700. g = 0.7 kg
1 g = 0.001 kg
[tex]K_f[/tex] of solvent X= [tex]6.19 ^oC/m[/tex]
i = ?
Depression in freezing point:[tex]13.3^oC[/tex]
[tex]13.3^oC=i\times 6.19^oC\times \frac{63.4 g}{162.5 g/mol\times 0.7 kg}[/tex]
Solving for i:
i = 3.85 ≈ 3.8
3.8 is the van't Hoff factor for iron(III) chloride in X.
A chemical engineer must calculate the maximum safe operating temperature of a high-pressure gas reaction vessel. The vessel is a stainless-steel cylinder that measures 31.0 cm wide and 37.2 cm high. The maximum safe pressure inside the vessel has been measured to be 5.90 MPa. For a certain reaction the vessel may contain up to 3.27 kg of chiorine pentafluoride gas. Calculate the maximum safe operating temperature the engineer should recommend for this reaction. Write your answer in degrees Celsius. Round your answer to 3 significant digits.
Answer:
523°C
Explanation:
Step 1:
Data obtained from the question.
Diameter (d) = 31.0 cm
Height (h) = 37.2 cm
Pressure (P) = 5.90 MPa
Mass of ClF5 = 3.27 kg
Temperature (T) =?
Step 2:
Determination of the volume of the stainless-steel cylinder.
Volume of cylinder = πr^2h = π/4d^2h
V = π/4 x (31)^2 x 37.2
V = 28077.36 cm3
Converting the volume to L, we have:
1 cm3 = 0.001 L
Therefore, 28077.36 cm3 = 28077.36 x 0.001 = 28.07736 L
Step 3:
Conversion of 5.90 MPa to atm.
1 MPa = 9.869 atm
Therefore, 5.90 MPa = 5.90 x 9.869
= 58.2271 atm
Step 4:
Determination of the number of mole of chiorine pentafluoride gas (ClF5). This is illustrated below:
Molar Mass of ClF5 = 35.5 + (19x5) = 35.5 + 95 = 130.5g/mol
Mass of ClF5 = 3.27kg =3.27x10000
Mass of ClF5 = 3270g
Number of mole = Mass /Molar Mass
Number of mole of ClF5 = 3270/130.5
Number of mole of ClF5 = 25.057 moles
Step 5:
Determination of the temperature.
Applying the ideal gas equation
PV = nRT, the temperature T, can be obtained as follow:
P = 58.2271 atm
V = 28.07736 L
n = 25.057 moles
R (gas constant) = 0.082atm.L/Kmol
T=?
PV = nRT
Divide both side by nR
T = PV /nR
T = (58.2271x28.07736)/(25.057x0.082)
T = 795.68 K
Step 6:
Conversion of Kelvin temperature to celsius temperature.
°C = K - 273
K = 795.68 K
°C = 795.68 - 273
°C = 523°C
Therefore, the maximum safe operating temperature the engineer should recommend is 523°C
The concentration of a saturated solution of an ionic compound at 25 oC is 7.00E-4. The formula for this compound is XQ3. What is the concentration of Q at equilibrium? Answer Submitted: Your final submission will be graded when the time limit is reached. Tries 1/98 Previous Tries What is the Ksp value for XQ3?
1. The concentration of[tex]\( \text{Q}^- \)[/tex] at equilibrium is[tex]\( 2.10 \times 10^{-3} \)[/tex] M.
2. The[tex]\( K_{sp} \)[/tex] value for [tex]\( \text{XQ}_3 \) is \( 6.48 \times 10^{-12} \).[/tex]
To find the concentration of [tex]\( \text{Q} \)[/tex] at equilibrium and the solubility product constant [tex](\( K_{sp} \))[/tex]for the ionic compound [tex]\( \text{XQ}_3 \)[/tex], follow these steps:
Step 1: Determine the Dissociation EquationThe compound [tex]\( \text{XQ}_3 \)[/tex] dissociates in water as follows:
[tex]\[ \text{XQ}_3(s) \rightleftharpoons \text{X}^{3+}(aq) + 3\text{Q}^-(aq) \][/tex]
Step 2: Set Up the Solubility RelationshipLet ( s ) be the solubility of [tex]\( \text{XQ}_3 \)[/tex] in mol/L, which is given as \[tex]7.00 \times 10^{-4} \) M.[/tex]
When [tex]\( \text{XQ}_3 \)[/tex] dissociates:
- The concentration of[tex]\( \text{X}^{3+} \)[/tex] ions will be ( s ).
- The concentration of [tex]\( \text{Q}^- \)[/tex] ions will be ( 3s ) (since three[tex]\( \text{Q}^- \)[/tex] ions are produced for each formula unit of[tex]\( \text{XQ}_3 \)).[/tex]
Step 3: Calculate the Concentration of [tex]\( \text{Q}^- \)[/tex]Given [tex]\( s = 7.00 \times 10^{-4} \) M:[/tex]
[tex]{Q}^-] = 3s = 3 \times 7.00 \times 10^{-4} = 2.10 \times 10^{-3} \text{ M} \][/tex]
Thus, the concentration of [tex]\( \text{Q}^- \)[/tex] at equilibrium is[tex]\( 2.10 \times 10^{-3} \)[/tex].
Step 4: Determine the Solubility Product Constant [tex](\( K_{sp} \))[/tex]The solubility product constant [tex]\( K_{sp} \) for \( \text{XQ}_3 \)[/tex] can be calculated using the equilibrium concentrations of the ions:
[tex][ K_{sp} = [\text{X}^{3+}][\text{Q}^-]^3 \][/tex]
Substitute the equilibrium concentrations:
[tex][ [\text{X}^{3+}] = s = 7.00 \times 10^{-4} \text{ M} \][/tex]
[tex][ [\text{Q}^-] = 3s = 2.10 \times 10^{-3} \text{ M} \][/tex]
Now, calculate[tex]K_{sp} \):[/tex]
[tex]\[ K_{sp} = (7.00 \times 10^{-4}) \times (2.10 \times 10^{-3})^3 \[/tex]
First, compute[tex]( (2.10 \times 10^{-3})^3 \)[/tex]
[tex]\[ (2.10 \times 10^{-3})^3 = 2.10^3 \times (10^{-3})^3 = 9.261 \times 10^{-9} \][/tex]
Then, multiply by[tex]7.00 \times 10^{-4} \):[/tex]
[tex][ K_{sp} = (7.00 \times 10^{-4}) \times (9.261 \times 10^{-9}) \][/tex]
[tex][ K_{sp} = 6.4827 \times 10^{-12} \][/tex]
Final Results1. The concentration of[tex]{Q}^- \)[/tex] at equilibrium is [tex]\( 2.10 \times 10^{-3} \) M.[/tex]
2. The [tex]K_{sp} \) value for {XQ}_3 \) is \( 6.4827 \times 10^{-12} \).[/tex]
Which of the following descriptions about standard addition and internal standards are NOT correct?
a) Internal standard is useful when the matrix in the unknown is complicated.
b) Standard addition could be a single standard addition or multiple standard additions to an unknown solution.
c) Standard addition is useful when the matrix in the unknown is complicated.
d) Internal standard is used when instrument response varies from run to run.
Final answer:
Internal standards and standard addition are techniques used in analytical chemistry to ensure accuracy. The incorrect statement is about the use of internal standards.
Explanation:
Internal standards and standard addition are both techniques used in analytical chemistry to ensure the accuracy and reliability of quantitative measurements. In internal standard method, a known amount of a compound is added to all samples and standards, which allows for compensation of errors that may occur during sample preparation and analysis.
Standard addition, on the other hand, involves adding known amounts of a standard solution to an unknown solution to determine the concentration of the analyte of interest. It can be a single standard addition or multiple standard additions, depending on the requirements of the analysis.
Based on the given options, the incorrect description is:
d) Internal standard is used when instrument response varies from run to run.
The correct answer is a) Internal standard is useful when the matrix in the unknown is complicated.
Let's analyze each option to understand why option (a) is incorrect:
a) Internal standard is useful when the matrix in the unknown is complicated.
This statement is incorrect because it confuses the roles of internal standards and standard addition.
An internal standard is a substance that is added in a constant amount to all samples, including the calibration standards and the unknowns. It is used to correct for any variations in the analytical procedure, such as changes in instrument response, sample preparation, and matrix effects that affect the analyte's signal.
However, it is not specifically used because the matrix is complicated; rather, it is used to account for variations in the analytical process. The complexity of the matrix is typically addressed by the standard addition method, which involves adding known quantities of the analyte to the sample to overcome matrix effects.
b) Standard addition could be a single standard addition or multiple standard additions to an unknown solution.
This statement is correct. Standard addition can involve adding a known amount of analyte to the sample once (single standard addition) or several times (multiple standard additions) to construct a calibration curve.
This method is used to compensate for matrix effects that might not be accounted for by using external calibration alone.
c) Standard addition is useful when the matrix in the unknown is complicated.
This statement is correct. The standard addition method is particularly useful for samples with complex matrices that can interfere with the analysis.
By adding known amounts of the analyte directly to the sample, the method allows for the determination of the analyte's concentration while accounting for the matrix effects.
d) Internal standard is used when instrument response varies from run to run.
This statement is correct. An internal standard is used to normalize the response of the analyte and to correct for any variations in the analytical procedure, including changes in instrument response over time.
It helps to ensure the accuracy and precision of the analytical results, regardless of the variations that occur between different runs of the analysis.
Therefore, the statement in option (a) is the one that is not correct, as it misrepresents the use of internal standards in the context of complex matrices.
When the epoxide 2-vinyloxirane reacts with lithium dibutylcuprate, followed by protonolysis, a compound A is the major product formed. Oxidation of A with PCC yields B, a compound that gives a positive Tollens test and has an intense UV absorption around 215 nm. Treatment of B with Ag2O, followed by catalytic hydrogenation, gives octanoic acid. Identify A and B.
Answer:
A --- (E)-oct-2-en-1-o1
B ----(E)-oct-2-enal
Explanation:
See the attached file for the structure.
Which gas is most abundant in Earth’s atmosphere?
oxygen
nitrogen
water vapor
carbon dioxide
Answer:
Nitrogen
Explanation: nitrogen is around 70% of the air hope this helps god bless
Calculate the amount of heat needed to convert 15.0 g of liquid water at 87 ºC to steam at 135. ºC.
45,765 J
815 J
33,900 J
35,807 J
Answer:
The total heat required is 35807 J
Explanation:
Step 1: Data given
Mass of water = 15.0 grams
Initial temperature of water = 87.0 °C
Temperature of steam = 135.0 °C
ΔHfus = 334 J/g
ΔHVap = 2260 J/g
Step 2: Calculate heat required to heat water from 85 to 100 °C
Q = m*c*ΔT
⇒Q = the heat required = TO BE DETERMINED
⇒m = the mass of water = 15.0 grams
⇒c= the specific heat of water = 4.18 J/g°C
⇒ΔT = the change of temperature = T2 - T1 = 100 - 87 = 13°C
Q = 15.0 * 4.18 J/g°C * 13 °C
Q = 815.1 J
Step 3: Calculate heat required to change water at 100 °C to steam
Q = m * ΔHVap
Q = 15.0 grams * 2260 J/g
Q = 33900 J
Step 4: Calculate heat required to heat steam from 100 °C to 135 °C
Q = m*c*ΔT
⇒Q = the heat required = TO BE DETERMINED
⇒m = the mass of water = 15.0 grams
⇒c= the specific heat of steam = 2.09 J/g°C
⇒ΔT = the change of temperature = T2 - T1 = 135 - 100 = 35°C
Q = 15.0 * 2.09 * 35 °C
Q = 1097.25 J
Step 5: Calculate the total heat required
Q = 35807 J
The total heat required is 35807 J
T
An atom has the following electron configuration 1s2 2s2 2p6 3s2 3p4 . How many valence electrons does this Atom have
Answer:
6
Explanation:
This atom is sulfur (if the electrons are equal to the protons/not an ion). You can tell the number of valence electrons by looking at the individual shell. The first shell (1s) can only hold 2 electrons. The second shell (2s and 2p) can hold 8 electrons. The third shell (3s and 3p), which is the valence shell, only has 6 out of its possible 8 electrons, so this atom has 6 valence electrons.
The number of valence electron that the atom with the electronic configuration of 1s²2s²2p⁶3s²3p⁴ has is 6
Valence electron(s) are the electrons located on the outermost shell of an atom.
With the above information in mind, we shall determine the number of valence electron that the atom has. This is illustrated as follow:
Electronic configuration => 1s²2s²2p⁶3s²3p⁴
Valence shell => 3s²3p⁴
Valence electron = 2 + 4
Valence electron = 6Therefore, the atom has 6 valence electrons.
See attachment image
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Describes the three-dimensional arrangement of the atoms in a molecule states that two negatively charged particles (electrons) will always repel one another with equal and opposite forces ensures bonding through shared valence electron pairs is the illustrated definition of the octet rule but not the duet rule
Answer:
When atoms other than hydrogen form covalent bonds, an octet is accomplished by sharing. The octet rule can be used to explain the number of covalent bonds an atom forms. This number normally equals the number of electrons that atom needs to have a total of eight electrons (an octet) in its outer shell
Explanation:
chemistry, the octet rule explains how atoms of different elements combine to form molecules. ... In a chemical formula, the octet rule strongly governs the number of atoms for each element in a molecule; for example, calcium fluoride is CaF2 because two fluorine atoms and one calcium satisfy the rule.
octet rule: Atoms lose, gain, or share electrons in order to have a full valence shell of eight electrons. Hydrogen is an exception because it can hold a maximum of two electrons in its valence level.
There is another rule, called the duplet rule, that states that some elements can be stable with two electrons in their shell. Hydrogen and helium are special cases that do not follow the octet rule but the duplet rule. ... They are stable in a duplet state instead of an octet state.
2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l) was studied with the following results: Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s) 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828 a. Determine the rate law for the reaction. b. Calculate the value of the rate constant with the proper units. c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.
Explanation:
2 ClO2 (aq) + 2OH- (aq)→ ClO3- (aq) + ClO2- + H2O (l)
The data is given as;
Experiment [ClO2] (M) [OH-] (M) Initial Rate (M/s)
1 0.060 0.030 0.0248
2 0.020 0.030 0.00276
3 0.020 0.090 0.00828
a) Rate law is given as;
Rate = k [ClO2]^x [OH-]^y
From experiments 2 and 3, tripling the concentration of [OH-] also triples the rate of the reaction. This means the reaction is first order with respect to [OH-]
From experiments 1 and 2, when the [ClO2] decreases by a factor of 3, the rate decreases by a factor of 9. This means the reaction is second order with respect to [ClO2]
Rate = k [ClO2]² [OH-]
b. Calculate the value of the rate constant with the proper units.
Taking experiment 1,
0.0248 = k (0.060)²(0.030)
k = 0.0248 / 0.000108
k = 229.63 M-2 s-1
c. Calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M.
Rate = 229.63 [ClO2]² [OH-]
Rate = 229.63 (0.100)²(0.050)
Rate = 0.1148 M/s
Final answer:
The rate law is rate = k[ClO2]^2[OH-], with k ≈ 22.2222 M^-2s^-1. The calculated rate with [ClO2] = 0.100 M and [OH-] = 0.050 M is approximately 0.111 M/s.
Explanation:
To determine the rate law for the reaction given, we should look at the changes in concentration and the effect on the initial rate.
Comparing experiment 1 and 2:
When [ClO2] is changed from 0.060 M to 0.020 M (decreased by a factor of 3), the initial rate goes from 0.0248 M/s to 0.00276 M/s (decreased by a factor of ~9).This suggests that the rate is proportional to [ClO2] squared ([ClO2]^2).Comparing experiment 2 and 3:
When [OH-] is increased from 0.030 M to 0.090 M (tripled), the rate increases from 0.00276 M/s to 0.00828 M/s (tripled).This indicates the rate is first order with respect to [OH-].Therefore, the rate law is: rate = k[ClO2]^2[OH-].
Using experiment 1 data to calculate the rate constant (k):
0.0248 M/s = k(0.060^2)(0.030)k = 0.0248 / (0.060^2 * 0.030)k ≈ 22.2222 M-2s-1To calculate the rate when [ClO2] = 0.100 M and [OH-] = 0.050 M:
rate = 22.2222 M-2s-1 * (0.100)^2 * 0.050rate = 0.111 M/sHow many grams of NaCl are needed to prepare 50.0 grams of 35.0% of salt solution?
Answer:
17.5 g
Explanation:
Given data
Mass of solution to be prepared: 50.0 gramsConcentration of the salt solution: 35.0%The concentration by mass of NaCl in the solution is 35.0%, that is, there are 35.0 grams of sodium chloride per 100 grams of solution. We will use this ratio to find the mass of sodium chloride required to prepare 50.0 grams of a 35.0% salt solution.
[tex]50.0gSolution \times \frac{35.0gNaCl}{100gSolution} = 17.5gNaCl[/tex]
To prepare a 50 gram 35% salt solution, you need 17.5 grams of NaCl.
Explanation:To find out how many grams of NaCl are needed to prepare 50.0 grams of a 35.0% salt solution, you use the definition of percent concentration by mass: (mass of solute/mass of solution) x 100%. The mass of the solution is the total mass, which is the sum of the mass of the solute (NaCl in this case) and the solvent (usually water). For a 35% solution, this means that 35 grams of NaCl are in every 100 grams of solution.
However, we want to prepare 50 grams of solution. So, you set up a ratio: (35 g NaCl/100 g solution) = (x g NaCl/50 g solution). Solving this equation, you find that x = 17.5. Therefore, 17.5 grams of NaCl are needed to prepare a 50 gram 35% salt solution.
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Nitrate salts (NO3-), when heated, can produce nitrites (NO2-) plus oxygen (O2). A sample of potassium nitrate is heated, and the O2 gas produced is collected in a 730 mL flask. The pressure of the gas in the flask is 2.5 atm, and the temperature is recorded to be 329 K. The value of R= 0.0821 atm L/(mol K)
How many moles of O2 gas were produced?
Answer:
0.0676 moles of oxygen gas is produced
Explanation:
Step 1: Data given
Volume of the flask = 730 mL = 0.730 L
The pressure of the gas in the flask is 2.5 atm
The temperature is recorded to be 329 K
Step 2: The balanced equation
2NO3- → 2NO2- + O2
Step 3: Calculate moles oxygen gas (O2)
p*V =n * R*T
⇒with p = the pressure of oxygen gas = 2.5 atm
⇒with V = the volume of the flask = 0.730 L
⇒with n = the number of moles O2 gas
⇒with R = the gas constant = 0.0821 L*atm/mol*K
⇒with T ⇒ the temperature = 329 K
n = (p*V) / (R*T)
n = (2.5 * 0.730 ) / (0.0821*329)
n = 0.0676 moles
0.0676 moles of oxygen gas is produced
Which statement best describes what is taking place? Copper is being oxidized. Copper is being reduced. Copper is losing electrons. Copper is a reducing agent.
Answer:
copper is being reduced
Explanation: I got 100% on the quiz good luck this class is hard lol
Answer:
The answer is the second option
Copper is being reduced
Explanation:
Have a good day just took test
a gas at 80kPa occupies a volume of 5mL. WHat volume will the gas occupy at 70kPa?
Answer:
Volume = 5.71mL
Explanation:
Applying Boyle's law
P1V1= P2V2
P1= 80kPa, V1= 5mL,P2= 70kPa, V2=?
Substitute into above formula
80×5= 70×V2
V2= (80×5)/70 = 5.71mL
Lithium has two naturally occurring isotopes: lithium−6 and lithium−7. Lithium−6 has a mass of 6.01512 relative to carbon−12 and makes up 7.47 percent of all naturally occurring lithium. Lithium−7 has a mass of 7.016 compared to carbon−12 and makes up the remaining 92.53 percent. According to this information, what is the atomic weight of lithium?
Answer:
6.941
Explanation:
Step 1:
Representation.
Let Lithium−6 be isotope A
Let Lithium−7 be isotope B
Let the abundance of isotope A (Lithium-6) be A%
Let the abundance of isotope B (Lithium-6) be B%
Step 2:
Data obtained from the question. This includes:
Mass of isotope A (Lithium−6) =
6.01512
Abundance of isotope A (Lithium−6) = A% = 7.47%
Mass of isotope B (Lithium−7) = 7.016
Abundance of isotope B (Lithium−7) = B% = 92.53%
Atomic weight of lithium =?
Step 3:
Determination of the atomic weight of lithium. This is illustrated below:
Atomic weight = [(Mass of AxA%)/100] + [(Mass of BxB%)/100]
Atomic weight = [(6.01512x7.47)/100] + [(7.016x92.53)/100]
Atomic weight = 0.449 + 6.492
Atomic weight of lithium is 6.941
Final answer:
The atomic weight of lithium is approximately 6.9423 u, which is the weighted average of the masses of its isotopes, lithium-6 and lithium-7, with respect to their natural abundances.
Explanation:
To calculate the atomic weight of lithium, we consider the relative abundances and masses of its naturally occurring isotopes, lithium-6 and lithium-7. The atomic weight is the weighted average of the atomic masses of an element's isotopes, based on their natural abundance. Using the provided information:
Lithium-6 has a mass of 6.01512 and an abundance of 7.47%.
Lithium-7 has a mass of 7.016 and an abundance of 92.53%.
The calculation is as follows:
(6.01512 × 0.0747) + (7.016 × 0.9253) = 0.4487 + 6.4936 = 6.9423
Therefore, the atomic weight of lithium is approximately 6.9423 u (atomic mass units).
A precipitate forms when mixing solutions of silver nitrate (AgNO,) and sodium chloride (NaCI). Complete the net ionic equation for this reaction by filling in the blanks. Do not include charges on any ions and do not include phase symbols. Provide your answer below (aq)+(aq)6)
The reaction between solutions of silver nitrate and sodium chloride results in a precipitate due to a double-replacement reaction. The solid precipitate formed is silver chloride (AgCl). The net ionic equation becomes: Ag+ (aq) + Cl- (aq) → AgCl (s).
Explanation:
When you mix solutions of silver nitrate (AgNO3) and sodium chloride (NaCl), a precipitate forms. This is because of a double-replacement reaction that occurs between the two compounds. You start with AgNO3 (silver nitrate) and NaCl (sodium chloride), and end up with NaNO3 (sodium nitrate) and AgCl (silver chloride). The silver chloride precipitates, or becomes solid, leaving the sodium nitrate in solution. This can be shown by the net ionic equation:Ag+ (aq) + Cl- (aq) → AgCl (s). This equation represents the formation of the silver chloride precipitate. The ions Na+ and NO3- do not take part in the reaction and so are not included in the net ionic equation.
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Consider the general reaction: 2 A + b B → c C and the following average rate data over a specific time period \Delta t: - \frac{ \Delta A}{\Delta t} = 0.0080 mol L-1 s-1 - \frac{ \Delta B}{\Delta t} = 0.0120 mol L-1 s-1 \frac{ \Delta C}{\Delta t} = 0.0160 mol L-1 s-1 Determine what the coefficient c could be in the balanced equation. Select one: 1 5 2 3 4
Answer:
c = 4
Explanation:
In general, for the reaction
a A + b B ⇒ c C + d D
the rate is given by:
rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt
this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.
For this question in particular we know the coefficient of A and need to determine the coefficient c.
- 1/2 ΔA/Δt = + 1/c ΔC/Δt
- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )
0.0040 mol L⁻¹s⁻¹ c = 0.0160 mol L⁻¹s⁻¹
∴ c = 0.0160 / 0.0040 = 4
A titration is performed to determine the amount of sulfuric acid, H2SO4, in a 6.5 mL sample taken from car battery. About 50 mL of water is added to the sample, and then it is titrated with 43.37 mL of a standard 0.5824 molar NaOH solution. You balanced this reaction in a previous problem. How what is the molar concentration of sulfuric acid in the original sample
Answer: The molar concentration of sulfuric acid in the original sample is 1.943 M
Explanation:
To calculate the molarity of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=2\\M_1=?\\V_1=56.5mL\\n_2=1\\M_2=0.5824M\\V_2=43.37mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 56.5=1\times 0.5824\times 43.37[/tex]
[tex]M_1=0.2235[/tex]
Now to calculate the molarity of original solution:
[tex]M_1\times 6.5=0.2235\times 56.5[/tex]
[tex]M_1=1.943[/tex]
Thus the molar concentration of sulfuric acid in the original sample is 1.943 M
Final answer:
The molar concentration of sulfuric acid in the original 6.5 mL sample is calculated to be approximately 1.9436 M, based on titration with a 0.5824 M NaOH solution.
Explanation:
Determining the Concentration of Sulfuric Acid in a Sample Using Titration:
To determine the molar concentration of sulfuric acid, H2SO4, we will use the data that 43.37 mL of a 0.5824 M NaOH solution was needed to titrate a 6.5 mL sample of the acid.
Firstly, we calculate the moles of NaOH used in the titration:
Moles of NaOH = Volume (in Liters) imes Molarity
Moles of NaOH = 0.04337 L imes 0.5824 mol/L
Moles of NaOH = 0.02526888 mol
According to the balanced chemical equation, H2SO4 + 2NaOH
ightarrow Na2SO4 + 2H2O, the stoichiometry of the reaction is 1:2. This means one mole of sulfuric acid reacts with two moles of sodium hydroxide. Thus, the moles of H2SO4 will be half of the moles of NaOH used.
Moles of H2SO4 = 0.02526888 mol / 2
Moles of H2SO4 = 0.01263444 mol
Now, we calculate the molarity of the sulfuric acid in the original sample:
Molarity of H2SO4 = Moles of H2SO4 / Volume of sample in liters
Molarity of H2SO4 = 0.01263444 mol / 0.0065 L
Molarity of H2SO4 = 1.9436 M
Therefore, the molar concentration of the sulfuric acid in the original sample is approximately 1.9436 M.
1. Choose the alkyl halide(s) from the following list of C6H13Br isomers that meet each criterion below. 1) 1-bromohexane 2) 3-bromo-3-methylpentane 3) 1-bromo-2,2-dimethylbutane 4) 3-bromo-2-methylpentane 5) 2-bromo-3-methylpentane a) the compound(s) that can exist as enantiomers b) the compound(s) that can exist as diastereomers c) the compound that gives the fastest SN2 reaction with sodium methoxide d) the compound that is least reactive to sodium methoxide in methanol e) the compound(s) that undergo an SN1 reaction to give rearranged products f) the compound that gives the fastest SN1 reaction
Answer:
See explanation
Explanation:
the compound that gives the fastest SN2 reaction with sodium methoxide- 1-bromohexane
the compound that gives the fastest SN1 reaction- 3-bromo-3-methylpentane
the compound(s) that undergo an SN1 reaction to give rearranged products- 1-bromo-2,2-dimethylbutane
the compound that is least reactive to sodium methoxide in methanol -
3-bromo-3-methylpentane
the compound(s) that can exist as diastereomers - 3-bromo-3-methylpentane
the compound(s) that can exist as enantiomers- 3-bromo-2-methylpentane
The reaction between nitrogen dioxide and carbon monoxide is NO2(g)+CO(g)→NO(g)+CO2(g)NO2(g)+CO(g)→NO(g)+CO2(g) The rate constant at 701 KK is measured as 2.57 M−1⋅s−1M−1⋅s−1 and that at 895 KK is measured as 567 M−1⋅s−1M−1⋅s−1. The activation energy is 1.5×102 1.5×102 kJ/molkJ/mol. Predict the rate constant at 525 KK .
Answer : The rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]701K[/tex] = [tex]2.57M^{-1}s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]525K[/tex] = ?
[tex]Ea[/tex] = activation energy for the reaction = [tex]1.5\times 10^2kJ/mol=1.5\times 10^5J/mol[/tex]
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = 701 K
[tex]T_2[/tex] = final temperature = 525 K
Now put all the given values in this formula, we get:
[tex]\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}][/tex]
[tex]K_2=0.0606M^{-1}s^{-1}[/tex]
Therefore, the rate constant at 525 K is, [tex]0.0606M^{-1}s^{-1}[/tex]
What is the molar mass of BaBrz?
Answer:
Molar mass= 297.1gmol-1
Explanation:
BaBr2 = 137.3+ (79.9*2)= 297.1gmol-1
1. A smoothie contains 1 banana (B), 4 strawberries (St), 1 container of yogurt (Y), and 3 ice cubes (Ic). Write a balanced equation to describe the relationship. *
Write a conversion factor to show the relationship between the number of ice cubes and the number of smoothies produced. *
How many strawberries would you need to make 12 smoothies? *
To calculate the atoms of an element in a given molecule, we need to multiply stoichiometry by the number that is written on the foot of that element. Therefore, the balanced equation is
1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]
What is Balanced equation?Balanced equation is the one in which the total number of atoms of a species on reactant side is equal to the total number of atoms on product side. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, displacement reaction.
The other characteristic of balanced reaction is that physical state should be written with each compound or molecule on reactant and product side. Physical state should be written in brackets. s means solid, l means liquid, g means gas.
The word equation is
1 banana+ 4 strawberries+1 container of yogurt+3 ice cubes[tex]\rightarrow[/tex] 1 smoothie
The balanced chemical equation is
1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]
Therefore, the balanced equation is
1B+4St+1Y+3Ic[tex]\rightarrow[/tex]1 BSt[tex]_4[/tex]YIc[tex]_3[/tex]
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