Answer:
False
False
True
False
False
False
Explanation:
The colour emitted by an excited atom depends on the emission Spectra of the atom rather than on the number of free electrons passing through the lamp.
When a free electrons hits an atom, the atom can be excited to various intermediate states other than the highest energy level depending on the amount of energy it absorbed.
Increase in voltage of the battery also increases the kinetic energy of the electron
The kinetic energy of the electron depends on the voltage difference not on the distance from the atom.
The Paschen, Brackett, and Pfund series of lines are due to transitions from higher-energy orbits to orbits with n = 3, 4, and 5, respectively; these transitions release substantially less energy, corresponding to infrared radiation.
Excited atoms can jump from a higher level to the ground state in a series of steps or directly to the ground state
A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate the pHpH after the addition of 13.0 mLmL of HNO3
Answer:
The pH is [tex]pH = 9.4[/tex]
Explanation:
From the question we are told that
The volume of [tex]NH_3[/tex] is [tex]V_N = 75mL = 75 *10^{-3} L[/tex]
The concentration of [tex]NH_3[/tex] is [tex]C_N = 0.200M[/tex]
The concentration of [tex]HNO_3[/tex] is [tex]C_H = 0.500 M[/tex]
The volume of [tex]HNO_3[/tex] added is [tex]V_H = 13mL = 13 *10^{-3 } L[/tex]
The base dissociation constant is [tex]K_b = 1.8*10^{-5}[/tex]
The number of moles of [tex]HNO_3[/tex] that was titrated can be mathematically represented as
[tex]n__{H}} = C_H * V_H[/tex]
substituting values
[tex]n__{H}} = 0.500* 13*10^{-3}[/tex]
[tex]n__{H}} = 0.0065 \ moles[/tex]
The number of moles of [tex]NH_3[/tex] that was titrated can be mathematically represented as
[tex]n__{N}} = C_N * V_N[/tex]
substituting values
[tex]n__{N}} = 0.200 * 75*10^{-3}[/tex]
[tex]n__{N}} = 0.015 \ mole[/tex]
So from the calculation above the limited reactant is [tex]HNO_3[/tex]
The chemical equation for this reaction is
[tex]NH_3 + HNO_3 ------> NH^{4+} + NO^{3+}[/tex]
From the chemical reaction
1 mole of [tex]HNO_3[/tex] is titrated with 1 mole of[tex]NH_3[/tex] to produce 1 mole of NH^{4+}
So
0.0065 moles of [tex]HNO_3[/tex] is titrated with 0.0065 mole of [tex]NH_3[/tex] to produce 0.0065 mole of [tex]NH^{4+}[/tex]
So
The remaining moles of [tex]NH_3[/tex] after the titration is
[tex]n = n__{N}} - n__{H}}[/tex]
=> [tex]n = 0.015 - 0.0065[/tex]
[tex]n = 0.0085 \ moles[/tex]
Now according to Henderson-Hasselbalch equation the pH of the reaction is mathematically represented as
[tex]pH = pK_a + log [\frac{NH_3}{NH^{4+}} ][/tex]
Where [tex]pK_b[/tex] is mathematically represented as
[tex]pK_a = -log K_a[/tex]
Now [tex]K_a = \frac{K_w}{K_b}[/tex]
Where [tex]K_w[/tex] is the ionization constant of [tex]NH_3[/tex] with value [tex]K_w = 1.0*10^{-14}[/tex]
Hence [tex]K_a = \frac{1.0*10^{-14}}{1.8 *10^{-5}}[/tex]
[tex]K_a = 5.556 * 10^{-10}[/tex]
Substituting this into the equation
[tex]pH = -log K_a + log [\frac{NH_3}{NH^{4+}} ][/tex]
[tex]pH = log [\frac{\frac{NH_3}{NH^{4+}} }{K_a} ][/tex]
substituting values
[tex]pH = log [\frac{\frac{0.0085}{0.0065} }{5.556*10^{-10}} ][/tex]
[tex]pH = 9.4[/tex]
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42 kJ/mol. Theoretically, to what temperature (°C) would one have to heat the solution so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction at 20 °C? Assume the frequency factor A is constant, and assume the initial concentrations are the same.
Answer:
T = 215.33 °C
Explanation:
The activation energy is given by the Arrhenius equation:
[tex] k = Ae^{\frac{-Ea}{RT}} [/tex]
Where:
k: is the rate constant
A: is the frequency factor
Ea: is the activation energy
R: is the gas constant = 8.314 J/(K*mol)
T: is the temperature
We have for the uncatalyzed reaction:
Ea₁ = 70 kJ/mol
And for the catalyzed reaction:
Ea₂ = 42 kJ/mol
T₂ = 20 °C = 293 K
The frequency factor A is constant and the initial concentrations are the same.
Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:
[tex] k_{1} = k_{2} [/tex]
[tex] Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}} [/tex] (1)
By solving equation (1) for T₁ we have:
[tex]T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C[/tex]
Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.
I hope it helps you!
Consider the following scenario
You are the manager of a chemical stockroom, and find a bottle containing approximately one liter of a clear and colorless solution of unknown identity and concentration. Your only clue to its identity is that it was found between bottles of silver fluoride and sodium fluoride, so it is likely an aqueous solution of one of those two compounds. You will need to develop a procedure to determine the following:
a) The identity of the unknown solution
b) The concentration of the unknown solution
Write out a precise procedure, which includes all glassware, reagents, and steps. You will also need to write the calculations that you would need to determine the concentration of the solution. Assume that you have access to all of the equipment that you used in the chemistry lab this semester and any reagent you might need. To complete this assignment, consider both the techniques learned in lab and the information learned in lecture.
Answer:
See explaination
Explanation:
Going by the clues that it is between Silver Flouride (AgF) and Sodium Fluoride (NaF) and since it is an aqueous solution , the 1 liter bottle is likely to be Sodium Chloride( NaCl). Going by the reaction,
AgF + NaCl= AgCl + NaF
Here, the color of AgCl is white, hence the solution cannot be AgCl.
Determination of NaCl
Determination of NaCl can be done by Mohr's Method or Volhard's method. But results in Volhard's method are more accurate . Its uses the method of back titration with Potassium Thiocynate which forms a AgCl precipitate . Prior to titration,excess AgNO3 ( The problem also has a clue that excess reagents are present in the lab ) is added to the NaCl solution so that all the Cl- ions react with Ag+. Fe3+ is then added as an indicator and the solution is titrated with KSCN to form a silver thiocyannite precipitate (AgSCN). Once all the silver has reacted, a slight excess of SCN- reacts with Fe3+ to form Fe(SCN)3 dark red complex. The concentration of Cl- is determined by subtracting the titer findings of Ag+ ions that reacted to form AgSCN from the Ag NO3 moles added to the solution. This is used because pH of the solution is acidic. If the pH of solution is basic, Mohr's method is used.
Reactions
Ag+ (aq)+ Cl-(aq) = AgCl(aq)
Ag+(aq) + SCN-(aq) = AgSCN(aq)
Fe3+(aq) + SCN-(aq) = [FeSCN]2- (aq)
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq), as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 235 mL Cl2(g)235 mL Cl2(g) at 25 °C and 805 Torr805 Torr?
Answer: 0.887 g of [tex]MnO_2[/tex] should be added to excess HCl(aq).
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 805 torr = 1.06 atm (760torr=1atm)
V = Volume of gas = 235 ml = 0.235 L
n = number of moles = ?
R = gas constant =[tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]25^0C=(25+273)K=298K[/tex]
[tex]n=\frac{PV}{RT}[/tex]
[tex]n=\frac{1.06atm\times 0.235L}{0.0820 L atm/K mol\times 298K}=0.0102moles[/tex]
[tex]MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)[/tex]
According to stoichiometry:
1 mole of chlorine is produced by = 1 mole of [tex]MnO_2[/tex]
Thus 0.0102 moles of chlorine is produced by = [tex]\frac{1}{1}\times 0.0102=0.0102[/tex] moles of [tex]MnO_2[/tex]
Mass of [tex]MnO_2[/tex] =[tex]moles\times {\text {Molar mass}}=0.0102mol\times 87g/mol=0.887g[/tex]
0.887 g of [tex]MnO_2[/tex] should be added to excess HCl(aq).
A steel container filled with H₂ gas is at a pressure of 6.5 atm and a temperature of 22°C. If the container is placed near a furnace and is heated to a temperature of 50°C, what would be the new pressure in the container?
Answer: The new pressure is 7.1 atm
Explanation:
To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=6.5atm\\T_1=22^0C=(22+273)K=295K\\P_2=?\\T_2=50^0C=(50+273)K=323K[/tex]
Putting values in above equation, we get:
[tex]\frac{6.5}{295}=\frac{P_2}{323}\\\\P_2=7.1[/tex]
Hence, the new pressure is 7.1 atm
Final answer:
The new pressure in the steel container filled with H₂ gas, after being heated from 22°C to 50°C, is approximately 7.18 atm, calculated using Gay-Lussac's law.
Explanation:
The question deals with the effect of temperature change on the pressure of a gas within a steel container. According to Gay-Lussac's law, the pressure of a gas is directly proportional to its absolute temperature when the volume and the amount of gas are held constant. This situation can be represented mathematically as P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature, respectively.
To find the new pressure in the container when the temperature is increased from 22°C to 50°C, we first need to convert these temperatures to Kelvin by adding 273. Thus, T1 = 22 + 273 = 295 K and T2 = 50 + 273 = 323 K. Substituting the initial conditions into the equation along with T2, we can solve for P2. Using P1 = 6.5 atm and rearranging the formula gives us P2 = P1 * (T2/T1) = 6.5 * (323/295) ≈ 7.18 atm. Therefore, the new pressure in the container after heating is approximately 7.18 atm.
Which system starts the process of breathing?
A. nervous
B. muscular
C. circulatory
Answer:
your nervous system C:
Answer:
It would be C.
(Circulatory)
Explanation:
Breathing starts when you inhale air into your nose or mouth. It travels down the back of your throat and into your windpipe, which is divided into air passages called bronchial tubes. For your lungs to perform their best, these airways need to be open. Knowing that it all starts from the inside is what you should remember.
I hope this helps. :)The major product of reacting the molecule above with sodium cyanide in di-methyl formamide (DMF) would be ?
Help please
Answer:
DMF is an aprotic polar solvent.
hence CN acts as a strong nucleophile in DMF to give SN2 reaction with the reactant resulting into inversion of configuration at stereo centers.
Explanation:
check the attached file for little explanation(diagram)
The correct answer is that the major product of reacting the given molecule with sodium cyanide in di-methyl formamide (DMF) would be the nitrile compound formed by the nucleophilic addition of the cyanide ion to the carbonyl carbon of the molecule.
To understand the reaction, let's consider the mechanism step by step:
1. The carbonyl group in the given molecule contains a carbon atom that is electrophilic due to the partial positive charge on it, resulting from the electron-withdrawing effect of the oxygen atom.
2. Sodium cyanide (NaCN) dissociates in the polar aprotic solvent DMF to give free cyanide ions (CN^-). The cyanide ion is a good nucleophile because of the negative charge on the carbon atom.
3. The cyanide ion attacks the electrophilic carbonyl carbon, leading to the formation of a tetrahedral intermediate.
4. The intermediate then undergoes elimination of a hydroxide ion (OH^-), which is a good leaving group, to form the nitrile.
5. The final product is the nitrile compound, where the carbonyl oxygen has been replaced by the cyanide group.
The reaction can be represented as follows:
In this reaction, R and R' represent the alkyl or aryl groups attached to the carbonyl carbon. The product is a nitrile, which has a carbon triple-bonded to a nitrogen (C}\equiv\text{N}), and this is the major product of the reaction under the given conditions.
It is important to note that the reaction conditions, such as the use of a polar aprotic solvent like DMF, favor the formation of the nitrile by stabilizing the transition state and the intermediate, and by not protonating the cyanide ion, which would reduce its nucleophilicity.
Laura has 3 beakers. Each contain 200cm3 of colourless liquid. Describe how Laura could determine which beakers contain pure water and which contain solutions
Answer:
Laura can look for a transparent and translucent liquid and hence determine which beaker has water and which has solution
Explanation:
Pure water is a compound that is transparent in color. However, a solution is a liquid mixture comprising of a solvent or a solute. The atoms of solute occupy space between the atoms of solvent and hence are said to dissolve in it. Water can be a solvent.
Thus, if the beaker has a transparent liquid in it, then it would be pure water while a beaker having a translucent liquid, then it would be a solution
Final answer:
Laura can differentiate pure water from solutions by using a conductivity test, indicators like pH paper or red cabbage water, measuring boiling and freezing points, or measuring the refractive index with a refractometer.
Explanation:
Laura can determine which beakers contain pure water and which contain solutions by conducting a series of tests that rely on the physical and chemical properties of the substances. Here are potential methods for differentiating between pure water and solutions:
Conducting a conductivity test, as pure water is a poor conductor of electricity while solutions with ions will conduct electricity.
Using indicators such as pH paper or red cabbage water, as solutions might be acidic or basic, changing the color of the indicator while pure water will not.
Examining the boiling and freezing points, as solutions have different boiling and freezing points compared to pure water.
Employing a refractometer or similar devices to measure the refractive index, which would differ between pure water and a solution.
To elaborate, a conductivity test can be set up by inserting electrodes into each beaker and connecting these to a circuit with a bulb or a conductivity meter. A color change when using red cabbage water as an indicator would signify the presence of an acidic or basic solution, since red cabbage juice changes color at different pH levels. Observing the boiling and freezing points would require heating or cooling the liquid and taking note of the temperature at which the change of state occurs. Pure water has specific boiling and freezing points (100°C and 0°C at standard atmospheric pressure), and deviations from these numbers would indicate a solution. Lastly, a refractometer could be used to compare the refraction of light through the liquids against known values for pure substances, revealing which beakers contain pure water and which contain solutions.
The small bags of silica gel you often see in a new shoe box are placed there to control humidity. Despite its name, silica gel is a solid. It is a chemically inert, highly porous, amorphous form of SiO2. Water vapor readily adsorbs onto the surface of silica gel, so it acts as a desiccant. Despite not knowing mechanistic details of the adsorption of water onto silica gel, from the information provided, you should be able to make an educated guess about the thermodynamic characteristics of the process. Predict the signs of ΔG, ΔH, and ΔS. Predict the sign of ΔG. ΔG = 0 ΔG < 0 ΔG > 0 Predict the sign of ΔH. ΔH < 0 ΔH > 0 ΔH = 0 Predict the sign of ΔS. ΔS > 0 ΔS < 0 ΔS = 0
Answer:
The sign of \Delta GΔG is negative.
The sign of \Delta HΔH is negative.
The sign of \Delta SΔS is negative.
Explanation:
The water vapor is adsorbed on silica gel due to strong hydrophilicity of silica get towards the water.
The thermodynamic properties of adsorbate and adsorbent changes after adsorption. Silica gel is very porous and hydrophilic, thus, it absorbs the water from the shoe box.
The adsorption process occurs spontaneously, therefore Delta G < 0.
When adsorption occurs, bonds are formed between water molecules and SiO2, and the bond formation process is exothermic (heat is released).
Thus, Delta H < 0.
The water molecules become immobilized on the surface when adsorption occurs, thus, entropy/disorder decreases.
So, Delta S < 0.
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . NaOH(aq). Calculate the amount of Ga ( s ) Ga(s) that can be deposited from a Ga ( III ) Ga(III) solution using a current of 0.880 A 0.880 A that flows for 30.0 min .
Answer:
0.382 g
Explanation:
Let's consider the reduction of gallium (III) to gallium that occurs in the electrolysis.
Ga³⁺ + 3 e⁻ → Ga
We can establish the following relations:
1 minute = 60 second1 Ampere = 1 Coulomb / secondThe charge of 1 mole of electrons is 96,468 Coulomb (Faraday's constant)1 mole of gallium is deposited when 3 moles of electrons circulate.The molar mass of gallium is 69.72 g/molWe will use this that to determine the mass of gallium deposited from a Ga(III) solution using a current of 0.880 A that flows for 30.0 min
[tex]30.0min \times \frac{60s}{1min} \times \frac{0.880c}{s} \times \frac{1mole^{-} }{96,468c} \times \frac{1molGa}{3 mole^{-}} \times \frac{69.72g}{1molGa} = 0.382 g[/tex]
Write a balanced net ionic equation to show why the solubility of NiCO3 (s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Consider only the FIRST STEP in the reaction with strong acid. Use the pull-down boxes to specify states such as (aq) or (s). + + + K =
Final answer:
The solubility of NiCO3 increases in the presence of a strong acid due to the neutralization reaction that occurs. The balanced net ionic equation for this reaction is NiCO3 (s) + 2H+ (aq) → Ni2+ (aq) + CO2 (g) + H2O (l). The equilibrium constant for the reaction can be considered as the solubility product constant (Ksp) for NiCO3, which is 1.36 x 10^-7.
Explanation:
The solubility of NiCO3 (s) increases in the presence of a strong acid due to the neutralization reaction that occurs. When a strong acid is added, it provides H+ ions which react with the carbonate ions of NiCO3 to form carbonic acid, H2CO3. Carbonic acid is unstable and decomposes into water and carbon dioxide, causing more NiCO3 to dissolve.
The balanced net ionic equation for this reaction is:
NiCO3 (s) + 2H+ (aq) → Ni2+ (aq) + CO2 (g) + H2O (l)
As for calculating the equilibrium constant, since the question only asks for the first step in the reaction with the strong acid, the equilibrium constant can be considered to be the solubility product constant (Ksp) for NiCO3. The solubility product constant is given as Ksp = 1.36 x 10-7 for NiCO3.
Experiment #1A melting point of an old sample of Naphthalene was completed and a melting range of 77-83 oC was observed and recorded. Complete the following calculations and show your work. 1. From the information and data in the JoVE video, calculate the percent error of the melting point compared to the literature value for the naphthalene sample.2. What does the melting point range and percent error suggest about the sample?
Answer:
%error = 0.32%
Explanation:
Let's answer both questions, by parts.
1. Percentage error:
In this case, I do not have the video, but I do have the reported melting point of naphtalene which is 80.26 °C.
The expression to calculate the percentage error is the following:
%Error = absolute error / actual percentage. (1)
And the absolute error is:
Abs error = actual value - experimental value (2)
But the experimental value is a range, so we just have to get a average of that:
Exp value = 77 + 83 / 2 = 80 °C
Now the absolute error:
Abs error = 80.26 - 80 = 0.26 °C
Finally the %error:
%error = (0.26 / 80.26) * 100
%error = 0.32%2. Meaning of melting point range and %error
The melting point range just means that the sample of naphtalene has impurities, and when a sample of any compound has impurities, melting point tends to be low. However, this decrease of temperature is a wider range. But usually a range of just 5° C means that compound has little traces of impurities but it can still be used for reactions.
The %error means that the impurities of the sample are really low, so the sample is practically pure with little traces of impurities.
Final answer:
The percent error for the melting point of naphthalene is calculated using the formula and considering the literature value of 80.2°C, yielding an approximate percent error of 0.25%. The broad melting range and low percent error suggest the sample is mostly pure with possible minor impurities.
Explanation:
The student's question is regarding the calculation of the percent error for the melting point of a sample of naphthalene and what the results suggest about the sample's purity.
Percent Error Calculation
To find the percent error, we use the formula:
Percent Error = (|Experimental Value - Literature Value| / Literature Value) x 100%
Assuming the literature value of naphthalene's melting point is approximately 80.2°C (the value will need to be verified as it can vary slightly in literature), we calculate percent error using the experimental value's mean (80°C, the midpoint of 77-83°C) as follows:
Percent Error = (|80°C - 80.2°C| / 80.2°C) x 100% = (0.2°C / 80.2°C) x 100% ≈ 0.25%
Suggestion About the Sample
The observed melting range of 77-83°C and the low percent error suggest that the sample is relatively pure but may contain minor impurities since a pure sample would have a narrower melting point range close to the literature value.
what is the molarity of a solution made by dissolving 21.2 g of sodium hydroxide in enough water to make 7.92 L of solution
Answer:
0.067M NaOH is the answer
What are the reactions that allow the conversion of cytosolic NADHNADH into NADPHNADPH during fatty acid biosynthesis? malate+NADP+⟶pyruvate+CO2+NADPHmalate+NADP+⟶pyruvate+CO2+NADPH glucose 6-phosphate+2NADP++H2O⟶ribulose 5-phosphate+2NADPH+2H++CO2glucose 6-phosphate+2NADP++H2O⟶ribulose 5-phosphate+2NADPH+2H++CO2 oxaloacetate+NADH+H+↽−−⇀malate+NAD+oxaloacetate+NADH+H+↽−−⇀malate+NAD+ pyruvate+CO2+ATP+H2O⟶oxaloacetate+ADP+Pi+2H+pyruvate+CO2+ATP+H2O⟶oxaloacetate+ADP+Pi+2H+ What enzymes are required? malic enzyme pyruvate carboxylase glucose 6‑phosphate dehydrogenase malate dehydrogenase What is the sum of these reactions?
Answer:
See explaination
Explanation:
Fatty acid synthesis is the creation of fatty acids from acetyl-CoA and NADPH through the action of enzymes called fatty acid synthases. This as a process usually takes place in the cytoplasm of the cell.
Check attachment for further solution of the given problem.
Which headings should be used to complete this table?
A. ether (left column); ester (right column)
B. ketone (left column); aldehyde (right column)
C. alkyl halide (left column); amine (right column)
D. alcohol (left column); carboxylic acid (right column)
Headings that should be used is alkyl halide (left column); amine (right column).
What are Organic Compounds?Chemical compounds in which carbon atoms are covalently linked to the other compounds like nitrogen, oxygen and hydrogen.Ether, Ester, Ketone, Aldehyde, Alkyl halide, Amine, Alcohol are some of the examples.Given: organic compounds
According to the table:
Left Column:
R-X, (X = F, Cl, Br, I) are the alkyl halide.
Alkyl halides are used as an organic solvent.
Right Column:
R-NH₂ are the amines.
Amines are the weak bases, they are used in many biochemical processes and they have foul smell.
Therefore, headings that should be used to complete the given table will be alkyl halide (left column); amine (right column).
Hence, option (C) is correct.
Learn more about the Organic Compounds here: https://brainly.com/question/704297
#SPJ2
Answer:
ether (left column); ester (right column)
ketone (left column); aldehyde (right column)
alkyl halide (left column); amine (right column)
alcohol (left column); carboxylic acid (right column)
Explanation:
The student titrated 10 ml of standered 0.15 M HCl with his sodium hydroxide solution. When the titration reached the equivalence point, the student found that he had used
10.3 ml of Sodium hydroxide solution. Calculate the molarity of the sodium hydroxide solution.
Answer:
0.15 M
Explanation:
Step 1: Write the neutralization reaction
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the moles of HCl that reacted
10 mL of 0.15 M HCl was used. The moles of HCl that reacted are:
[tex]0.010L \times \frac{0.15mol}{L} = 1.5 \times 10^{-3} mol[/tex]
Step 3: Calculate the moles of NaOH that reacted
The molar ratio of NaOH to HCl is 1:1. Then, the moles of NaOH that reacted are 1.5 × 10⁻³ moles.
Step 4: Calculate the concentration of NaOH
1.5 × 10⁻³ moles of NaOH are in 10.3 mL of solution. The molarity of NaOH is:
[tex]\frac{1.5 \times 10^{-3} mol}{10.3\times 10^{-3}L} =0.15 M[/tex]
How is the Gobi Desert different from the Sahara Desert
Explanation: The Gobi actually gets snow and frost! ... The Sahara is also seven times larger, taking up half a continent: Almost ten million square kilometres, compared to only 1.3 million for the Gobi. Both are expanding, though; the Gobi is gaining another 3.5 thousand square km a year, and the Sahara a commensurate amount
Answer:
I have the same question same
Explanation:
Solve the following problem:
Answer:
Option 3.
Explanation:
Isomerism is a phenomenon where by two or more compounds have the same molecular formula but different structural patterns.
Geometric Isomerism is a type of Isomerism that occurs within a double bond i.e Geometric isomers have different arrangement within the double bond.
Considering the options given above,
The 1st option is exactly the same as the compound only, it is inverted.
The 2nd option is still the same as the compound, only it is laterally inverted.
The 3rd option satisfy geometric Isomerism as the arrangement differ from the compound in the double bond.
The 4th option is entirely a saturated compound in which geometric Isomerism is not possible.
A solution has a hydroxide ion concentration of 1 × 10–5 M. What is the hydrogen ion concentration of the solution?
(a) 1 × 10–1 M
(b) 1 × 10–5 M
(c) 1 × 10–9 M
(d) 1 × 10–14 M
Answer:
The correct answer is (c) 1 × 10⁻⁹ M
Explanation:
Hydroxide ion = OH⁻
Hydrogen ion = H⁺
The autoionization equilibrium of water at 25ºC has a water constant Kw which is expressed as follows:
Kw = [H⁺] x [OH⁻]= 1 x 10⁻¹⁴
If we know the concentration of hydroxide ion ([OH⁻]) we can calculate the hydrogen ion concentration ([H⁺]) as follows:
[H⁺] = Kw/[OH⁻]= (1 x 10⁻¹⁴)/1 x 10⁻⁵ M= 1 x 10⁻⁹ M
The hydrogen ion concentration of a solution with a hydroxide ion concentration of 1 × 10-5 M is 1 × 10-9 M.
Explanation:In chemistry, the product of the concentration of hydrogen ions [H+] and the concentration of hydroxide ions [OH-] in a solution always equals 1 × 10–14 M² at a temperature of 25°C (this is known as the ion product of water).
Since the concentration of hydroxide ions [OH-] in your solution is given as 1 × 10-5 M, you can calculate the concentration of hydrogen ions [H+] by dividing the ion product of water, 1 × 10-14 M², by the concentration of hydroxide ions [OH-]. Therefore, the hydrogen ion concentration [H+] in your solution would be 1 × 10-14 M² / 1 × 10-5 M = 1 × 10-9 M.
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Does the hydrogen molecule obey the octet rule?
Explanation:
Hydrogen does not obey the octet rule. Boron does not always
obey the octet rule and in fact forms Lewis acids such as BF3 which
only has 6 electrons.
You are given 1.000 grams of hydrated Magnesium Sulfate (MgSO4*XH20). You place it into a crucible and heat it up past 100 C so that the water evaporates from the hydrated salt. Once you are done, the mass of the dry anhydrous Magnesium Sulfate is 0.488 grams.
How many molecules of water are attached to each atom of Magnesium?
Answer:
There would be seven mols of water per every mol of MgSO4.
Explanation:
To find out how many molecules of water are attached to each atom of Magnesium in the hydrated Magnesium Sulfate, we can calculate the number of moles of water lost during heating. From there, we can use the mole ratio in the formula of the hydrated salt to determine the number of water molecules attached to each atom of Magnesium.
Explanation:To determine how many molecules of water are attached to each atom of Magnesium, we need to calculate the ratio of the mass of water to the mass of the anhydrous Magnesium Sulfate.
From the given information, the mass of the hydrated Magnesium Sulfate is 1.000 grams and the mass of the anhydrous Magnesium Sulfate is 0.488 grams. Therefore, the mass of water lost during heating is 1.000 grams - 0.488 grams = 0.512 grams.
To convert grams of water to molecules, we can use Avogadro's number. The molar mass of water is 18.015 g/mol. So, the number of moles of water lost is 0.512 grams / 18.015 g/mol = 0.0284 moles.
Next, we can calculate the number of molecules of water using the mole ratio between water and Magnesium Sulfate in the formula of the hydrated salt. The formula is MgSO4 * XH20, where X represents the number of water molecules. Assuming the formula is MgSO4 * 7H20, the mole ratio is 1:7. Therefore, the number of water molecules attached to each atom of Magnesium is 0.0284 moles * 7 = 0.199 moles. Now, we multiply this by Avogadro's number to get the number of molecules, which is 0.199 moles * 6.022 x 10^23 molecules/mol = 1.20 x 10^23 molecules of water.
For a redox reaction to occur, there must be a transfer of *
A. protons
B. neutrons
C. electrons
D. ions
Answer:
The answer is C.
Explanation:
Oxidation and Reduction are determine by the transfer of electrons .
if you hang a bird feeder, fill it with food but no birds come to it what would the hypothesis be?
Answer:
You could have many hypothesis...
Explanation:
A hypothesis is a guess as to what is going to happen in a situation. Without knowing the exact question you are developing a hypothesis for I can't give you an exact answer and the hypothesis takes place before the experiment. So you wouldn't know why the birds didn't come.
But if I had to give a bs answer I would say that the conditions weren't ideal for the birds (ie not right location or type of food)
The hypothesis could be that the type of food, the feeder's visibility, or its location is not attracting the birds. You could test this hypothesis by trying to change these variables.
Explanation:If you filled a bird feeder with food and hung it but noticed no birds coming to it, the hypothesis might be that either the birds are not attracted to the type of food in the feeder, they are not aware of the feeder's presence, or the location of the feeder is not suitable or safe for them.
To test this hypothesis, one could try changing the food type, moving the feeder to a different location, or drawing attention to the feeder somehow, then observing if there are changes in bird visitation.
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Match the following aqueous solutions with the appropriate letter from the column on the right. Assume complete dissociation of electrolytes. 1. 0.17 m NH4CH3COO ---- A. Lowest freezing point 2. 0.18 m MnSO4 ---- B. Second lowest freezing point 3. 0.20 m CoSO4 ---- C. Third lowest freezing point 4. 0.42 m Ethylene glycol (nonelectrolyte) ---- D. Highest freezing point
Answer: 0.17 m [tex]CH_3COONH_4[/tex] : Highest freezing point
0.20 m [tex]CoSO_4[/tex]: Second lowest freezing point
0.18 m [tex]MnSO_4[/tex]: Third lowest freezing point
0.42 m ethylene glycol: Lowest freezing point
Explanation:
Depression in freezing point is a colligative property which depend upon the amount of the solute.
[tex]\Delta T_f=i\times k_f\times m[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
i= vant hoff factor
[tex]k_f[/tex] = freezing point constant
m = molality
a) 0.17 m [tex]CH_3COONH_4[/tex]
For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =2 for [tex]CH_3COONH_4[/tex], thus total concentration will be 0.34 m
b) 0.18 m [tex]MnSO_4[/tex]
For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for [tex]MnSO_4[/tex], thus total concentration will be 0.36 m
c) 0.20 m [tex]CoSO_4[/tex]
For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for [tex]CoSO_4[/tex], thus total concentration will be 0.40 m
d) 0.42 m ethylene glycol
For non electrolytes undergoing no dissociation, vant hoff factor is equal to 1 . Thus i = 1 for ethylene glycol, thus concentration will be 0.42 m
The more is the concentration, the highest will be depression in freezing point and thus lowest will be freezing point.
PLEASE HELP ASAP!
Hydrogen reacts with nitrogen to form ammonia according to equation 3 H2(g) + N2(g) → 2 NH3(g)
A. How many grams of NH3 can be produced from 4.27 mol of N2 and excess H2?
B. How many grams of H2 are needed to produce 13.01g of NH3?
C. How many molecules (not moles) of NH3 are produced from 0.0235 g of H2?
Answer:
A. 145.2 g NH3
B. 0.76 g H2
C. 1.41 x 10^22 molecules NH3
Explanation:
A. 1 mol N2 -> 2 mol NH3
4.27 mol N2 -> x
x= (4.27 mol N2 * 2 mol NH3)/1 mol N2 x= 8.54 mol NH3
1 mol N2 -> 17 g
8.54 mol N2 -> x x= 145.2 g NH3
B.
2 g H2 -> 34 g NH3
x -> 13.01 g NH3
x= 0.76 g H2
C.
2 g H2 -> 34 g NH3
0.0235 g H2 -> x
x= 0.39 g NH3
0.39 g NH3 (1 mol NH3/17 g NH3)(6.023 x 10^23 molecules/1 mol NH3) =
1.41 x 10^22 molecules NH3
Give the formula for the alkene containing 15 carbons.
The general formula for alkanes is CnH₂n+2. If it's containing 15 carbon, it contains 32 hydrogen atoms. The formula would be C₁₅H₃₂.
What are alkenes?Alkenes are defined as either branched or unbranched hydrocarbons with a general formula of CnH2n with at least one carbon-carbon double bond (CC).
Alkenes are unsaturated hydrocarbons with carbon-carbon double bonds and have the chemical formula CnH2n. The molecular structure of this is the same as that of cycloalkanes.
The naming convention for alkenes is the same as that for alkanes, with the exception that the suffix is now -ene.
Substitute 15 for n.
The number of hydrogen atoms is 2×15+2=32.
The formula for an acyclic alkane with 15 carbon atoms is C₁₅H₃₂.
Therefore, CnH₂n+2 is the standard formula for alkanes. 32 hydrogen atoms are present if it has 15 carbon atoms. The equation would be C₁₅H₃₂.
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A cell is set up with copper and lead electrodes in contact with CuSO4(aq) and Pb(NO3)2(aq), respectively, at 25°C. The standard reduction potentials are Pb2+ + 2e–? Pb E° = –0.13 V Cu2+ + 2e–? Cu E° = +0.34 V
If sulfuric acid is added to the Pb(NO3)2 solution, forming a precipitate of PbSO4, the cell potential
a. It is impossible to tell what will happen.
b. decreases.
c. is unchanged.
d. increases.
Answer:
increases
Explanation:
Now look at the matter closely. Given the values of electrode potential stated in the question, one can see that lead will function as the anode and copper as the cathode since lead has a more negative electrode potential.
This implies that at the anode, the half reaction going on is this;
Pb(s) -------> Pb^2+(aq) + 2e.
There will be a build up of Pb^2+ in the anode compartment. The addition of H2SO4 and formation of PbSO4 favours the removal of the Pb^2+ ions;
Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)
As this continues, Pb^2+ concentration begins to decrease, in order to maintain equilibrium, more Pb^2+ is formed thereby increasing the current and voltage flowing in the cell as more electrons are transferred from anode to cathode(more current flows).
In the case when the sulfuric acid is added to the Pb(NO3)2 solution increases so the cell potential is increased.
Impact on cell potential:Here the anode and copper treated as the cathode because it contains more negative electrode potential. Due to this, it should be half reaction
Pb(s) -------> Pb^2+(aq) + 2e.
Also, there should be the addition of H2SO4 and creation of the PbSO4 that eliminates the Pb^2+ ions.
So,
Pb^2+(aq) + SO4^2-(aq) -------> PbSO4(s)
Here the concentration of Pb^2+ should reduced for maintaining the equilibrium.
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Simply by living, an 87.5 kg human being will consume approximately 20.0 mol of O 2 per day. To provide energy for the human, the O 2 is reduced to H 2 O duing food oxidation by the reaction O 2 + 4 H + + 4 e − − ⇀ ↽ − 2 H 2 O Determine the current generated by the human per day. In this case, the current is defined as the flow of electrons ( e − ) to O 2 from the food the human consumes.
Answer:
The current generated by the human per day = 89.34 Amperes
Explanation:
Given that ;
the human consumes 20.0 mole of O₂ per day
The Food oxidation is given by the reaction :
[tex]O_2 + 4 H^+ + 4 e^- ----> 2H_2O[/tex]
So; the human will produce 20.0 × 4 moles of e⁻
Thus; moles of e⁻ produced is = 80
Charge on 1 e⁻ = 1.602 × 10⁻¹⁹ C
e⁻ in 1 mole = 6.023 × 10²³
The total charge per day = [tex]\frac{1.602*10^{-19}*6.023*10^{23}*80}{24*60*60}[/tex]
= 89.34 Amperes
A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by O3(g)+NO(g)⟶O2(g)+NO2(g) O3(g)+NO(g)⟶O2(g)+NO2(g) The rate law for this reaction is rate of reaction=k[O3][NO] rate of reaction=k[O3][NO] Given that k=4.09×106 M−1⋅s−1k=4.09×106 M−1⋅s−1 at a certain temperature, calculate the initial reaction rate when [O3][O3] and [NO][NO] remain essentially constant at the values [O3]0=5.84×10−6 M[O3]0=5.84×10−6 M and [NO]0=8.65×10−5 M,[NO]0=8.65×10−5 M, owing to continuous production from separate sources.
Answer:
Initial rate of reaction is [tex]2.07\times 10^{-3}M.s^{-1}[/tex].
Explanation:
It is a second order reaction.
Initial rate of reaction = [tex]k[O_{3}]_{0}[NO]_{0}[/tex] , where k is rate constant, [tex][O_{3}]_{0}[/tex] is the initial concentration of [tex]O_{3}[/tex] and [tex][NO]_{0}[/tex] is the initial concentration of NO.
Here, k = [tex]4.09\times 10^{6}M^{-1}.s^{-1}[/tex], [tex][O_{3}]_{0}=5.84\times 10^{-6}M[/tex] and [tex][NO]_{0}=8.65\times 10^{-5}M[/tex]
So, initial rate of reaction = [tex](4.09\times 10^{6}M^{-1}.s^{-1})\times (5.84\times 10^{-6}M)\times (8.65\times 10^{-5}M)[/tex]
= [tex]2.07\times 10^{-3}M.s^{-1}[/tex]
So, initial rate of reaction is [tex]2.07\times 10^{-3}M.s^{-1}[/tex]
In the halogenation reaction shown, ethane and chlorine gas yield chloroethane and hydrogen chloride. How do the properties of chloroethane compare to those of ethane?
Chloroethane is denser and has a lower boiling point
Chloroethane is denser and has a higher boiling point
Chloroethane is less dense and has a lower boiling point.
Chloroethane is less dense and has a higher boiling point.
Answer:
Chloroethane is denser and has a higher boiling point
Explanation:
The density of a gas depends directly on the molar mass of the gas. This means that as the molar mass increases, density increases and vice versa.
Having said that, we can easily see that the molar mass of chloroethane (64.51 g/mol) is greater than the molar mass of ethane (30.07 g/mol). Hence we expect that chloroethane is denser than ethane as established above.
In the absence of other strong intermolecular forces, the higher the molecular mass of a substance the greater its boiling point. Thus the boiling point of chloroethane is higher than that of Ethane since they both have weak Van der Waals forces holding their molecules together in the gaseous state.