Answer:
b
Explanation:
Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.
The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.
The correct option is b.
Beginning with Na, record the number of energy levels, number of protons, and atomic radius for each element in period 3.
Answer:
Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.
Explanation:
There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm. The atomic radius of aluminium atom is 143 ppm. The atomic radius of silicon atom is 111 ppm. The atomic radius of phosphorus atom is 98 ppm. The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.
In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270.
Use the information she obtained to determine the Ka for this acid.
Ka(experiment) = _____
Answer:
Ka = 6.87x10⁻⁵
Explanation:
The equilibrium of benzoic acid in water is:
C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The equilibrium constant, Ka, is:
Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]
The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.
X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.
Concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - X[C₆H₅COO⁻] = X [H₃O⁺] = XpH is defined as -log [H₃O⁺]. As pH = 2.270
2.270 = -log [H₃O⁺]
10^-2.270 = [H₃O⁺]
5.37x10⁻³M = [H₃O⁺] = X.
Replacing, concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M
[C₆H₅COO⁻] = 5.37x10⁻³M
[H₃O⁺] = 5.37x10⁻³M
Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]
Ka = 6.87x10⁻⁵Nitric oxide (NO) can be formed from nitrogen, hydrogen and oxygen in two steps. In the first step, nitrogen and hydrogen react to form ammonia: (g) (g) (g) In the second step, ammonia and oxygen react to form nitric oxide and water: (g) (g) (g) (g) Calculate the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest .
Answer: [tex]\Delta H = -272.25kJ[/tex] for 1 mole of NO.
Explanation: Hess' Law of Constant Summation or Hess' Law states that the total enthalpy change of a reaction with multiple stages is the sum of the enthalpies of all the changes.
For this question:
1) [tex]N_{2}_{(g)} + 3H_{2}_{(g)}[/tex] => [tex]2NH_{3}_{(g)}[/tex] [tex]\Delta H=-92kJ[/tex]
2) [tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
Amonia ([tex]NH_{3}_{(g)}[/tex]) appeares as product in the first equation and as reagent in the 2 reaction, so when adding both, there is no need to inverse reactions. However, in the 2nd, there are 4 moles of that molecule, so to cancel it, you have to multiply by 2 the first chemical equation and enthalpy:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
Now, adding them:
[tex]2N_{2}_{(g)} + 6H_{2}_{(g)}[/tex] => [tex]4NH_{3}_{(g)}[/tex] [tex]\Delta H=-184kJ[/tex]
[tex]4NH_{3}_{(g)}+5O_{2}_{(g)}[/tex] => [tex]4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H=-905kJ[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -185-905[/tex]
[tex]2N_{2}_{(g)}+6H_{2}_{(g)}+5O_{2}_{(g)}=>4NO_{(g)}+6H_{2}O_{(g)}[/tex] [tex]\Delta H = -1089kJ[/tex]
Note net enthalpy is for the formation of 4 moles of nitric oxide.
For 1 mole:
[tex]\Delta H = \frac{-1089}{4}[/tex]
[tex]\Delta H=-272.25kJ[/tex]
To form 1 mol of nitric oxide from nitrogen, oxygen and hydrogen, net change in enthalpy is [tex]\Delta H=-272.25kJ[/tex].
If the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?
Answer:
14.297 g
Explanation:
From the question;
1 mo of the compound requires 1320.0 kJ
From the molar mass;
1 ml of the compound weighs 30.55g
How many grams requires 617.30kJ?
1 ml = 1320
x mol = 617.30
x = 617.30 / 1320
x = 0.468 mol
But 1 mol = 30.55
0.468 mol = x
x = 14.297 g
2NH3 → N2 + 3H2 If 2.22 moles of ammonia (NH3) decomposes according to the reaction shown, how many moles of hydrogen (H2) are formed? A) 2.22 moles of H2 B) 1.11 moles of H2 C) 3.33 moles of H2 D) 6.66 moles of H2
Answer:
C
Explanation:
According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.
Answer:
I think it's C
Explanation:
Please, tell me if I'm incorrect.
105/22 • (1.251 - 0.620)=
Answer:
105/22*(1.251-0.620)
105/22*0.631
4.772*0.631
3.011132
Hope it helps
Answer:
3.0
Explanation:
First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).
The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.
How are Math, Physics, Chemistry, and Biology all related?
Answer:
- you have to do maths in all 3
- atoms make up everything even parts of a cel and theyre studied in chem and physics
- chemistry is used in biology by finding out what different substances are eg cytoplasm in a cell
A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of___________ bases along its backbone. a. 146 b. 292 c. 438 d. 20 e. 3
Answer:
Since the relationship between amino acid and codon bases would be the values of 3 nitrogenous bases per 1 amino acid.
knowing this relationship what you would do is simply multiply 146 x 3 to find the number of codon bases which would be C. 438.
A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.
What is messenger RNA?mRNA or messenger RNA is a single stranded RNA molecule. It is complementary to the DNA and carries genetic information present in the DNA. It is translated to form proteins. The genetic codes (triplet) present on mRNA get translated to amino acids, giving rise to the functional product of a gene.
So mRNA really is a form of nucleic acid, which helps the human genome which is coded in DNA to be read by the cellular machinery. mRNA is actually the translated form of DNA that the machinery can recognize and use to assemble amino acids into proteins.
Each strand has 3 bases, so 146 × 3 = 438 bases
Therefore, A strand of messenger RNA coding the synthesis of a protein with 146 amino acids must have a string of 438 bases along its backbone and the correct option is option C.
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If 1 mol of a pure triglyceride is hydrolyzed to give 2 mol of RCOOH, 1 mol of R'COOH, and 1 mol of glycerol, which of the following compounds might be the triglyceride?
CHOC(O)R
A. CHOC(O)R
CHOC(O)R
CH,OC(O)R
B. CHOC(O)R
CH2OC(O)R
CHOC(O)R
C. CHOC(O)R
CHOC(O)R
CHOC(O)R
D. CHOC(O)R
CHOC(O)R
Answer:
The correct option is C.
Note the full question and structure of the moleculesis found in the attachment below.
Explanation:
Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.
The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.
During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.
From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.
The correct option therefore, is C
Calculate the amount of heat needed to melt of ice () and bring it to a temperature of . Round your answer to significant digits. Also, be sure your answer contains a unit symbol.
Answer:
your answer is given below
Explanation:
Take it stepwise, and you are going to have to look up the various heat values.
You have 36.5grams of ice, presumably at 0C. You will need to add heat to take the ice at 0 C to water at 0 C. (Latent heat of melting) Then you add in a different heat value to take the water at 0 C to water at 82.3 C. (Specific heat of water)
Add the two heat amounts together.
According to specific heat capacity, to calculate the amount of heat needed to melt ice given mass of ices is multiplied by specific heat of ice and the temperature change which it undergoes.
What is specific heat capacity?Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.
It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.
It is given by the formula ,
Q=mcΔT
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11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. How many grams per kilogram of body weight is a lethal dose for 50% of domestic chickens?
Answer:
[tex]0.033g[/tex]
Explanation:
Hello,
In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:
[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]
Thus, we obtain:
[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]
That in grams is:
[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]
Regards.
The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?
Answer:
[tex]Ksp=1.07x10^{-8}[/tex]
Explanation:
Hello,
In this case, the dissociation reaction is:
[tex]PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)[/tex]
For which the equilibrium expression is:
[tex]Ksp=[Pb^{2+}][I^-]^2[/tex]
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
[tex]Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M[/tex]
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
[tex][Pb^{2+}]=1.39x10^{-3}M[/tex]
[tex][I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M[/tex]
Thereby, the solubility product results:
[tex]Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}[/tex]
Regards.
Solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
The dissociation reaction for lead (II) iodide
[tex]\bold {Pb I^2 (s) \leftrightharpoons Pb^2^+ + 2I^- }[/tex]
Solubility product constant at equilibrium.
[tex]\bold {Ksp = [Pb^2^++[I^-]^2}[/tex]
The molar solubility of the substance can be calculated by using the molar mass,
[tex]\bold {s = \dfrac {0.064}{100 mL} \times 461.2 g/mol = 1.39x10^-^3}[/tex]
Molar ratio between between PbI to lead and iodide ions is 1:1 and 1:2 respectively.
Thus Ksp will be,
[tex]\bold {Ksp =(1.39x10^-^3)(2.78x10^-^3 )^2}\\\\\bold {Ksp = 1.07x 10^-^8}[/tex]
Therefore, solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
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At standard temperature and pressure conditions, the volume of an ideal gas contained in a jar is 55.3 L. How many molecules are in the jar. This question is to be answered in scientific notation.(eg. 1.5 e5)
Answer:
1.49e24
Explanation:
Standars temperature and pressure are 273.15K and 1atm, respectively.
Using ideal gas law, we can find moles of an ideal gas if we know its pressure, temperature and volume as follows:
PV = nRT
PV / RT = n
Where P is pressure (1atm), V is volume (55.3L), R is gas constant (0.082atmL/molK), T is temperature (273.15K) and n moles of the ideal gas.
Replacing:
PV / RT = n
1atm*55.3L / 0.082atmL/molK*273.15K = n
2.47 moles = n
Now, the question is about the number of molecules in the jar. By definition, 1 mole = 6.022x10²³ molecules.
As we have 2.47 moles:
2.47 mol × (6.022x10²³ molecules / 1 mole) =
1.49x10²⁴ molecules that are in the jar
In scientific notation:
1.49e24During a titration, a known concentration of _____ is added to a _____ of an unknown concentration g
Explanation:
The whole process of titration involves finding the concentration of a solution (usually an acid or base) by adding (titrating) it to a solution(acid or base) with a known concentration.
The solution of unknown concentration (the analyte) is usually placed in an flask, while the solution of known concentration (titrant) is placed in a burette and slowly added to the flask.
A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia
Answer:
0.327 M
Explanation:
Step 1: Write the balanced equation
2 NH₃(g) ⇄ N₂(g) + 3H₂(g)
Step 2: Make an ICE chart
2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)
I x 0 0
C -2y +y +3y
E x-2y y 3y
Step 3: Find the value of y
The concentration of N₂ at equilibrium is 0.0161 M. Then,
y = 0.0161
Step 4: Find the value of x
The concentration of NH₃ at equilibrium is 0.295 M. Then,
x-2y = 0.295
x-2(0.0161) = 0.295
x = 0.327
Which of the following goes through the largest volumetric change? Question 4 options: A) Water when it's heated from 1oC to 99oC B) Water when it freezes into ice C) Ice when it melts into water D) Water when it boils into steam
Answer:
Water when it freezes into ice
Explanation:
Most liquids expand when heated and contract when cooled, water behaves in an anomalous fashion. Water rather expands when cooled and contracts when heated.
Water usually contracts on cooling from any temperature until 4°C, after 4°C, the water begins to expand rapidly. Hence water has its least volume at 4°C and increases rapidly afterwards.
Thus the largest volume change for water occurs during freezing since it expands when cooled.
what is the lewis structure for OP(N3)3
Explanation:
this is the ans
hope this helps
The complex ion Fe(CN)63- is paramagnetic with one unpaired electron. The complex ion Fe(SCN)63- has five unpaired electrons. Where does SCN- lie in the spectrochemical series with respect to CN-?
Answer:
SCN- is a weak field ligand while CN- is a strong field ligand
Explanation:
The spectrochemical series is an arrangement of ligands according to their magnitude of crystal field splitting. Ligands that cause only a small degree of crystal field splitting are called weak field ligands while ligands that cause large crystal field splitting are called strong field ligands.
Strong field ligands often lead to the formation of low spin complexes with the least number of unpaired electrons while high spin complexes are formed by weak field ligands.
CN- is a strong field ligand as it lies towards the right hand side of the spectrochemical series.
SCN- is a weak field ligand hence it forms a high spin complex having the maximum number of unpaired electrons for Fe^3+, hence the answer.
SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.
CN⁻ is a strong field ligand with a large splitting constant, and it is high up in the spectrochemical series.
Conversely, SCN⁻ is a weak field ligand with a low splitting constant, and it is lower in the spectrochemical series.
Hence, SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.
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What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?
Answer:
8.68
Explanation:
pOH = 8.68
all you need is contained in the sheet
Answer:
Approximately [tex]8.68[/tex].
Explanation:
The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:
[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].
On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:
[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].
At [tex]25 \; ^\circ \rm C[/tex], [tex]K_{\text{w}} \approx 1.0 \times 10^{-14}[/tex]. For this particular [tex]25 \; ^\circ \rm C[/tex] solution, [tex]\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}[/tex].Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:
[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].
Therefore, the [tex]\rm pOH[/tex] of this solution would be:
[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].
Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].
For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.
what is ammonium nitrate
Answer:
Ammonium nitrate is a chemical compound with the chemical formula NH₄NO₃. It is a white crystalline solid consisting of ions of ammonium and nitrate.
Testbank Question 47 Consider the molecular orbital model of benzene. In the ground state how many molecular orbitals are filled with electrons?
Answer:
There are fifteen molecular orbitals in benzene filled with electrons.
Explanation:
Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;
There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.
There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons
The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.
All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.
A student determines the value of the equilibrium constant to be 1.5297 x 107 for the following reaction: HBr(g) + 1/2 Cl2(g) --> HCl(g) +1/2 Br2(g) Based on this value of Keq, calculate the Gibbs free energy change for the reaction of 2.37 moles of HBr(g) at standard conditions at 298 K.
Answer:
[tex]\Delta G=-97.14kJ[/tex]
Explanation:
Hello,
In this case, the relationship between the equilibrium constant and the Gibbs free energy of reaction is:
[tex]\Delta G=-RTln(K)[/tex]
Hence, we compute it as required:
[tex]\Delta G=-8.314\frac{J}{mol\times K}*298K*ln(1.5297x10^7)\\\\\Delta G=-40.99kJ/mol[/tex]
And for 2.37 moles of hydrogen bromide, we obtain:
[tex]\Delta G=-40.99kJ/mol*2.37mol\\\\\Delta G=-97.14kJ[/tex]
Best regards.
Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
Calculate the equilibrium constant K c for the following overall reaction: AgCl(s) + 2CN –(aq) Ag(CN) 2 –(aq) + Cl –(aq) For AgCl, K sp = 1.6 × 10 –10; for Ag(CN) 2 –, K f = 1.0 × 10 21.
Answer:
1.6x10¹¹ = Kc
Explanation:
For the reaction:
AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)
Kc is defined as:
Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²
Ksp of AgCl is:
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
Where Ksp is:
Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰
In the same way, Kf of Ag(CN)₂⁻ is:
Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹
The multiplication of Kf with Ksp gives:
[Ag⁺] [Cl⁻] * [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf
[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf
Obtaining the same expression of the first reaction
That means Ksp*Kf = Kc
1.6x10⁻¹⁰*1.0x10²¹ = Kc
1.6x10¹¹ = KcCalculate the pH of a solution formed by mixing 250.0 mL of 0.15 M NH4Cl with 200.0 mL of 0.12 M NH3. The Kb for NH3 is 1.8 × 10-5.
Answer:
The pH of the solution is 9.06.
Explanation:
The reaction of the dissociation of NH₃ in water is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq) (1)
[NH₃] - x [NH₄⁺] + x x
The concentration of NH₃ and NH₄⁺ is:
[tex] [NH_{3}] = \frac{n_{NH_{3}}}{V_{T}} = \frac{C_{i}_{(NH_{3})}*Vi_{NH_{3}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.12 M*0.2 L}{0.2 L + 0.25 L} = 0.053 M [/tex]
[tex] [NH_{4}^{+}] = \frac{C_{i}_{(NH_{4}^{+})*V_{NH_{4}^{+}}}}{V_{NH_{3}} + V_{NH_{4}^{+}}} = \frac{0.15 M*0.25 L}{0.2 L + 0.25 L} = 0.083 M [/tex]
From equation (1) we have:
[tex]Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}[/tex]
[tex] 1.8 \cdot 10^{-5} = \frac{(0.083 + x)*x}{0.053 - x} [/tex]
[tex] 1.8 \cdot 10^{-5}(0.053 - x) - (0.083 + x)*x = 0 [/tex]
By solving the above equation for x we have:
x = 1.15x10⁻⁵ = [OH⁻]
The pH of the solution is:
[tex] pOH = -log([OH^{-}]) = -log(1.15 \cdot 10^{-5}) = 4.94 [/tex]
[tex] pH = 14 - pOH = 14 - 4.94 = 9.06 [/tex]
Therefore, the pH of the solution is 9.06.
I hope it helps you!
We discussed the different types of intermolecular forces in this lesson, which can affect the boiling point of a substance.
1. Which of these has the highest boiling point?
A) Ar
B) Kr
C) Xe
D) Ne
2. Which substance has the highest boiling point?
A) CH4
B) He
C) HF
D) Cl2
Answer:
1, C, Xe 2, B,He
Explanation:
1, cause as u go down a group the boiling point increases.
2, boiling point of single element is greater than a compound
According to periodic trends in periodic table boiling point increases down the group and hence Xe has highest boiling point and more amount of heat is required to boil an element hence He has highest boiling point.
What is periodic table?
Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.
It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.
Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.
Learn more about periodic table,here:
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The gas with an initial volume of 24.0 L at a pressure of 565 torr is compressed until the volume is 16.0 L. What is the final pressure of the gas, assuming the temperature and amount of gas does not change
Answer:
848 torr
Explanation:
The only variables are the pressure and the volume, so we can use Boyle's Law.
p₁V₁ = p₂V₂
Data:
p₁ = 565 torr; V₁ = 24.0 L
p₂ = ?; V₂ = 16.0 L
Calculations:
[tex]\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{565 torr} \times \text{24.0 L} & = & p_{2} \times \text{16.0 L}\\\text{13 560 torr} & = & 16.0p_{2}\\p_{2} & = & \dfrac{\text{13 560 torr}}{16.0}\\\\& = &\textbf{848 torr}\\\end{array}\\\text{The final pressure of the gas is $\large \boxed{\textbf{848 torr}}$}[/tex]
Which of the following happens to a molecule of an object when the object is heated? (1 point)
Answer:
They get more energy, so they vibrate!
Explanation:
I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.
Answer:
THE FINAL TEMPERATURE OF WATER IS -4.117 °C
Explanation:
Mass of the aluminium = 50 g
c = 0.88 J/g C
Initial temperature of aluminium = 225 °C
Volume of water = 100 ml
Density of water = 1 g/ml
Mass of water = density * volume of water
Mass of water = 1 * 100 = 100 g of water
Initial temperature of water = 20 C
It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,
So therefore,
heat lost by aluminium = heat gained by water
H = mass * specific heat capacity * temeprature change
So:
m c ( T2- T1) = m c (T2-T1)
50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)
44 ( T2 - 225 ) = 418 ( T2 - 20)
44 T2 - 9900 = 418 T2 - 8360
-9900 + 8360 = 418 T2 - 44 T2
-1540 = 374 T2
T2 = - 4.117
So therefore the final temperature of water is -4.117 °C
i) Briefly discuss the strengths and weaknesses of the four spectroscopy techniques listed below. Include in your answer the specific structural information you get from each method.
IR
UV-VIS
NMR
Mass Spec
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