Answer:
AT STOICHIOMETRIC POINT, THE VOLUME OF ACID ADDED IS 0.01 L
AT HALF-WAY POINT, THE VOLUME OF ACID IS 0.0050 L
Explanation:
In solving titration problems, you must remember this formula;
MaVa = MbVb
Since M a= 0.005 M
Mb = 0.010 M
Vb = 5 mL = 5 /1000 = 0.005 L
Va = unknown.
Solving for Va, we have:
Va = MbVb / Ma
Va = 0.010 * 0.005 / 0.005
Va = 0.01 L
So therefore, the volume of acid added at:
1. the stoichiometric point is 0.01 L
2. half-way point of titration is 0.01 /2 = 0.0050 L
For the pH:
Since HCl is a strong acid, it dissociate into {H30}+ ion.
First calculate the number of moles of hydronium ion
number of mole = concentration of hydronium ion {H30}+ * Volume
n = 0.005 * 0.01 = 0.00005 moles
A. At initial point of the titration, the volume of base added is 0 L
{H30]+ = n(H+)/ V = 0.00005 / 0.01 = 0.005 M
pH = - log {0.005}
pH = 2.3
B. At the final point, since the volumes and concentrations of acid and base are the same, the pH is equal to 7.
n(H+) = n(OH^-)
pH = 7
To determine the absolute age of rocks and fossils, geologists use _____.
Answer:
The rates of decay of radioactive elements
Explanation:
The age of a rock in years is called its absolute age. Geologists find absolute ages by measuring the amount of certain radioactive elements in the rock. When rocks are formed, small amounts of radioactive elements usually get included.
Each of the insoluble salts below are put into 0.10 M hydrobromic acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution ?
a. Calcium sulfite
b. Calcium fluoride
c. Silver bromide
Answer:
A. Solubility of calcium sulfite increases
B. Solubility of calcium fluoride increases
C. Solubility of Silver bromide decreases
Explanation:
The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.
Case 1. Calcium sulfite
The dissociation reaction of CaSO3 is given below:
CaSO3 ----> Ca²+ + SO3²-
SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;
H+ + SO3²- ---> HSO3-
This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.
Case 2. Calcium fluoride
The dissociation reaction of calcium fluoride (CaF2) is shown below.
CaF2 ----> Ca²+ + 2F-
Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:
H+ + F- ---->. HF
Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.
Case 3: Silver bromide
The dissociation reaction of AgBr is as follows:
AgBr ----> Ag+ + Br-
The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.
The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.
Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.
Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.
However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.
Learn more: https://brainly.com/question/6505878
which element causes burning when me mix it with oxygen
Answer:
Hydrogen peroxide is ans
Consider a cobalt-silver voltaic cell that is constructed such that one half-cell consists of the cobalt, Co, electrode immersed in a Co(NO3)3 solution, and the other half-cell consists of the silver, Ag, electrode immersed in a AgNO3 solution. The two electrodes are connected by a copper wire. The Co electrode acts as the anode, and the Ag electrode acts as the cathode. To maintain electric neutrality, you add a KNO3 salt bridge separating the two half-cells. Use this information to solve Parts B, C, and D.
A. The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell is the site of the reduction reaction.
Type the half-cell reaction that takes place at the anode for the cobalt-silver voltaic cell. Indicate the physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.
B. The half-cell is a chamber in the voltaic cell where one half-cell is the site of an oxidation reaction and the other half-cell is the site of a reduction reaction.
Type the half-cell reaction that takes place at the cathode for the cobalt-silver voltaic cell. Indicate physical states using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. Use (aq) for an aqueous solution. Do not forget to add electrons in your reaction.
Answer:
Anode half reaction;
Co(s) ----> Co^2+(aq) + 2e
Cathode half reaction;
2Ag^+(aq) + 2e-------> 2Ag(s)
Explanation:
A voltaic cell is an electrochemical cell that spontaneously produces electrical energy from chemical reactions. A voltaic cell comprises of an anode (where oxidation occurs) and a cathode (where reduction occurs). The both electrodes are connected with a wire . A salt bridge ensures charge neutrality in the anode and cathode compartments. Electrons flow from anode to cathode.
For the cell referred to in the question;
Anode half reaction;
Co(s) ----> Co^2+(aq) + 2e
Cathode half reaction;
2Ag^+(aq) + 2e-------> 2Ag(s)
How many molecules of water are in 5 moles
Answer:
3.011x10^24 molecules
Explanation:
1 mole=6.022x10^23 molecules
5 moles*6.022x10^23 molecules/mole=3.011x10^24 molecules
Choose the INCORRECT statement. A. Temperatures of two bodies are equal when the average kinetic energies of the two bodies become the same. B. The heat capacity is the quantity of heat required to change the temperature of the system by one degree. C. The specific heat is the heat capacity for one mole of substance. D. Most metals have low specific heats, as metals can be heated quickly. E. The law of conservation of energy can be written: qsystem qsurroundings
Answer:
Option C
The specific heat is the heat capacity for one mole of a substance.
Explanation:
The incorrect statement is The specific heat is the heat capacity for one mole of a substance.
This is because the specific heat capacity is the amount of heat required to raise the temperature of 1 gramme of a substance by 1 degree Celcius.
Note that the unit in question here for the specific heat capacity of the substance is in grammes.
The definition given in the options is actually for the molar heat capacity of the substance, not the specific heat capacity.
1- A volumen constante un gas ejerce una presión de 880 mmHg a 20º Celsius dentro de una olla a presión ¿Qué temperatura habrá si el marcador de presión muestra un valor de 1050 mmHg?
Answer:
In this problem the correct thing would be to use the ideal gas equation.
Explanation:
Well in this exercise we will use the following equation:
(P x V) / T = (p x v) / t
On the right side of the equation we will find the initial values, that is, the values with which the reaction begins and in general they are always the first to write in the problems.
Instead on the left side of the equation, the letters that are in lowercase are the final values, that is to say at the end of the reaction that the values of pressure, temperature and volume are reached.
P is pressing, just like p, T and t are temperature, and V and v are volume.
We use this equation so we consider the behavior of said gas to be an IDEAL gas, a constant volume.
That is why the given pressures require an atmosphere to pass, which is another unit used to press the pressure ... Much needed in this equation! An atmosphere is equivalent to 760 millimeters of mercury ...
Then the final and initial pressures would be:
initial pressure: 1.15 atm
final pressure: 1.38 atm
In this way you already have the values to be able to solve in the equation your unknown that would be the final temperature:
Considering that the volume is constant, we cancel it from the equation, 1.15 atm would be in the value of P and 1.38 in the value of p ... In this way it considers that 20 degrees Celsius is the initial temperature or ses T, we would only have to clear the t.
Calculate Keq for these reactions and predict if the equilibrium will lie to the right or to the left as written. (You may enter your answer in scientific notation, e.g. 1.0*10^-9. Enter your answer to two significant figures.) Reaction 1: + + pKa = 9 pKa = 38 Keq = Equilibrium position = _______ Reaction 2: + + pKa = 35 pKa = 25 Keq = Equilibrium position = _______
Complete Question
The complete question is shown on the first uploaded image
Answer:
For reaction 1
[tex]K_{eq} = 10^{29}[/tex]
The equilibrium position is to the right
For reaction 2
[tex]K_{eq} = 10^{-6.66}[/tex]The equilibrium position is to the left
Explanation:
Generally [tex]pKa[/tex] is mathematically evaluated as
[tex]pKa = pKa _ \ {left }} - pKa _ \ {right }}[/tex]
And equilibrium position [tex]K_a[/tex] is mathematically evaluated as [tex]K_{eq} = 10^\ {-pK_a}[/tex]
From the question we are told that
For reaction 1
[tex]pKa_\ {left}} \ = 9[/tex]
[tex]pKa_\ {right }} \ = 38[/tex]
So
[tex]pKa = 9-38[/tex]
[tex]pKa =-29[/tex]
So [tex]K_{eq} = 10^{-(-29)}[/tex]
[tex]K_a = 10^{29}[/tex]
This implies that the equilibrium position is to the right
For reaction 2
[tex]pKa_\ {left}} \ = 15.9[/tex]
[tex]pKa_\ {right }} \ = 9.24[/tex]
So
[tex]pKa = 15.9-9.24[/tex]
[tex]pKa = 6.66[/tex]
So [tex]K_{eq} = 10^{-(6.66)}[/tex]
[tex]K_{eq} = 10^{-6.66}[/tex]
This implies that the equilibrium position is to the left
Describe, in detail, to a freshman undergraduate how to make 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media. [Include things like how many grams of each component that you use, how much antibiotic (in ml) to add (stock solutions – 100 mg/ml ampicillin, 25 mg/ml kanamycin), and in what type of container you perform the sterilization step.] Show your calculations.
Answer:
Explanation:
The objective here is to prepare 1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.
Given that :
the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)
it is required that we convert stock concentration in μg/ml since 100 mg/ml = 100000 μg/ml
However, using formula C₁V₁=C₂V₂ (Ampicilin),
where:
C₁ = 100000 μg/ml,
V₁=?,
C₂= 50 μg/ml,
V₂=1000 ml
100000 μg/ml × V₁ = 50 μg/ml × 1000 ml
V₁ = 50 μg/ml × 1000 ml/100000 μg/ml
V₁ = 0.5 ml
Given that:
the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)
it is required that we convert stock concentration in μg/ml , 25 mg/ml = 25000 μg/ml
Now by using formula C₁V₁=C₂V₂ (Kanamycin),
C₁ = 25000 μg/ml,
V₁=?,
C₂= 100 μg/ml,
V₂=1000 ml
25000 μg/ml × V₁ = 100 μg/ml × 1000 ml
V₁ = 100 μg/ml × 1000 ml/25000 μg/ml
V₁ = 4 ml
Thus; in 1 lite of Lb+ Kan+Amp preparation;
0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.
Finally;
Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.
Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid water H2O. Suppose 62. g of sulfuric acid is mixed with 33.8 g of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
Approximately [tex]21\; \rm g[/tex].
Explanation:
[tex]\rm H_2SO_4[/tex] (a diprotic acid) reacts with [tex]\rm NaOH[/tex] (a monoprotic base) at a one-to-two ratio:
[tex]\rm 2\; NaOH\, (s) + H_2SO_4\, (aq) \to Na_2SO_4\; (aq) + 2\; H_2O\, (l)[/tex].
In other words, if [tex]n(\mathrm{NaOH})[/tex] and [tex]n(\mathrm{H_2SO_4})[/tex] represent the number of moles of the two compounds reacted, then:
[tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex].
Look up the relative atomic mass data on a modern periodic table:
[tex]\rm H[/tex]: [tex]1.008[/tex].[tex]\rm S[/tex]: [tex]32.06[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm Na[/tex]: [tex]22.990[/tex].Calculate the (molar) formula mass of [tex]\rm H_2SO_4[/tex] and [tex]\rm NaOH[/tex]:
[tex]M(\mathrm{H_2SO_4}) = 2 \times 1.008 + 32.06 + 4 \times 15.999 = 98.072\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{NaOH}) = 22.990 + 15.999 + 1.008 = 39.997\; \rm g \cdot mol^{-1}[/tex].
Calculate the number of moles of formula units in that [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:
[tex]\begin{aligned}n(\mathrm{NaOH}) &= \frac{m(\mathrm{NaOH})}{M(\mathrm{NaOH})} \\ &= \frac{33.8\; \rm g}{39.997\; \rm g \cdot mol^{-1}} \approx 0.845\; \rm mol\end{aligned}[/tex].
Apply the ratio [tex]\displaystyle \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} = \frac{1}{2}[/tex] to find the (maximum) number of moles of [tex]\rm H_2SO_4[/tex] that would react with the [tex]33.8\; \rm g[/tex] of [tex]\rm NaOH[/tex]:
[tex]\begin{aligned}n(\mathrm{H_2SO_4}) &= \frac{n(\mathrm{H_2SO_4})}{n(\mathrm{NaOH})} \cdot n(\mathrm{NaOH})\\ &= \frac{1}{2} \times 0.845 \approx 0.4225\; \rm mol\end{aligned}[/tex].
Calculate the mass of that [tex]0.4225\; \rm mol[/tex] of [tex]\rm H_2SO_4[/tex]:
[tex]\begin{aligned}m(\mathrm{H_2SO_4}) &= n(\mathrm{H_2SO_4}) \cdot M(\mathrm{H_2SO_4})\\ &= 0.4225 \; \rm mol \times 98.072\; \rm g \cdot mol^{-1} \approx 41.435\; \rm g \end{aligned}[/tex].
When the maximum amount of [tex]\rm H_2SO_4[/tex] is reacted, the minimum would be in excess. Hence, the minimum mass of
[tex]62\; \rm g - 41.435\; \rm g \approx 21\; \rm g[/tex] (rounded to two significant figures.)
Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated by extraction with ether. The solubility of malononitrile in ether at room temperature is 20.0 g/100 mL, and in water is 13.3 g/100 mL. What weight of malononitrile would be recovered by extraction with (a) three 100-mL portions of ether and (b) one 300-mL portion of ether
The given question is not complete, the complete question is:
Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated by extraction with ether. The solubility of malononitrile in ether at room temperature is 20.0 g/100 mL, and in water is 13.3 g/100mL. The ratio of these quantities is equal to the partition coefficient, k, which equals What weight of malononitrile would be recovered by extraction of (a) three 100-mL portions of ether and (b) one 300-mL portion of ether? SHOW WORK (Can be written in pen and attached to report). Suggestion: For each extraction, let x equal the weight extracted into the ether layer. In part (a), for the first of the three extractions, the concentration of malononitrile in the ether layer is x/100 and in the water layer is (30-x)/100.
Answer:
The correct answer is 10 grams and 18 grams.
Explanation:
Based on the given question, 20 gram per 100 ml is the solubility of malononitrile in ether, and 13.3 gram per 100 ml is the solubility of malononitrile in water.
Thus, the ration of the solubility is,
Solubility in water/solubility in ether = 20/13.3 = 1.50
a) Let w be the weight of malononitrile extracted into water in every extraction. Then the concentration of the ether layer will be w/100. The concentration in the water layer will be 30-w/300. Now the ratio will be,
Ratio = w/100 / (30-w)/300
1.50 = w/100 * 300 (30-w)
w = 10
Hence, the weight of malononitrile recovered by extraction is 10 grams.
b) The concentration in the ether layer will be w/300. The concentration in the water layer will be (30-w) / 300. Now the ratio will be,
Ratio = w/300 / (30-w) / 300
1.50 = w/300 * 300 (30-w)
w = 18
Hence, 18 grams is the weight of malononitrile recovered by extraction.
A scientist mixed 25.00 mL of 2.00 M KOH with 25.00 mL of 2.00 M HBr. The temperature of the mixed solution rose from 22.7 oC to 31.9 oC. Calculate the enthalpy change for the reaction in kJ/mol HBr, assuming that the calorimeter loses negligible heat, that the volumes are additive, and that the solution density is 1.00 g/mL, and that its specific heat is 4.184 J/g.oC.
Answer:
38.493 KJ/mol
Explanation:
Equation of reaction; HBr + KOH ---> KBr + H2O
Heat evolved = mass * specific heat capacity * temperature rise
Mass of solution = density * volume
Mass = 1.00 g/ml*50 ml = 50g
Temperature rise = 31.9 - 22.7 = 9.2 °C
Heat evolved = 50 * 4.184 * 9.2 = 1924.64 J
From the equation of reaction, 1 mole of HBr reacts with 1mole of KOH to produce 1 mole of H20
Number of moles of HBr involved in the reaction = molar concentration * volume (L)
Molar concentration = 2.0 M, volume = 25 ml = 0.025 L
Number of moles = 2.0 M * 0.025 L= 0.05 moles
Therefore, 0.05 moles of HBr reacts with 0.05 moles of KOH to produce 0.05 moles of H20
Enthalpy change per mole of HBr = 1924.64 J/0.05 moles = 38492.8 J/mol = 38.493 KJ/mol
A temperature of 50°F is equal to °C.
Answer:
CONVERT IT:
50°F is equal to 10°C
Answer:
10 degrees Celsius
Explanation:
(50°F − 32) × 5/9 = 10°C
What mass of salt would you need to add to 1.00kg of water to achieve a freezing point of -5 degrees C
Answer:
The type of salt to be added to the water is not known from the question but no worries, I will try to give you the step by step procedure to answer any type of question similar to this.
To answer this question, we should know some facts.
1. the molar freezing point depression constant of water (Kf) = 1.86 K kg/mol
2. the molar mass of the salt if NaCl = 58.5 g/mol ; KCl = 74.5 g/mol
3. since the salt can dissociate if NaCl or KCl into two ions, the Van't Hoff factor ( i )= 2
Note that: the change in freezing point, molarity, deepression constant and van't Hoff factor are related by this formula;
ΔTf = i Kf m
So lets take NaCl as the salt:
Molar mass = 58.5 g/mol
Van't Hoff factor = 2
1. calculate the number of moles
So we can calculate the molarity of the salt NaCl from the formula;
m = ΔTf / i Kf
m = 5 / 2 * 1.86
m = 5 / 3.72
m = 1.344 mol/kg
2. calculate the number of moles of the salt required
Next is to multiply the molarity by the mass of water. Density of water = 1kg/L
number of moles = 1.344 mol/kg * 1 Kg/L * 1 kg water
number of moles = 1.344 moles.
3. calculate the mass of the salt.
numner of moles = mass / molar mass
mass = number of moles * molar mass
mass = 1.344 * 58.5
mass = 78.624 g of NaCl salt.
You can follow these steps to solve for the type of salt you are given in the question.
how many molecules (not moles) of NH3 are produced from 5.25x10^-4 g of H2?
due in a few, please help. will mark as brainliest
Answer:
not 100% but i think its 1.57x10^20
Explanation:
5.25x10^-4g / 2.016g
2.60x10^-4 x 6.022x10^23= 1.56x10^20 molecules
Volume of water is 35 cm3 and mass of water is 60 gram, what is the density of the water.
Answer:
p = 1.714 g/cm3
Explanation:
Density Equation
p=mV
Where:
p = density
m = mass = 60g
V = volume = 35cm3
p = 60g x 35cm3
p = 1.714 g/cm3
p=1.714g/cm³
Explanation:
v=35cm³
m=60g
P=mass/volume (density formula)
=60/35
=1.714g/cm³
Which best explains how heat plays a role in the movement of materials within Earth's interior?
Answer:
Hot material near the core is less dense and rises, when it cools, it becomes more dense and sinks.
Explanation:
This best explains how heat plays a role in the movement of materials within Earth's interior because it's how convection works. Convection is the circular motion that happens when warmer air or liquid which has faster moving molecules, making it less dense rises, while the cooler air or liquid drops down. Convection currents within the earth move layers of magma, and convection in the ocean creates currents.
Earth’s interior is like a closed chamber, hot material near Earth's surface is more denser and sinks, and when it cools, it becomes less dense and Rises.
What is Convection?Convection is the process of heat transfer by the bulk movement of molecules within fluids such as gases and liquids. The heat transfer take place from solid to no solid material. like air and water.
As we take example of home. In home heat take nearer the roof and cold air will be nearer to the flour. It is because of hot is less denser then the cold air. Hot air will become less denser and move towords upside. Cold air become denser move towards down.
Similarly, on earth. Nearer to earth hot materials are more dense and sinks. When it cools, it become less dense and rises.
To find more about Heat flow, refer the link below:
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Consider the addition of an electron to the following atoms from the third period. Rank the atoms in order from the most negative to the least negative electron affinity values based on their electron configurations.
Atom or ion Electron configuration
Br 1s22s22p63s23p64s23d104p5
Ge 1s22s22p63s23p64s23d104p2
Kr 1s22s22p63s23p64s23d104p6
Answer:
The ranks is
Ge: 3d10 4s2 4p2 (6 electrons in the outer shell)
Br: 3d10 4s2 4p5 (7 electrons in the outer shell)
Kr: 3d10 4s2 4p6 (8 electrons in the outer shell)
Explanation:
Electronic configuration reffers to the distribution of electrons of an atom or molecule in atomic or molecular orbitals. It gives us the understanding of the shape and energy of its electrons. The electronic configuration explain the The electron affinity or propensity to attract electrons
It Should be noted that the most stable configuration in an electronic configuration is attributed to when the last shell is full, i.e. when the last shell has 8 electrons.
When an atom is closer to reach the 8 electrons in the outer shell, then it's electron affinity big.
Considering the given three configuration of the elements above, we can see that "Br"needs requires only 1 electron to have 8 electrons in the outer shell, therefore, it is considered to have the biggest electron affinity among them which is reffers to as the LEAST NEGATIVE.
Ge: with the electronic configuration 3d10 4s2 4p2 has 6 electrons in the outer shell which means it still requires 2 electrons to complete 8 electrons in its outer shell, so it can be deducted that it posses an atom that is more negative than Br.
Kr: with the electronic configuration 3d10 4s2 4p6 which is a noble gas has 8 electrons in the outer shell cannot add more electrons to its outer shell because the 8 electrons is complete posses the least electron affinity among the three elements and it is the MOST NEGATIVE
A salt solution was found to contain 1.50 g of salt dissolved in 50 mL of water. On evaporation, the recovered salt weighed 1.47 g. What percent of salt was recovered?
A) 20.4%
B) 107%
C) 98%
D) 20.0%
Answer:
C = 98%
Explanation:
Hello,
To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.
Mass of salt recovered = 1.47g
Initial mass of salt = 1.50g
Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100
Percentage of salt recovered = (1.47 / 1.50) × 100
Percentage of salt recovered = 0.98 × 100
Percentage of salt recovered = 98%
The percentage of salt recovered is equal to 98%
4-Methylphenol, CH3C6H4OH (pKa 10.26), is only slightly soluble in water, but its sodium salt, CH3C6H4O-Na , is quite soluble in water. Describe the solubility of 4-methylphenol in solutions of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3). The pKa values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3) are 15.7, 6.36, and 10.33, respectively.
Explanation:
We know that more is the [tex]pK_{a}[/tex] value, weaker will be the acid. Also, an acid completely dissociates into ions in an aqueous base solution when [tex]pK_{a}[/tex] of conjugate acid of base is greater than acid.
4-methylphenol [tex](CH_{3}C_{6}H_{4}OH)[/tex] ([tex]pK_{a} = 10.26[/tex]) is quite soluble in its sodium salt. In NaOH, the dissociation will be as [tex]Na^{+}[/tex] and [tex]OH^{-}[/tex] ions as NaOH is a strong base.
Therefore, 4-methylphenol will readily dissolve in NaOH solution.
As, [tex]NaHCO_{3}[/tex] is not a strong base but as 4-Methylphenol forms a sodium salt hence, it will have a low solubility as compared to NaOH.
Whereas [tex]Na_{2}CO_{3}[/tex] is not a base but when dissolved in water it shows basic character as it produces NaOH (strong base) and [tex]H_{2}CO_{3}[/tex] (weak acid). As a result, the solution gets basic. Hence, 4-methylphenol will readily dissolve in [tex]Na_{2}CO_{3}[/tex].
One method of experimentally determining whether a species is paramagnetic is to weigh it in an instrument called a magnetic susceptibility balance. This is a balance with a strong electromagnet placed next to the sample holder. If the species is paramagnetic, the mass reading of the balance will increase when the field is switched on.Classify these species as paramagnetic or diamagneticWhich species will have the strongest mass shift on a magnetic susceptibility balance?
Answer:
Diamagnetic have paired electrons while paramagnetic have at least one unpaired electron.
Explanation:
F2, C2 and N2 are diamagnetic while O2 and B2 are paramagnetic. Diamagnetic are those atoms which have paired electrons while paramagnetic are those atoms which contain at least one unpaired electron so we can say that F2, C2 and N2 have paired electrons while O2 and B2 have unpaired electrons. When diamagnetic materials are allowed to contact with external magnetic field so they will be repelled while paramagnetic materials are attracted by magnetic field due to the presence of unpaired electrons.
Iron oxide (FeO) is the strongest paramagnetic material having the value of 720.
The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound containing C, H, and O exhibits intense absorption at 1720 cm-1. No additional information is available. List possible classes for which there is positive evidence.
Relative absorption intensity: (s)=strong, (m)=medium, (w)=weak.
What functional class(es) does the compound belong to?
List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly.
Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm-1.
a. alkane (List only if no other functional class applies.)
b. alkene h. amine
c. terminal alkyne i. aldehyde or ketone
d. internal alkyne j. carboxylic acid
e. arene k. ester
f. alcohol l. nitrile
g. ether
Answer:
The class of this compound is aldehyde or ketone (i).
Explanation:
Absorption peak at 1720 cm-1 shows the presence of a carbonyl group, possibly an aldehyde or ketone with C=O bond.
Further information on molecular formula would be required for structural elucidation.
Why is tape attracted to my skin? Give explanation
Answer:
Since the tape has extra electrons, it has a negative charge. When you move your finger close to the tape, electrons in your skin are repelled and move away. This makes the skin on your finger tip have a slight positive charge. Since positive and negative attract, the tape moves toward your finger.
What is the predicted order of first ionization energies from highest to lowest for lithium (Li), sodium (Na), potassium (K), and rubidium (Rb)?
Rb > K > Na > Li
K > Rb > Na > Li
Li > Na > K > Rb
Rb > K > Li > Na
Answer:
Li>Na>K>Rb
Explanation:
Answer:
c. Li > Na > K > Rb
Explanation:
edge 2021
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Amylase is the enzyme that controls the breakdown of starch to glucose. Describe how the student could investigate the effect of pH on the breakdown of starch by amylase.
Answer:
Explanation:
You will investigate the breakdown of starch by amylase at different pHs.
The different pHs under investigation will be produced using buffer solutions. Buffer solutions produce a particular pH, and will maintain it if other substances are added.
The amylase will break down the starch.
A series of test tubes containing a mixture of starch and amylase is set up at different pHs.
A sample is removed from the test tubes every 10 seconds to test for the presence of starch. Iodine solution will turn a blue/black colour when starch is present, so when all the starch is broken down, a blue-black colour is no longer produced. The iodine solution will remain orange-brown.
A control experiment must be set up - without the amylase - to make sure that the starch would not break down anyway, in the absence of an enzyme. The result of the control experiment must be negative - the colour must remain blue-black - for results with the enzyme to be valid.
When the starch solution is added:
Start timing immediately.Remove a sample immediately and test it with iodine solution.Sample the starch-amylase mixture continuously, for example every 10 seconds.For each pH investigated, record the time taken for the disappearance of starch, ie when the iodine solution in the spotting tile remains orange-brown.
The time taken for the disappearance of starch is not the rate of reaction.
It will give us an indication of the rate, but is the inverse of the rate - the shorter the time taken, the greater the rate of the reaction.
We can calculate the rate of the reaction by calculating \frac{1}{t}, obtaining a measure of the rate of reaction by dividing one by the time taken for the reaction to occur.
A similar experiment can be carried out to investigate the effect of temperature on amylase activity.
Set up a series of test tubes in the same way and maintain these at different temperatures using a water bath - either electrical or a heated beaker of water.
Depending on the chemical reaction under investigation, you might monitor the reaction in a different way. If investigating the effect of temperature on the breakdown of lipid by lipase, you could monitor pH change - lipids are broken down into fatty acids and glycerol. As the reaction begins, the release of fatty acids will mean that the pH will decrease.
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An unknown compound, B, has the molecular formula C7H12. On catalytic hydrogenation 1 mol of B absorbs 2 mol of hydrogen and yields 2-methylhexane. B has significant IR absorption band at about 3300 and 2200 cm-1. Which compound best represents B?
Answer:
5-methylhex-1-yne//5-methylhex-2-yne//2-methylhex-3-yne
Explanation:
We have to start with the information on the IR spectrum. The signal at 3300 is due to a C-H bend sp carbon and the signal in 2200 is due to the stretch carbon-carbon. Therefore we will have an alkyne. Now if we have 2-methylhexane as the product of hydrogenation we have several options to put the triple bond. Between carbons 1 and 2 (5-methylhex-1-yne), between carbons 2 and 3 (5-methylhex-2-yne) and between carbons 3 and 4 (2-methylhex-3-yne). On carbon 5 we have a tertiary carbon therefore we dont have any other options.
See figure 1
I hope it helps!
What is the molar mass of P2O5?
Answer:
142 grams
Explanation:
To find the molar mass of a molecule or compound, you simply need to add together the molar masses of all of the atoms that comprise it. Phosphorus has a molar mass of about 31, while oxygen has one of about 16, meaning that the molar mass of this molecule is:
2(31)+5(16)=62+80=142
Hope this helps!
A new non-electrolyte molecule is discovered. When 241 mg of the molecule is dissolved in 250.0 mL of water, it has an osmotic pressure of 0.072 atm at 25 oC.What is the molar mass of the molecule
Answer:
327.89g/mol
Explanation:
Step 1:
The following data were obtained from the question:
Van 't Hoff factor (i) = 1 (since the molecule is non-electrolyte)
Temperature (T) = 25°C = 25°C + 273 = 298K
Gas constant (R) = 0.0821 atm.L/Kmol
Mass of molecule = 241mg
Volume of water = 250mL
Molarity (M) =?
Osmotic pressure (Π) = 0.072 atm
Step 2:
Determination of the molarity of the molecule.
This can be obtained as follow:
Π = iMRT
0.072 = 1 x M x 0.0821 x 298
Divide both side by 0.0821 x 298
M = 0.072 / (0.0821 x 298)
M = 2.94×10¯³ mol/L
Step 3:
Determination of the number of mole of the molecule. This can be obtained as follow:
Molarity = 2.94×10¯³ mol/L
Volume of water = 250mL = 250/1000 = 0.25L
Mole of molecule =..?
Molarity = mole /Volume
2.94×10¯³ = mole / 0.25
Cross multiply
Mole of molecule = 2.94×10¯³ x 0.25
Mole of molecule = 7.35×10¯⁴ mole.
Step 4:
Determination of the molar mass of the molecule.
Mole of molecule = 7.35×10¯⁴ mole.
Mass of molecule = 241mg = 241×10¯³g
Molar mass of molecule =..?
Mole = Mass /Molar Mass
7.35×10¯⁴ = 241×10¯³/ Molar Mass
Cross multiply
7.35×10¯⁴ x molar mass = 241×10¯³
Divide both side by 7.35×10¯⁴
Molar Mass = 241×10¯³/7.35×10¯⁴
Molar Mass = 327.89g/mol
Therefore, the molar mass of the molecule is 327.89g/mol
A sample of 6.022 x 1023 particles of gas has a volume of 22.4 L at 0°C and a pressure of 1.000 atm. Although it may seem silly to contemplate, what volume would 1 particle of gas occupy?
pv=nRT
Answer:
1 particle of the gas would occupy a volume of 3.718*10⁻²³L
Explanation:
Hello,
1. The sample has a particle of 6.022×10²²particles
2. Volume of the sample = 22.4L
3. Temperature of the sample = 0°C = (0 +273.15)K = 273.15K
4. Pressure of the sample = 1.0atm
What volume would 1 particle of the gas occupy?
But we remember that 1 mole of any substance = 6.022×10²² molecules or particles or atoms
What would be the number of moles for 1 particule?
1 mole = 6.022×10²² particles
X moles = 1 particle
X = (1 × 1) / 6.022×10²² particles
X = 1.66×10⁻²⁴ moles
Therefore, 1 particle contains 1.66×10⁻²⁴ moles
Since we know our number of moles, we can proceed to use ideal gas equation,
Ideal gas equation holds for all ideal gas and is defined as
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles of the gas
R = ideal gas constant = 0.082 L.atm / mol.K
T = temperature of the gas
PV = nRT
Solving for V,
V = nRT/ P
We can now plug in our values into the above
equation.
V = (1.66*10⁻²⁴ × 0.082 × 273.15) / 1
V = 3.718*10⁻²³L
Therefore, 1 particule of the gas would occupy a volume of 3.718*10⁻²³L.
when a car drives down a street, what evergy conversions are happening?