Answer:
a) C Cu = 1588.6 mg/L
b) C Cu = 1588600 μg/L
c) this concentration exceeds the acute freshwater criteria:
∴ C Cu = 1588.6 mg/L >> 0.065 mg/L
Explanation:
∴ C Cu = 2.5 E-2 mol/L
∴Mw Cu = 63.546 g/mol
a) C Cu = 2.5 E-2 mol/L * 63.546 g/mol = 1.5886 g/L
⇒ C Cu = 1.5886 g/L * ( 1000 mg/g ) = 1588,6 mg/L
b) C Cu = 1588.6 mg/L * ( μg / 0.001 mg ) = 1588600 μg/L
c) C Cu = 65 ppb * ( ppm / 1000ppb ) = 0.065 ppm = 0.065 mg/L
∴ ppm ≡ mg/L
⇒ C Cu = 1588.6 mg/L >> 0.065 mg/L; this concentration exceeds the acute freshwater criteria
A 25L tank of nitrogen has a pressure of 6.7 kpa. Calculate the volume of nitrogen if the pressure is decreased to 3.4 kPa while maintaining constant temperature.
Answer:
49.26L
Explanation:
Hello,
Considering Boyle's law:
[tex]P_1V_1=P_2V_2[/tex]
The volume in the second state is given by (solving for it):
[tex]V_2=\frac{P_1V_1}{P_2} =\frac{25L*6.7kPa}{3.4kPa} \\V_2=49.26L[/tex]
Best regards.
Final answer:
By applying Boyle's Law, we can calculate that when the pressure of the nitrogen gas in a 25 L tank is decreased from 6.7 kPa to 3.4 kPa, the volume increases to 49.26 L.
Explanation:
The subject of the question involves applying Boyle's Law, which is a principle in Chemistry related to the behavior of gases under pressure. Boyle's Law states that for a given mass of gas at constant temperature, the volume of the gas is inversely proportional to the pressure. To calculate the new volume when the pressure is decreased, we can set up the equation as follows:
P1 × V1 = P2 × V2
Given that P1 = 6.7 kPa and V1 = 25 L (the initial state), and P2 = 3.4 kPa (the final pressure), we want to find V2, the final volume. Using Boyle's Law, we can calculate the new volume:
V2 = (P1 × V1) / P2
V2 = (6.7 kPa × 25 L) / 3.4 kPa
V2 = 49.26 L
Therefore, when the pressure of the nitrogen gas is decreased to 3.4 kPa, the volume increases to 49.26 L.
How does the presence of a catalyst affect the activation energy of a reaction? It depends on whether you are talking about the forward or the reverse reaction. A catalyst decreases the activation energy of a reaction. A catalyst increases the activation energy of a reaction. A catalyst does not affect the activation energy of a reaction.
Answer:
A catalyst decreases the activation energy of a reaction.
Explanation:
A catalyst allows a reaction to go through a new pathway that has a lower activation energy. Therefore, it will recquire less energy to reach the transition state (activated complex) which increases the rate of the reaction. This is the reason why adding a catalyst makes a reaction faster.
The figure attached shows the graphical representation of the effect of a catalyst on the activation energy.
Answer:
c. A catalyst reduces the activation energy required for a reaction.
Explanation:
plato
5.0 g of NaCl is dissolved in water, and the solution is brought to a final volume of 150 mL. What is the molarity of this solution?
Answer:
Molarity is a sort of concentration for solutions. When you talk about it, means mols of solute, that are in 1000 ml of solution. The molarity at this is 0.57M
Explanation:
As you have the solution in a volume of 150ml with 5 g of solute, in 1000 ml how much solute, do u have? The answer is 33.333g so now, you have to take the molar mass of NaCl and get the mols. Mass/molar mass, you will get the moles, so 33,3333 g / 58,44 g/m is 0.570 moles. That's M
If the volame ofa gas coetainer at 32°C changes froem 1.55 L to 753 ml, what will the final temperature be? Assume the pressure doesn't change and the amount of gas in the coatainer doesn't change. Remember that there are 1,000 ml in 1 L. O a. 149°C Ob. 353 C Oc. 273C O d.-125 C Oe None of the above.
Answer : The final temperature of gas will be, 149 K
Explanation :
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 1.55 L
[tex]V_2[/tex] = final volume of gas = 753 ml = 0.753 L
[tex]T_1[/tex] = initial temperature of gas = [tex]32^oC=273+32=305K[/tex]
[tex]T_2[/tex] = final temperature of gas = ?
Now put all the given values in the above formula, we get the final temperature of the gas.
[tex]\frac{1.55L}{0.753L}=\frac{305K}{T_2}[/tex]
[tex]T_2=149K[/tex]
Therefore, the final temperature of gas will be, 149 K
Draw a bond-line structure for each of the following compounds: 2.55 a) CH2-CHCH2C(CH3)3 (b) (CH3CH2)2CHCH2CH2OH (d) CH3CH2OCH2CH2OCH2CH3 (c) CH COCH2CH(CH3)2 (f) (CH3)2C=CHCH3 (e) (CH3CH2)3CBr
Answer:
See attachment
Explanation:
Bond-line structures are representations of molecules, where lines are drawn to represent the bonds between carbon atoms or between carbon atoms and heteroatoms (atoms other than C or H). Hydrogen atoms are not represented. Heteroatoms are indicated by their symbol but carbon atoms are not. Carbon atoms are located at the intersection of two lines.
A single bond is represented by one parallel line, a double bond by two parallel lines, and a triple bond by three parallel lines.
For (c), the formula is assumed to CH₃COCH₂CH(CH₃)₂
The question asks for bond-line structures of various compounds which are simplified drawings of molecules with the endpoints and intersections of each line representing carbon atoms and hydrogen atoms are understood to be filling any remaining free bonds, not represented in the diagram. An example is provided.
Explanation:The question asks for bond-line structures of various compounds. A bond-line structure (also known as a line-angle diagram or skeletal formula) is a type of molecular structural formula that simplifies and reduces the drawing of a molecule to its basics, without including non-carbon and non-hydrogen atoms, or any multiple bonds. The endpoints and intersections of each line are carbon atoms and hydrogen atoms are understood to be filling any remaining free bonds, not represented in the diagram.
For example, for compound (CH₃CH₂)₂CHCH₂CH₂OH), the bond-line structure would look like a series of zig-zag lines, representing the carbon backbone of the molecule, with an OH group attached to the end. The CH₃ and CH₂ groups are understood without being explicitly drawn. The same principle applies to all the specified compounds. Due to the textual limitations of this platform, it's not possible to sketch the structures in the response.
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A student obtained a 0.4513g sample containing aspirin. This sample was analyzed through a titration with NaOH and phenolphthalein was used as the indicator. The endpoint (pH around 9) for the reaction was reached after the addition of 15.22 mL of 0.1105M NaOH. Molar mass for aspirin = 180.16g/mol.
Calculate the % purity for the sample. Show all your work
*Hint: Aspirin reacts with NaOH in a 1:1 ratio.
% Purity = (Actual moles of aspirin in sample based on titration÷ MAXIMUM moles of aspirin in sample) × 100%
AnswerAnswer:
The purity of the sample is 67.14 %
Explanation:
The titration reaction is as follows:
NaOH + aspirin-H → Na⁺ + aspirin⁻ + H₂O
When no more aspirin-H is left, the addition of more NaOH raises the pH and the color of the indicator turns, in this case to a pink color. This is the reaction at the endpoint that indicates that no more aspirin-H is left:
NaOH + aspirin⁻ → Na⁺ + aspirin⁻ + OH⁻
Then, the moles of NaOH added until the turn of the indicator must be equal to the number of moles of aspirin present in the solution since NaOH reacts with aspirin in a 1:1 ratio.
Then:
moles of aspirin in the solution = moles of added NaOH
moles of aspirin in the solution = Concentration of NaOH * volume
moles of aspirin in the solution = 0.1105 mol/l * 0.01522 l = 1.682 x 10⁻³ mol
Knowing the molar mass of aspirin, we can calculate the mass of aspirin present in the solution:
1.682 x 10⁻³ mol aspirin *(180.16 g / mol) = 0.3030 g.
Since the sample contained 0.4513 g, the percent of aspirin in the sample will be: 0.3030 g * (100 % / 0.4513 g) = 67.14 %
We will get the same result if we convert the mass of the sample to mol and calculate the purity using moles instead of mass:
moles of aspirin in the sample:
0.4513 g * ( 1 mol / 180.16g) = 2.505 x 10⁻³ mol aspirin
The purity will be then:
1.682 x 10⁻³ mol * ( 100 % / 2.505 x 10⁻³ mol) = 67.14 %
In the lab you weigh out 76.02 g of Iron (Fe). How many moles of Iron do you have in the sample. (Your answer must have a unit...please use the abbreviation).
Answer:
We have 1.361 moles in the sample
Explanation:
Mass of iron = 76.02g
Molar mass of iron = 55.845 g/ mole ( This we can find in the periodic table, and menas that 1 mole of iron has a mass of 55.845 g).
To calculate the number of moles we will use following formula:
moles (n) = mass / molar mass
moles iron = 76.02g / 55.845 g/ mole
moles iron = 1.36127 moles
To use the correct number of significant digits we use the following rule for multiplication and division :
⇒ the number with the least number of significant figures decides the number of significant digits.
⇒76.02 has 4 digits ( 2 after the comma) and 55.845 has 5 digits (3 after the comma).
⇒ this means 1.361 moles
We have 1.361 moles in the sample
Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO2 and 0.1159 g of H2O. Menthol has a molar mass of 156.27 g/mol. What is the molecular formula of menthol?
Answer: The molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=0.2829g[/tex]
Mass of [tex]H_2O=0.1159g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 0.2829 g of carbon dioxide, [tex]\frac{12}{44}\times 0.2829=0.077g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 0.1159 g of water, [tex]\frac{2}{18}\times 0.1159=0.0129g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 gTo formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles[/tex]
Step 2: Calculating the mole ratio of the given elements.For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00066 moles.
For Carbon = [tex]\frac{0.0064}{0.00066}=9.69\approx 10[/tex]
For Hydrogen = [tex]\frac{0.0129}{0.00064}=19.54\approx 20[/tex]
For Oxygen = [tex]\frac{0.00066}{0.00066}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 10 : 20 : 1
Hence, the empirical formula for the given compound is [tex]C_{10}H_{20}O_1=C_{10}H_{20}O[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 156.27 g/mol
Mass of empirical formula = 156 g/mol
Putting values in above equation, we get:
[tex]n=\frac{156.27g/mol}{156g/mol}=1[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]
Thus, the molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]
Answer:
[tex]\boxed{\text{C$_{10}$H$_{20}$O}}[/tex]
Explanation:
In a combustion experiment, all the carbon ends up as CO₂, and all the hydrogen ends up as water.
Data:
Mass of menthol = 0.1005 g
Mass of CO₂ = 0.2829 g
Mass of H₂O = 0.1159 g
Calculations:
(a) Mass of each element
[tex]\text{Mass of C} = \text{0.2829 g CO$_{2}$} \times \dfrac{\text{12.01 g C}}{\text{44.01 g CO$_{2}$}} = \text{0.077 20 g C}\\\\\text{Mass of H} = \text{0.1159 g H$_{2}$O} \times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.012 97 g H}\\\\\text{Mass of O} = \text{mass of menthol - mass of C - mass of H}\\= \text{0.1005 - 0.077 20 - 0.01297}= \text{0.010 33 g O}[/tex]
(b) Moles of each element
[tex]\text{Moles of C} = \text{0.077 20 g C} \times \dfrac{\text{1 mol C}}{\text{12.01 g C}} = 6.428 \times 10^{-3}\text{ mol C}\\\\\text{Moles of H} = \text{0.012 97 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{ 0.012 86 mol H}\\\\\text{Moles of O} = \text{0.010 33 g O} \times \dfrac{\text{1 mol O }}{\text{16.00 g O}} = 6.458 \times 10^{-4}\text{ mol O}[/tex]
(c) Molar ratios
Divide all moles by the smallest number of moles.
[tex]\text{C: } \dfrac{6.428 \times 10^{-3}}{6.458 \times 10^{-4}} = 9.954\\\\\text{H: } \dfrac{0.012 86}{6.458 \times 10^{-4}} = 19.92\\\\\text{O: } \dfrac{6.458 \times 10^{-4}}{6.458 \times 10^{-4}} = 1[/tex]
(d) Round the ratios to the nearest integer
C:H:O = 10:20:1
(e) Write the empirical formula
The empirical formula is C₁₀H₂₀O.
(f) Calculate the empirical formula mass
10 × C = 10 × 12.01 = 120.1 u
20 × H = 20 × 1.008 = 20.16 u
1 × O = 1 × 16.00 = 16.00 u
EF mass = 156.3 u
(g) Divide the molecular mass by the empirical formula mass.
[tex]n = \dfrac{\text{MM}}{\text{EFM}} = \dfrac{156.27}{156.3} = 0.9998 \approx 1[/tex]
(h) Determine the molecular formula
[tex]\text{MF} = \text{(EF)}_{n} = \rm (C_{10}H_{20}O)_{1} = \textbf{C$_{10}$H$_{20}$O}\\\text{The molecular formula of menthol is } \boxed{\textbf{C$_{10}$H$_{20}$O}}[/tex]
Wilma dissolves 10,00 grams of glucose with water and the final volume of solute and solvent is 66.67 mL. What is the concentration of glucose in her solution? a. 10% (m/v) O b.15.% (mv) O c.25 % (m/v) d. 30.% (m/v) e. None of the above.
Answer:
b.15 % ( m / v )
Explanation:
The percent concentration ( m / v ) =
mass of the solute / volume of the solution * 100
From the question , the mass of glucose = 10.00 g
The total volume of the solution = 66.67 mL
using the above formula ,and putting the respective values , the percent concentration ( m / v ) is calculated as -
The percent concentration ( m / v ) = 10.00 g / 66.67 mL * 100 = 15 (m/v)%
What is the average density of a hexane/octane mixture comprised
of 10 kg of hexane and 30 kg of octane? The density of hexane is
0.66 and that of octane is 0.703.
Explanation:
The given data is as follows.
Mass of hexane = 10 kg, Mass of octane = 30 kg
Formula to calculate average density of mixture is as follows.
[tex]\sum x_{i} \rho_{i}[/tex]
where, [tex]x_{i}[/tex] = mass fraction of i component in mixture
[tex]\rho_{i}[/tex] = density of i component
Hence, calculating the mass fraction of both hexane and octane as follows.
[tex]x_{hexane}[/tex] = [tex]\frac{10}{(10 + 30)}[/tex] = 0.25
[tex]x_{octane}[/tex] = [tex]\frac{30}{(10 + 30)}[/tex] = 0.75
Therefore, calculate the average density as follows.
[tex]\rho_{mixture}[/tex] = [tex](0.26 \times 660) + (0.75 \times 703)[/tex]
= 692.25 [tex]kg/m^{3}[/tex]
Thus, we can conclude that the average density of given hexane/octane mixture is 692.25 [tex]kg/m^{3}[/tex].
Which of the following choices has the compounds correctly arranged in order of increasing solubility in water? (least soluble to most soluble) Which of the following choices has the compounds correctly arranged in order of increasing solubility in water? (least soluble to most soluble) LiF < NaNO3 < CHCl3 CH3OH < CH4 < LiF CH4 < NaNO3 < CHCl3 CCl4 < CHCl3 < NaNO3 CH3OH < Cl4 < CHCl3
Answer: Option (d) is the correct answer.
Explanation:
As it is known that like dissolves like. So, water being a polar compound is able to dissolve only polar compounds.
Hence, a compound that is ionic or polar in nature will readily dissolve in water. Whereas non-polar compounds will be insoluble in water.
As [tex]CCl_{4}[/tex] is a non-polar compound. Hence, it is insoluble in water.
On the other hand, [tex]CHCl_{3}[/tex] is a polar compound due to difference in electronegativity of chlorine and carbon atom there will be development of partial charges. Hence, there will be dipole-dipole forces existing between them.
Whereas [tex]NaNO_{3}[/tex] is an ionic compound and it will readily dissociate into ions when dissolved in water. Also, there will be ion-dipole interactions between sodium and nitrate ions.
Hence, [tex]NaNO_{3}[/tex] will readily dissolve in water.
Thus, we can conclude that the compounds correctly arranged in order of increasing solubility in water are [tex]CCl_{4}[/tex] < [tex]CHCl_{3}[/tex] < [tex]NaNO_{3}[/tex].
The correct order of compounds from least soluble to most soluble in water is CH4 < NaNO3 < CHCl3. CH4 is nonpolar and doesn't dissolve well in water, NaNO3 is an ionic compound and readily soluble, and CHCl3 is more soluble than CH4 but less than NaNO3 due to its polar bonds.
Explanation:Looking at your options, the correct arrangement of compounds in order of increasing solubility in water from least to most soluble is CH4 < NaNO3 < CHCl3. The solubility of a compound in water depends on its molecular structure. CH4 (Methane) is a nonpolar compound and therefore, it doesn't dissolve well in water, a polar solvent.
On the other hand, NaNO3 (Sodium nitrate) is an ionic compound and can dissociate into its ions in water, making it highly soluble. CHCl3 (Chloroform) is a polar compound due to the presence of polar C-Cl and C-H bonds, and it is more soluble in water than CH4 but less soluble than NaNO3.
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Carbon dioxide (1.100g) was introduced into a 1L flask which contained some pure oxygen gas. The flask was warmed to 373K and the pressure was then found to be 608mmHg. If CO2 and O2 were the only gases present, what was the mass of the oxygen in the flask?
Answer : The mass of oxygen present in the flask is 0.03597 grams.
Explanation :
First we have to determine the moles of [tex]CO_2[/tex] gas.
[tex]\text{ Moles of }CO_2=\frac{\text{ Mass of }CO_2}{\text{ Molar mass of }CO_2}=\frac{1.100g}{44g/mole}=0.025moles[/tex]
Now we have to calculate the moles of the oxygen gas.
Using ideal gas equation:
[tex]PV=nRT[/tex]
As, the moles is an additive property. So,
[tex]PV=(n_{O_2}+n_{CO_2})RT[/tex]
where,
P = pressure of gas = 608 mmHg = 0.8 atm
(conversion used : 1 atm = 760 mmHg)
V = volume of gas = 1 L
T = temperature of gas = 373 K
[tex]n_{O_2}[/tex] = number of moles of oxygen gas = ?
[tex]n_{CO_2}[/tex] = number of moles of carbon dioxide gas = 0.025 mole
R = gas constant = [tex]0.0821L.atmK^{-1}mol^{-1}[/tex]
Now put all the given values in the ideal gas equation, we get:
[tex](0.8atm)\times (1L)=(n_{O_2}+0.025)mole\times (0.0821L.atmK^{-1}mol^{-1})\times (373K)[/tex]
[tex]n_{O_2}=0.001124mole[/tex]
Now we have to calculate the mass of oxygen gas.
[tex]\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2[/tex]
[tex]\text{Mass of }O_2=0.001124mole\times 32g/mole=0.03597g[/tex]
Therefore, the mass of oxygen present in the flask is 0.03597 grams.
Final answer:
To find the mass of oxygen in a flask with CO2, convert the pressure to atm, calculate moles of CO2, and use the ideal gas law to find CO2's partial pressure. Subtract this from the total to find oxygen's partial pressure, calculate its moles, and use its molar mass to find the oxygen's mass.
Explanation:
To find the mass of the oxygen in the flask, we can apply the ideal gas law and use Dalton's Law of Partial Pressures. Given that 1.100g of CO2 was introduced into a 1L flask containing oxygen gas at 373K with a total pressure of 608mmHg, if CO2 and O2 were the only gases present, we have enough information to solve for the mass of the oxygen gas.
First, convert the total pressure to atm (608 mmHg = 0.8 atm) because standard gas law constants are typically given in atmospheres. The first step in solving for the mass of oxygen involves finding the moles of CO2 introduced, using its molar mass (44.01 g/mol), and applying the ideal gas law:
Calculate moles of CO2: Moles = mass / molar mass = 1.100g / 44.01 g/mol = 0.025 moles of CO2.
Apply Ideal Gas Law for CO2: Use P = nRT/V, where R = 0.0821 L atm/mol K, to find the partial pressure of CO2.
Since the total pressure is known, we can subtract the partial pressure of CO2 to find the partial pressure of O2.
Finally, use the ideal gas law again with the partial pressure of O2 to find the moles of O2 present, and then use the moles of O2 along with its molar mass (32.00 g/mol) to calculate the mass of oxygen in the flask.
This approach accounts for the behavior of the gas mixture under the given conditions, utilizing pressure, volume, temperature, and the molar mass of gases to solve for the unknown mass of oxygen.
Select the statement that best describes a buffer. View Available Hint(s) Select the statement that best describes a buffer. A buffer prevents the pH of a solution from changing when an acid or base is added. Buffered solutions are always neutral, with a pH of 7. A buffer stabilizes the pH of a solution by preventing acids or bases from dissociating. A buffer causes acidic solutions to become alkaline, and alkaline solutions to become acidic. A buffer resists change in pH by accepting hydrogen ions when acids are added to the solution and donating hydrogen ions when bases are added.
The statement that best describes a buffer is: A buffer resists change in pH by accepting hydrogen ions when acids are added to the solution and donating hydrogen ions when bases are added.
Why?
A buffer is a solution made by combining either:
A weak acid (HA) and its conjugate base (A⁻).A weak base (B) and its conjugate acid (HB⁺)The purpose of a buffer is to resist changes in pH when a strong acid or base is added to the solution.
If the buffer is composed of HA and A⁻ and a strong acid (e.g. HCl) is added, the buffer accepts hydrogen ions in the following way:
A⁻+HCl → HA+Cl⁻
If a strong base (e.g. NaOH) is added, the buffer donates hydrogen ions in the following way:
HA + NaOH → NaA + H₂O
The pH of the buffer at any given moment can be found by using the Henderson-Hasselbach equation, based on the equilibrium HA + H₂O ⇄ H₃O⁺ + A⁻
[tex]pH=pKa+log\frac{[A^{-}] }{[HA]}[/tex]
Have a nice day!
Answer:
A buffer resists change in pH by accepting hydrogen ions when acids are added to the solution and donating hydrogen ions when bases are added.
Explanation:
i took it
The ideal gas equation is PV=nRT where P is pressure, V is volume, n is the number of moles, R is a constant, and T is temperature. You are told that a sample of gas has a pressure of P = 859 torr , a volume of V = 8960 mL , and a temperature of T = 304 K . If you use R = 8.206×10−2 L⋅atm/(K⋅mol) , which of the following conversions would be necessary before you could find the number of moles of gas, n, in this sample?
Answer:
Take a look to R, where the units are L . atm/K . mol.. your pressure is in Torr...so make the conversion to atm. (760 Torr is 1 atm) and then take the volume... as you have mL, remember that R is with L, so convert mL to L by making the division /1000. Pressure and volume are those you have to convert
To calculate the number of moles, convert the pressure from torr to atm and the volume from mL to L, in order to match the given gas constant's units of L⋅atm/(K⋅mol). Then, use the ideal gas equation.
Explanation:To find the number of moles of gas, n, in the given sample using the ideal gas equation PV=nRT, the pressure and volume units must match the units of the gas constant R. The given R is 8.206x10^-2 L⋅atm/(K⋅mol) meaning that P should be in atmospheres (atm) and V should be in liters (L).
The necessary conversions you need are:
Convert pressure from torr to atm. 1 atm is approximately equivalent to 760 torr, so P (in atm) can be found by dividing the given pressure P by 760.Convert volume from mL (milliliters) to L (liters). 1 L is equal to 1000 mL, so V (in L) can be calculated by dividing the given volume V by 1000.
After these conversions are carried out, the ideal gas equation can be used to calculate the number of moles, n.
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which functional group is more electronegative, amine C-NH, or alcohol C-OH, in assigning carbon environment?
Answer:
The correct answer is: alcohol (C-OH) functional group
Explanation:
Electronegativity is described as the ability or the tendency of an element to attract the electron density or shared bonding electrons towards itself.
The electronegativity value of oxygen atom, nitrogen atom and, carbon atom is 3.44, 3.04, and 2.55.
From this data we can conclude that oxygen is more electronegative than nitrogen. Also, the electronegativity difference of oxygen and carbon (0.89) is greater than the electronegativity difference of nitrogen and carbon (0.49).
Therefore, the electronegativity of the oxygen containing alcohol functional group (C-OH) will be greater than the electronegativity of the nitrogen (C-NH) containing amine functional group.
A mileage test is conducted for a new car model. Thirty randomly selected cars are driven for a month and the mileage is measured for cach. The mean mileage for the sample is 28.6 miles per gallon (mpg) and the sample standard deviation is 2.2 mpg Estimate a 95% confidence interval for the mean mpg in the entire population of that car model.
Answer: [tex](27.81,\ 29.39)[/tex]
Explanation:
Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .
Significance level : [tex]\alpha: 1-0.95=0.05[/tex]
Critical value: [tex]z_{\alpha/2}=1.96[/tex]
Sample mean : [tex]\overline{x}=28.6[/tex]
Standard deviation : [tex]\sigma=2.2[/tex]
The formula to find the confidence interval is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
i.e. [tex]28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}[/tex]
i.e. [tex]28.6\pm 0.787259889321[/tex]
[tex]\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)[/tex]
Hence, the 95% confidence interval for the mean mpg in the entire population of that car model = [tex](27.81,\ 29.39)[/tex]
4. Al2O3 (s) + 6HCl (aq) → 2AlCl3 (aq) + 3H20(1) Find the mass of AlCl3 that is produced when 10.0 grams of Al2O3 react with 10.0 g of HCI moar mass AlqO3 102 gm Imol Al2O3 = 0.098 moles molar mass of HCl = 36, 5gr/mol #of moles #6l =0,274 moles mole of Al2 Oz 6 mol of HC 01274 X2 = 0.0913 moles AlC3=13,5 6 I Mass ALCO3 = 12, 193gm. 5. How many grams of the excess reagent in question 4 are left over?
Answer:
Are produced 12,1 g of AlCl₃ and 5,33 g of Al₂O₃ are left over
Explanation:
For the reaction:
Al₂O₃ (s) + 6 HCl (aq) → 2AlCl₃ (aq) + 3H₂0(l)
10,0g of Al₂O₃ are:
10,0g ₓ[tex]\frac{1mol}{102g}[/tex] = 0,0980 moles
And 10,0g of HCl are:
10,0 gₓ[tex]\frac{1mol}{36,5g}[/tex] = 0,274 moles
For a total reaction of 0,274 moles of HCl you need:
0,274×[tex]\frac{1molesAl_{2}O_3}{6 mole HCl}[/tex] = 0,0457 moles of Al₂O₃
Thus, limiting reactant is HCl
The grams produced of AlCl₃ are:
0,274 moles HCl ×[tex]\frac{2 moles AlCl_{3}}{6 moles HCl}[/tex] × 133[tex]\frac{g}{mol}[/tex] = 12,1 g of AlCl₃
The moles of Al₂O₃ that don't react are:
0,0980 moles - 0,0457 moles = 0,0523 moles
And its mass is:
0,0523 molesₓ[tex]\frac{102g}{1mol}[/tex] = 5,33 g of Al₂O₃
I hope it helps!
A measure of the number of possibilities for a given system is o entropy O enthalpy O kinetics thermodynamics
Answer: Entropy
Explanation:
Entropy is the measure of randomness or disorder. It is a thermodynamic state function corresponding to the number of available microstates of a system.
Enthalpy is the difference between the energy of products and the energy of reactants.
Kinetics is the the branch of chemistry which deals with rates of chemical reactions.
Thermodynamics is the branch of chemistry which deals with energy and energy conversions.
Christina is studying a sugar molecule known as ribose. The molar mass of ribose is 150r/mole. What is the mass of 5.0 moles of ribose? a. 150.23 O 6.13 x 10-33 O c. 148.8 g O d.306 750 g Oe
Answer : The mass of 5.0 moles of ribose is 750 grams.
Explanation : Given,
Moles of ribose = 5.0 moles
Molar mass of ribose = 150 g/mole
Formula used :
[tex]\text{Mass of ribose}=\text{Moles of ribose}\times \text{Molar mass of ribose}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Mass of ribose}=5.0mole\times 150g/mole=750g[/tex]
Therefore, the mass of 5.0 moles of ribose is 750 grams.
Calculate the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20 Calculate the number of grams of BaCrO4 that would have to be dissolved and diluted to 100ml to prepare a 0.200M solution.
Final answer:
To calculate the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20, you need to determine the molar mass of the compound and then divide the mass of the sample by the molar mass. Finally, multiply the number of moles by 1000 to convert to millimoles.
Explanation:
To calculate the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20, we first need to determine the molar mass of FeSO4•C2H4(NH3)2SO4.4H20.
The molar mass of FeSO4 is 55.85 g/mol. The molar mass of C2H4(NH3)2SO4.4H20 can be calculated by adding the molar masses of each element (12.01 g/mol for C, 1.008 g/mol for H, 14.01 g/mol for N, 32.06 g/mol for S, 16.00 g/mol for O, and 1.008 g/mol for H).
Next, we convert 500mg to grams. 500mg is equal to 0.5g.
Then, we divide the mass of the sample by the molar mass to calculate the number of moles. Finally, we multiply the number of moles by 1000 to convert to millimoles.
Therefore, the number of millimoles contained in 500mg of FeSO4•C2H4(NH3)2SO4.4H20 can be calculated as follows:
Number of millimoles = (0.5g / molar mass) * 1000
A 300-gallon anaerobic digester will be loaded daily with a feedstock that contains two parts dairy manure and one-part water by volume. The feedstock contains 6% volatile solids (VS) by weight and has a density of 37.5 lb/ft'. What volume of feedstock would be required each day to maintain an organic loading rate (OLR) of 2.0 kg VS/m/day? What is the hydraulic retention time (HRT) in the anaerobic digester tank for a loading rate of 2.0 kg VS/m/day?
Answer:
The volume of feedstock needed to mantain an organic load rate of 2 kgVS/day is 0.055 m3/day of feedstock.
The HRT is 20.6 days.
Explanation:
First, we calculate how many kg is 1 m3 of feedstock. We know the density, so we can calculate the mass:
[tex]M=\rho*V=37.5\frac{lb}{ft^3}*1m^3*(\frac{3.281ft}{1m}) ^3=1324.5lb=600.7 kg[/tex]
If the VS are 6% in weight,
[tex]M_{vs}=0.06*M=0.06*600.7\,kg/m^3=36,0kgVS/m3[/tex]
The volume per day needed to feed 2 kg of VS/day is:
[tex]V=\frac{2kg}{36kg/m^3}= 0.055m3/day=5.5litres/day[/tex]
The HRT depends on the volume of the tank and the flow. Its equation is
[tex]HRT=\frac{V}{Q}=\frac{300gal}{0.055 m^3/day}*\frac{1m^3}{264.172gal}\\ \\HRT=20.6\,days[/tex]
You need to make 10 mL of 2 mg/ml solution of protein and you have 25 mg/mL solution. How much protein solution and water do you need to mix in order to make the required solution? the problem cannot be solved, as we do not know the molecular weight of the protein 8 mL of protein solution, 92 mL of water 0.8 mL of protein solution, 9.2 mL of water 2.5 mL of protein solution, 7.5 mL of water 8 mL of protein solution, 2 mL of water
Answer:
0.8 mL of protein solution, 9.2 mL of water
Explanation:
The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:
C₁V₁ = C₂V₂
We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.
V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL
Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:
(10 mL - 0.8 mL) = 9.2 mL
Complete the following operation and then enter your answer as a decimal with the correct number of significant figures. (17.543 + 2.19) × 1.04821 = (17.543+2.19)×1.04821= ?
Answer : The answer will be 20.68
Explanation :
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
The rule apply for the multiplication and division is :
The least number of significant figures in any number of the problem determines the number of significant figures in the answer.
The rule apply for the addition and subtraction is :
The least precise number present after the decimal point determines the number of significant figures in the answer.
The given expression is:
[tex](17.543+2.19)\times 1.04821[/tex]
In the given expression, 17.543 has 5 significant figures and 2.19 has 3 significant figures. From this we conclude that least precise number present after the decimal point is 2. So, the answer will be:
[tex](19.73)\times 1.04821[/tex]
In the given expression, 19.73 has 4 significant figures and 1.04821 has 6 significant figures. From this we conclude that 4 is the least significant figures in this problem. So, the answer should be in 4 significant figures.
[tex]\Rightarrow 20.68[/tex]
Thus, the answer will be 20.68
A solution contains 0.10 M Pb2+ and 0.10 M Cu2.. Which cation will precipitate first when a solution of NazS is slowly added to the mixture? Refer to the information sheet for solubility constants. P A) Pb2+ B) Cu2+ C) impossible to tell D) both cations
Answer:
b) Cu2+
Explanation:
information sheet for solubility constants:Ksp PbS = 3.4 E-28
Ksp CuS = 6.0 E-37
PbS ↔ Pb2+ + S2-∴ Ksp = 3.4 E-28 = [ Pb2+ ] * [ S2- ]
∴ [ Pb2+ ] = 0.10 M
⇒ [ S2- ] = 3.4 E-28 / 0.10 = 3.4 E-27 M
CuS ↔ Cu2+ + S2-∴ Ksp = 6.0 E-37 = [ Cu2+ ] * [ S2- ]
∴ [ Cu2+ ] = 0.10 M
⇒ [ S2- ] = 6.0 E-37 / 0.10 = 6.0 E-36 M
we have:
(1) [ S2- ] PbS >> [ S2- ] CuS
(2) Ksp PbS >> Ksp CuS
from (1) and (2) it can determined, that separation can be carried out and also the cation that precipitates first is the Cu2+
Magnetic fields are _____.
weakest at the poles
strongest in the middle of the magnet
strongest at the poles
Answer:
Magnetic fields are the strongest at the poles
Explanation:
The strength of the field varies depending on its location around the magnet.
Magnetic fields are the strongest in either both pole of the magnet. The magnetic fields are equally strong in north and south pole.
This because, near the poles, magnetic field lines are very close to each other. In other words , near the poles the magnetic flux density is maximum so the magnetic field is stronger at that position.
Those flux lines are the mathematical representation of the magnetic field and are continuous - a given line must always form a closed loop, that want to return to the opposite pole via air. Density of those lines is the so called magnetic flux density - the number of lines per unit square - determines how strong the field is in a given area of space. As they spread out, their density drops, so does the magnetic strength. In actual open-circuited bar magnet you would see the strongest field on the edges of the pole, this is where the flux lines bend and concentrate.
Halfway between the poles, in the middle of the magnet, the magnetic fields are the weakest.
Answer:
Magnetic fields are the strongest at the poles
Explanation:
Citric acid is often used as an acidity regulator in hot water canning of tomatoes. At 25°C it has Kat 7.4x10- and A.Hº = +4.1 kl/mol for the acid dissociation process. What is the value of the Kat of citric acid at 100°C? к ТС о
Answer:
The rate constant at T = 100 C is 1.0*10⁻³
Explanation:
The Arrhenius equation relates two rate constants K1 and K2 measured at temperatures T1 and T2 as shown below:
[tex]ln\frac{K_{2}}{K_{1}}=\frac{\Delta H^{0}rxn}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]
here, ΔHrxn = standard enthalpy change of the reaction
R = gas constant
From the given information:
K1 = 7.4*10^-4
T1 = 25 C = 25+273 = 298 K
T2 = 100 C = 100+273 = 373K
ΔH°=4.1kJ/mol
[tex]ln\frac{K_{2}}{7.4*10^{-4}}=\frac{4.1 kJ/mol}{0.08314kJ/mol.K}(\frac{1}{298}-\frac{1}{373})K[/tex]
K2 = 1.03*10⁻³
Unit conversion Given 20 meq/L K2CO3, find the following units:
a. mol/L
b. mg/L
c. ppm
d. %
Answer:
a. 0.01 mol/L
b. 1382 mg/L
c. 1382 ppm
d. 0.138%
Explanation:
Hello,
I'm attaching two photos showing the numerical procedure for this exercise.
Note: since we don't have the density of the solution and the solvent isn't specified, we could assume it is water (universal solvent).
Best regards.
The element germanium was once an important component
oftransistors. It can be made by heating the ore germanite
withhydrogen chloride, distilling of the germanium chloride to
theoxide to the metal.
When 1.00g germanite as treated this way, the germanium
presentedwas completely converted into 0.177g of a chloride
containing 33.9%by mass of germanium.
a) Calculate the percentage of germanium in germatite.
Answer:
The percentage of germanium in germatite is 6.0003%.
Explanation:
Mass of an ore of germanite = 1.00 grams
Mass of germanium chloride = 0.177 grams
Percentage of germanium in germanium chloride = 33.9%
Let the mass of germanium present in germanium chloride be x.
Percentage of an element in a compound:
[tex]\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of element}}\times 100[/tex]
[tex]33.9\%=\frac{x}{0.177 grams}\times 100[/tex]
x = 0.060003 grams
Percentage of germanium in an ore of germanite:
[tex]=\frac{0.060003 gram}{1 gram}\times 100=6.0003\%[/tex]
The percentage of germanium in germatite is 6.0003%.
A bowling ball (mass = 7.2 kg, radius = 0.10 m) and a billiard
ball (mass = 0.35 kg, radius = 0.028 m) may each be treated
asuniform spheres. What is the magnitude of the maximum
gravitationalforce that each can exert on the other?
Answer:
Maximum gravitational Force: [tex]F_{gmax} = 1,026*10^{-08} N[/tex]
Explanation:
The maximum gravitational force is achieved when the center of gravity are the closer they can be. For the spheres the center of gravity is at the center of it, so the closer this two centers of gravity can be is:
bowling ball radius + billiard ball radius = 0,128 m
The general equation for the magnitude of gravitational force is:
[tex]F_{gmax} = G \frac{M*m}{r_{min}^{2} }[/tex]
Solving for:
[tex]G = 6,67*10^{-11} \frac{Nm^{2}}{kg^{2}}[/tex]
[tex]M = 7,2 kg[/tex]
[tex]m = 0,35 kg[/tex]
[tex]r_{min} = 0,128 m[/tex]
The result is:
[tex]F_{gmax} = 1,026*10^{-08} N[/tex]
Define "Enantiomer" and "Diastereomer"
Answer:
Enantiomers are the non-superimposable mirror images of each other.
Diastereomers are the stereisomers that are not a reflection or mirror images of each other.
Explanation:
Stereoisomers are the chemical molecules having the same molecular formula and bond connectivity but different arrangement of atoms in space.
Stereoisomers are of two types: Enantiomers and Diastereomers
Enantiomers are the non-superimposable mirror images of each other. Enantiomers are also called optical isomers.
Diastereomers are the stereisomers that are not a reflection or mirror images of each other. Diastereomers include E-Z isomers, cis–trans isomers, meso compounds, non-enantiomeric optical isomers.