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]
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
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
Give the number of protons and electrons in each of these ions or atoms (show your work) Cs. Ba , .s
Explanation:
The sum of total number of protons present in an element is known as atomic number of the element.
As atomic number of Cs is 55.And, it is known that for a neutral atom the number of protons equal to the number of electrons.
Since, no charge in present on given Cs atom it means that it is neutral in nature. Hence, number of protons and electrons present in Cs are 55.
For Ba, it is also neutral in nature and atomic number of barium is 56. Hence, number of protons and electrons present in Ba are 56.For S, there is no charge on it so it is also neutral in nature. Atomic number of S is 16. Hence, number of protons and electrons present in S are 16.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 %
Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Use a molecular weight of 28.9 kg/kmol for air. Give answer in kg/m3
Answer:
[tex]d=0.92\frac{kg}{m^{3}}[/tex]
Explanation:
Using the Ideal Gas Law we have [tex]PV=nRT[/tex] and the number of moles n could be expressed as [tex]n=\frac{m}{M}[/tex], where m is the mass and M is the molar mass.
Now, replacing the number of moles in the equation for the ideal gass law:
[tex]PV=\frac{m}{M}RT[/tex]
If we pass the V to divide:
[tex]P=\frac{m}{V}\frac{RT}{M}[/tex]
As the density is expressed as [tex]d=\frac{m}{V}[/tex], we have:
[tex]P=d\frac{RT}{M}[/tex]
Solving for the density:
[tex]d=\frac{PM}{RT}[/tex]
Then we need to convert the units to the S.I.:
[tex]T=100^{o}C+273.15[/tex]
[tex]T=373.15K[/tex]
[tex]P=1bar*\frac{0.98atm}{1bar}[/tex]
[tex]P=0.98atm[/tex]
[tex]M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}[/tex]
[tex]M=0.0289\frac{kg}{mol}[/tex]
Finally we replace the values:
[tex]d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}[/tex]
[tex]d=9.2*10^{-4}\frac{kg}{L}[/tex]
[tex]d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}[/tex]
[tex]d=0.92\frac{kg}{m^{3}}[/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
pH indicator. A dye that is an acid and that appears as different colors in its protonated and deprotonated forms can be used as a pH indicator. Suppose that you have a 0.001 M solution of a dye with a p Ka of 7.2. From the color, the concentration of the protonated form is found to be 0.0002 M. Assume that the remainder of the dye is in the deprotonated form. What is the pH of the solution? Berg, Jeremy M.. Biochemistry (p. 24). W. H. Freeman. Kindle Edition.
Answer:
pH = 7.8
Explanation:
The Henderson-Hasselbalch equation may be used to solve the problem:
pH = pKa + log([A⁻] / [HA])
The solution of concentration 0.001 M is a formal concentration, which means that it is the sum of the concentrations of the different forms of the acid. In order to find the concentration of the deprotonated form, the following equation is used:
[HA] + [A⁻] = 0.001 M
[A⁻] = 0.001 M - 0.0002 M = 0.0008 M
The values can then be substituted into the Henderson-Hasselbalch equation:
pH = 7.2 + log(0.0008M/0.0002M) = 7.8
Final answer:
Using the Henderson-Hasselbalch equation with the provided values, the pH of the dye solution is calculated to be approximately 7.8.
Explanation:
To calculate the pH of the solution, we can use Henderson-Hasselbalch equation which relates pH, pKa, and the ratio of the concentrations of the deprotonated (In−) to protonated (HIn) forms of the indicator.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([In−]/[HIn])
Given that the pKa of the dye is 7.2 and the concentration of the protonated form ([HIn]) is 0.0002 M, and the total concentration of the dye is 0.001 M, we can infer that the concentration of the deprotonated form ([In−]) is 0.001 M - 0.0002 M = 0.0008 M. Using these values in the Henderson-Hasselbalch equation:
pH = 7.2 + log(0.0008/0.0002)
pH = 7.2 + log(4)
pH = 7.2 + 0.6021
pH = 7.8021
Therefore, the pH of the solution is approximately 7.8.
PROCESS MASS and ENERGY BALANCES A solid material with 15% water by weight is to be dried to 7% water. Fresh air is mixed with recycled air and blown over the solid. Fresh air contains 0.01 kg moisture per kg of dry air and recycled air, which is part of the air leaving the drier, contains 0.1 kg moisture per kg of dry air. Mixed air entering the drier contains 0.03 kg moisture per kg of dry air. Determine the following: 1) (a) The amount of water removed per 100 kg of wet material fed to the drier. (b) The amount of dry air in fresh air per 100 kg of wet material. (c) The amount of dry air in recycled air per 100 kg of wet material.
Answer:
a) Water removed = 8.6 kg
b) Dry air in the fresh air = 95.6 kg
c) Dry air in the recycled air = 27.3 kg
Explanation:
To solve this problem we have to make mass balances of the different streams.
1) Material balance for the dry solid
For every 100 kg of feed, we have 85 kg of dry solid and 15kg of water.
If the exit material has 7% of moisture content, the total dry solid represents 93% of the mass exiting the drier.
If the dry solid is 85 kg and represents 93% of the total exit material, the total amount of exit material is 85/0.93=91.4 kg. The difference (7%) is water, weighting (91.4-85)=6.4 kg.
The water removed for every 100 kg of feed is (15-6.4)=8.6 kg.
2) Material balance for the water
The water entering the system has to be the same that exit the system.
Let da be the amount of dry air. Then the water entering the drier is (15+0.01*da) and the water exiting the drier is (6.4+0.1*da). We can calculate the amount of dry air:
[tex]15+0.01*da=6.4+0.1*da\\(15-6.4)=(0.1-0.01)*da\\da=8.6/0.09=95.6[/tex]
For every 100 kg of feed, 95.6 kg of dry air is entering the drier.
3) Recycled air
Let rda be the amount of dry air in the recycled stream. We can balance the water content like:
water in the fresh air + water in the recycled air = water in the air entering the drier
[tex]0.01*da+0.1*rda=0.03*(da+rda)\\\\0.1*rda-0.03*rda=0.03*da-0.01*da\\\\0.07*rda=0.02*da\\\\rda=(0.02/0.07)*da=0.286*da=0.286*95.6=27.3 kg[/tex]
The amount of dry air in the recycled stream is 27.3 kg.
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.
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]
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.
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!
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
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.
The Haber-Bosch process is a very important industrial process. In the Haber-Bosch process, hydrogen gas reacts with nitrogen gas to produce ammonia according to the equation 3H2(g)+N2(g)→2NH3(g) The ammonia produced in the Haber-Bosch process has a wide range of uses, from fertilizer to pharmaceuticals. However, the production of ammonia is difficult, resulting in lower yields than those predicted from the chemical equation. 1.15 g H2 is allowed to react with 9.93 g N2, producing 1.12 g NH3. Part A What is the theoretical yield in grams for this reaction under the given conditions? Express your answer to three significant figures and include the appropriate units.
The theoretical yield of NH3 in the Haber-Bosch process under the given conditions is 12.036 grams.
Explanation:Theoretical yield refers to the maximum amount of product that can be obtained in a chemical reaction according to the balanced chemical equation. To calculate the theoretical yield of ammonia in this reaction, you need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be obtained.
In this case, you have 1.15 g of H2 and 9.93 g of N2. To determine the limiting reactant, you can compare the moles of H2 and N2 using their molar masses:
Moles of H2 = (1.15 g H2) / (2 g/mol H2) = 0.575 mol H2
Moles of N2 = (9.93 g N2) / (28 g/mol N2) = 0.354 mol N2
Since the coefficients in the balanced equation are in a 1:1 ratio for H2 and N2, it is clear that the limiting reactant is N2 because there are fewer moles of N2 available.
Now you can use the limiting reactant to calculate the theoretical yield of NH3. According to the balanced equation, the stoichiometric ratio between N2 and NH3 is 1:2. Therefore, moles of NH3 = 2 * moles of N2 = 2 * 0.354 mol = 0.708 mol NH3. Finally, you can convert moles of NH3 to grams using the molar mass of NH3:
Mass of NH3 = (0.708 mol NH3) * (17 g/mol NH3) = 12.036 g NH3
Therefore, the theoretical yield of NH3 under the given conditions is 12.036 grams.
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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]
Consider the following metabolic reaction:
3-Phosphoglycerate → 2-Phosphoglycerate ΔG°’ = +4.40 kJ/mol
What is the ΔG for this reaction when the concentration of 2-phosphoglycerate is 0.290 mM and the concentration of 3-phosphoglycerate is 2.90 mM at 37°C?
Answer:
ΔG = -1.53 kJ/mol
Explanation:
The given reaction is:
3-Phosphoglycerate → 2-Phosphoglycerate
The standard Gibbs free energy, ΔG°=+4.40 kJ
[2-Phosphoglycerate] = 0.290 mM
[3-Phosphoglycerate] = 2.90 mM
Temperature T = 37 C = 310 K
The standard Gibbs free energy, ΔG° is related to the free energy change ΔG at a given temperature by the following equation:
[tex]\Delta G =\Delta G^{0}+RTlnQ[/tex]
In this reaction:
[tex]\Delta G =\Delta G^{0}+RTln\frac{[2-Phosphoglycerate]}{[3-Phosphoglycerate]}[/tex]
[tex]\Delta G = 4.40kJ/mol +0.008314 kJ/mol-K*310Kln\frac{[0.290]}{[2.90]}=-1.53 kJ/mol[/tex]
10 kg of saturated solution of a highly soluble component A at 80°C is cooled to 30°C Calculate the amount of an-hydrous crystals are coming out of the solution Solubility of A at 80*C is 0.8 kg of A 1 kg of water and at 30°C is 0.3 kg of A 1 kg of water a) 2.73 kg b) 5.73 kg c) 4.73 kg d) 3.73 kg
Answer:
The amount of anhydrous crystal are coming out of the solution when this is cooled from 80°C to 30°C are 5 kg of A
Explanation:
A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent, and knowing the solubility of component A at 80°C it is possible to know their amount, thus:
10Kg of water ×[tex]\frac{0,8 kg A}{1 kgWater}[/tex] = 8 kg of A
The maximum concentration that water can dissolve at 30°C is:
10Kg of water ×[tex]\frac{0,3 kg A}{1 kgWater}[/tex] = 3 kg of A
Thus, the amount of anhydrous crystal are coming out of the solution when this is cooled from 80°C to 30°C are:
8 kg of A - 3 kg of A = 5 kg of A
I hope it helps!
In one stroke of a reciprocating compressor, helium is isothermally and reversibly
compressed in a piston/cylinder from 298 K and 15 bar to 150 bar. Compute the heat
removal and work required.
Explanation:
It is known that in reversible isothermal compression, relation between work and pressure is as follows.
w = -2.303 RT log [tex]\frac{P_{2}}{P_{1}}[/tex]
= [tex]-2.303 \times 8.314 J/mol K \times log \frac{150 bar}{15 bar}[/tex]
= [tex]-5705.85 J /mol \times log (10)[/tex]
= -5705.85 J /mol
According to first law of thermodynamics, q = -w
Hence, q = -(-5705.85 J /mol)
= 5705.85 J /mol
As 1000 J = 1 kJ. Hence, convert 5705.85 J/mol into kJ/mol as follows.
[tex]\frac{5705.85}{1000} kJ /mol[/tex]
= 5.7058 kJ/mol
Thus, we can conclude that heat removal is 5.7058 kJ/mol and work required is -5705.85 J /mol.
Final answer:
In an isothermal and reversible compression of helium gas from 15 bar to 150 bar at 298 K, the work done can be calculated using the formula W = nRT ln([tex]P_1/P_2[/tex]), where n is the number of moles. The heat removed is equal to the work done, but with opposite sign.
Explanation:
Isothermal Compression of Helium Gas
When helium is isothermally and reversibly compressed in a piston/cylinder at a constant temperature of 298 K from an initial pressure of 15 bar to a final pressure of 150 bar, the work (W) done on the gas is found using the formula for isothermal processes for an ideal gas:
W = nRT ln([tex]P_1/P_2[/tex])
Where n is the number of moles of helium, R is the ideal gas constant (8.314 J/(mol K)), T is the temperature in Kelvin, and [tex]P_1[/tex] and [tex]P_2[/tex] are the initial and final pressures.
However, without knowing the number of moles of helium, we cannot compute the exact value of the work. As for the heat removal, for an isothermal process in an ideal gas, the amount of heat removed (Q) is equal to the work done on the gas:
Q = -W
Therefore, the heat removed would be numerically equal to the work required but opposite in sign since the work is done on the gas and the heat is released by the gas.
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
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
For the following systems (as underlined), determine which of the following conditions apply: open, closed, adiabatic, isolated, isothermal, isobaric, isochoric, or steady-state. a. An ice cube inside a freezer where it has already been for an extended period of time, during which the freezer door is never opened. (3 pts) b. The air inside the tire of a Nascar during the first minute of driving in a race. (3 pts) c. Your body over the last week.
Answer:
a. An ice cube inside a freezer where it has already been for an extended period of time, during which the freezer door is never opened. It is Closed because the freezer only exchanges energy, Isothermal since the freezer maintain the temperature constant and Isothermal and Isobaric because the ice cube remains with volume and pressure constant.
b. The air inside the tire of a Nascar during the first minute of driving in a race. Closed because the tire only exchange energy at first and Isochoric since the volume of the tire remain constant.
c. Your body over the last week. Open because the body exchange matter and energy.
Explanation:
The open, closed, adiabatic and isolated systems are defined considering if exchange matter or energy, as the definitions below:
- An open system exchange matter and energy.
- A closed system exchange only energy.
- An adiabatic system only exchange matter.
- An isolated system not exchange matter and energy
The isothermal, isobaric, isochoric, or steady-state are defined as follows:
- Isothermal is a process at a constant temperature.
- Isobaric is a process at constant pressure.
- Isochoric is a process at a constant volume.
- A steady-state refers to a reaction in which the concentrations of the reactants, intermediaries, and products don't change over time.
A liquid mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a) 20 b) 25 c) 6.75 d) 9.25
Answer: The correct answer is Option c.
Explanation:
We are given:
Mass percentage of [tex]CH_4[/tex] = 20 %
So, mole fraction of [tex]CH_4[/tex] = 0.2
Mass percentage of [tex]C_2H_4[/tex] = 30 %
So, mole fraction of [tex]C_2H_4[/tex] = 0.3
Mass percentage of [tex]C_2H_2[/tex] = 35 %
So, mole fraction of [tex]C_2H_2[/tex] = 0.35
Mass percentage of [tex]C_2H_2O[/tex] = 15 %
So, mole fraction of [tex]C_2H_2O[/tex] = 0.15
We know that:
Molar mass of [tex]CH_4[/tex] = 16 g/mol
Molar mass of [tex]C_2H_4[/tex] = 28 g/mol
Molar mass of [tex]C_2H_2[/tex] = 26 g/mol
Molar mass of [tex]C_2H_2O[/tex] = 48 g/mol
To calculate the average molecular mass of the mixture, we use the equation:
[tex]\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}[/tex]
where,
[tex]\chi_i[/tex] = mole fractions of i-th species
[tex]m_i[/tex] = molar masses of i-th species
[tex]n_i[/tex] = number of observations
Putting values in above equation:
[tex]\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}[/tex]
[tex]\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75[/tex]
Hence, the correct answer is Option c.
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.
Which of the following is spontaneous at SATP? O H2(g)—2H(9) O Hg(1)—-Hg(9) O N2(g)+2O2(g)+9 kJ—N204(9) O CO2(s)-CO2(g)
Answer: Option (a) is the correct answer.
Explanation:
A spontaneous reaction is defined as the reaction which occurs in the given set of conditions without any disturbance from any other source.
A spontaneous reaction leads to an increase in the entropy of the system. This means that degree of randomness increases in a spontaneous reaction.
For example, [tex]H_{2}(g) \rightarrow 2H(g)[/tex]
Here, 1 mole of hydrogen is giving 2 moles of hydrogen. This means that degree of randomness is increasing on the product side due to increase in number of moles.
Hence, there will also be increase in entropy.
Whereas in the reaction, [tex]CO_{2}(s) \rightarrow CO_{2}(g)[/tex] here number of moles remain the same. Hence, the reaction is not spontaneous.
Thus, we can conclude that the reaction [tex]H_{2}(g) \rightarrow 2H(g)[/tex] is spontaneous at STP.
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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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].
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+
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.
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.