There are approximately 6.011 × 10²⁴ oxygen atoms in the 120.0 g sample of glucose.
The chemical formula for glucose is C₆H₁₂O₆, indicating 6 carbon, 12 hydrogen, and 6 oxygen atoms. The molar mass of glucose may be used to calculate the number of oxygen atoms in 120.0 g of glucose:
Glucose molecule mass (C₆H₁₂O₆) = (6 * carbon) + (12 * hydrogen) + (6 * oxygen)
≈ (6 * 12.01 g/mol) + (12 * 1.008 g/mol) + (6 * 16.00 g/mol)
The molar mass may be used to compute glucose moles in the 120.0 g sample:
To calculate the number of moles of glucose, divide the sample mass by the molar mass: 120.0 g / 72.06 g/mol = 1.664 moles.
Since each glucose molecule has 6 oxygen atoms, we may compute the total:
The total quantity of oxygen atoms is calculated by multiplying the number of glucose moles by the number of oxygen atoms per molecule: 1.664 moles * 6 = 9.984 moles.
Oxygen atoms total roughly 9.984 × 6.022 × 10²³ atoms/mole
= 6.011 × 10²⁴ atoms.
In a 120.0 g glucose sample, there are roughly 6.011 × 10²⁴ oxygen atoms.
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Final answer:
To calculate the number of oxygen atoms in a 120.0 g sample of glucose, use the mole ratio to find the number of moles of oxygen. Then, convert moles of oxygen to atoms using Avogadro's number.
Explanation:
To calculate the number of oxygen atoms in a 120.0 g sample of glucose, we need to determine the number of moles of glucose and use the mole ratio to find the number of moles of oxygen. The molar mass of glucose is 180.16 g/mol, so the number of moles is 120.0 g / 180.16 g/mol = 0.6664 mol. According to the balanced chemical equation, one mole of glucose reacts with 6 moles of oxygen to yield 6 moles of CO2 and 6 moles of H2O. Therefore, the number of moles of oxygen is 0.6664 mol x 6 mol O2 / 1 mol glucose = 3.9984 mol O2. Finally, to convert moles of oxygen to atoms, we use Avogadro's number (6.022 × 10^23 atoms/mol), so the number of oxygen atoms is 3.9984 mol O2 x 6.022 × 10^23 atoms/mol = 2.406 × 10^24 oxygen atoms.
The surface temperature on Venus may approach 757 K. What is this temperature in degrees Celsius? 757 K = The temperature on Mercury may drop to -261 °F at night. What is this temperature in degrees Celsius? -261 °F =
Answer:
757 K = 484 °C
-261 °F = -163 °C
Explanation:
The formula to convert Kelvin to degrees Celsius is:
°C = K - 273.15 = 757 - 273.15 = 484 °C
The formula to convert °F to °C is:
°C = 5/9 (°F -32) = 5/9 (-261 - 32) = -163
An aqueous solution of sulfuric acid has a composition of 25wt% and a SG of 1.22. Calculate the Volume of the solution that has 245 kg of sulfuric acid.
Answer: The volume of solution is [tex]8.03\times 10^5mL[/tex]
Explanation:
The relationship between specific gravity and density of a substance is given as:
[tex]\text{Specific gravity}=\frac{\text{Density of a substance}}{\text{Density of water}}[/tex]
Specific gravity of sulfuric acid solution = 1.22
Density of water = 1.00 g/mL
Putting values in above equation we get:
[tex]1.22=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.22\times 1.00g/mL)=1.22g/mL[/tex]
We are given:
25% (m/m) sulfuric acid solution. This means that 25 g of sulfuric acid is present in 100 g of solution
Conversion factor: 1 kg = 1000 g
Mass of solution having 254 kg or 245000 g of sulfuric acid is calculated by using unitary method:
If 25 grams of sulfuric acid is present in 100 g of solution.
So, 245000 grams of sulfuric acid will be present in = [tex]\frac{100}{25}\times 245000=980000g[/tex]
To calculate volume of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of solution = 1.22 g/mL
Mass of Solution = 980000 g
Putting values in above equation, we get:
[tex]1.22g/mL=\frac{980000g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{980000g}{1.22g/mL}=8.03\times 10^5mL[/tex]
Hence, the volume of solution is [tex]8.03\times 10^5mL[/tex]
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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Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrical response proportional to the concentration of hydrogen ion. To convert these responses to a pH reading, the electrode must be calibrated against standard solutions of known H+ concentration. Determine the weight in grams of sodium dihydrogen phosphate (NaH2PO4 · H2O; FW 138) and disodium hydrogen phosphate (Na2HPO4; FW 142) needed to prepare 1 L of a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M
To prepare a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M, you will need to calculate the amount of sodium dihydrogen phosphate (NaH2PO4 · H2O) and disodium hydrogen phosphate (Na2HPO4) needed.
Explanation:To prepare a standard buffer at pH 7.00 with a total phosphate concentration of 0.100 M, you will need to calculate the amount of sodium dihydrogen phosphate (NaH2PO4 · H2O) and disodium hydrogen phosphate (Na2HPO4) needed.
Step 1: Calculate the individual concentrations of NaH2PO4 and Na2HPO4.
Using the molecular weight, you can calculate the number of moles of NaH2PO4 and Na2HPO4 needed to achieve a total phosphate concentration of 0.100 M in 1 L of solution.
NaH2PO4: (0.100 M) * (1 L) = x mol
Na2HPO4: (0.100 M) * (1 L) = y mol
Step 2: Calculate the weight of NaH2PO4 and Na2HPO4.
Using the number of moles calculated in step 1, you can calculate the weight of NaH2PO4 and Na2HPO4 needed.
NaH2PO4: (x mol) * (138 g/mol) = weight in grams
Na2HPO4: (y mol) * (142 g/mol) = weight in grams
By following these steps, you will be able to determine the weight in grams of NaH2PO4 · H2O and Na2HPO4 needed to prepare 1 L of the standard buffer solution.
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.
What is the volume (in dm3 of 1 mole of oxygen at 5 MPa and 200 K?
Answer: The volume of oxygen gas is [tex]0.332dm^3[/tex]
Explanation:
To calculate the volume of the gas, we use the equation given by ideal gas equation:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 5 MPa = 5000 kPa (Conversion factor: 1 MPa = 1000 kPa)
V = Volume of gas = 3.34 L
n = number of moles of oxygen gas = 1 mole
R = Gas constant = [tex]8.31dm^3\text{ kPa }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = 200 K
Putting values in above equation, we get:
[tex]5000kPa\times V=1mol\times 8.31dm^3\text{ kPa }mol^{-1}K^{-1}\times 200K\\\\V=0.332dm^3[/tex]
Hence, the volume of oxygen gas is [tex]0.332dm^3[/tex]
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.
What is the thermodynamic equilibrium constant under standard conditions for the following balanced redox reaction? Zr(s) + O2(g) - ZrO2 (s) Een=2.463 V
Answer:
Equilibrium constant = [tex]2.23 \times 10^{83}[/tex]
Explanation:
[tex]Zr(s) + O_2(g) \rightarrow ZrO_2(s)[/tex]
[tex]E^0_{cell}[/tex] = 2.463 V
Equilibrium constant is related with [tex]E^0_{cell}[/tex] as
[tex]E_{cell}=E^0_{cell} - \frac{2.303 RT}{nF} ln k_{eq}[/tex]
In standard condition,
T = 25 °C = 25 + 273 = 298 K
F = 96500 C mol^-1
R = 8.314 [tex]J\ K^{-1}mol^{-1}[/tex]
On substituting values, the above expression becomes:
[tex]E_{cell}=E^0_{cell} - \frac{0.059}{n} log k_{eq}[/tex]
n = 2
At equilibrium, [tex]E_{cell}= 0[/tex]
[tex]0=E^0_{cell} - \frac{0.059}{2} log k_{eq}[/tex]
[tex]log k_{eq}=\frac{2 \times 2.463}{0.059}[/tex]
= 83.35
[tex]K_{eq} = antilog 83.35 = 2.23 \times 10^{83}[/tex]
The activation energy for the reaction NO2 (g )+ CO (g) ⟶ NO (g) + CO2 (g) is Ea = 217 kJ/mol and the change in enthalpy for the reaction is ΔH = -293 kJ/mol .
What is the activation energy for the reverse reaction?
Enter your answer numerically and in terms of kJ/mol.
Answer : The activation energy for the reverse reaction is 510 kJ/mol.
Explanation :
Activation energy : It is defined as the minimum amount of energy given to the reactant so that it gets converted into products.
The relation between the activation energy for forward and backward reaction and change in enthalpy of reaction for exothermic reaction is:
When activation energy for forward reaction is less than the activation energy for backward reaction then the reaction will be exothermic. In exothermic reaction the enthalpy change will be negative.
[tex]Ea^b=Ea^f+|\Delta H|[/tex]
The relation between the activation energy for forward and backward reaction and change in enthalpy of reaction for endothermic reaction is:
When activation energy for forward reaction is more than the activation energy for backward reaction then the reaction will be endothermic. In endothermic reaction the enthalpy change will be positive.
[tex]Ea^f=Ea^b+|\Delta H|[/tex]
where,
[tex]Ea^f[/tex] = activation energy for forward reaction
[tex]Ea^b[/tex] = activation energy for backward reaction
[tex]\Delta H[/tex] = change in enthalpy of reaction
As per question, the value of enthalpy change is -293 kJ/mol that means the reaction will be exothermic reaction. So,
[tex]Ea^b=Ea^f+|\Delta H|[/tex]
Given:
[tex]Ea^f[/tex] = activation energy for forward reaction = 217 kJ/mol
[tex]Ea^b[/tex] = activation energy for backward reaction = ?
[tex]\Delta H[/tex] = change in enthalpy of reaction = -293 kJ/mol
Now put all the given values in above relation, we get:
[tex]Ea^b=217kJ/mol+|-293kJ/mol|[/tex]
[tex]Ea^b=217kJ/mol+293kJ/mol[/tex]
[tex]Ea^b=510kJ/mol[/tex]
Therefore, the activation energy for the reverse reaction is 510 kJ/mol.
Define ""green chemistry"" and ""ecological footprints"".
Explanation:
Green chemistry
It is the process of designing a chemical compound via reducing or eliminating the use or generation of the hazardous substances .
It is a eco - friendly method , which does not harm the nature .
Ecological footprints
It is the tool to measure the demand of humans on nature , the quantity of nature humans require to support economy .
It tracks the demand of the humans via ecological accounting system .
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]
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
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]
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!
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!
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.
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.What volume of concentrated nitric acid (15.0 M) is
requiredfor the preparationof 2.00 L of 0.001M nitric acid
solution?
Answer:
130 μL
Explanation:
The dilution formula is used to calculate the volume V₁ required:
C₁V₁ = C₂V₂
V₁ = (C₂V₂)/C₁ = (2.00L)(0.001M)/(15.0M) = 1.3 x 10⁻⁴ L or 130 μL
(1.3 x 10⁻⁴ L)(10⁶ μL/L) = 130 μL
A well-insulated, closed device claims to be able to compress 100 mol of propylene, acting as a Soave- Redlich-Kwong gas and with Cp* = 100 J/(mol·K), from 300 K and 2 m^3 to 800 K and 0.02 m^3 by using less than 5 MJ of work. Is this possible?
Explanation:
The given data is as follows.
Moles of propylene = 100 moles, [tex]C_{p}[/tex] = 100 J/mol K
[tex]T_{i}[/tex] = 300 K, [tex]T_{f}[/tex] = 800 K
[tex]V_{i}[/tex] = 2 [tex]m^{3}[/tex], [tex]V_{f}[/tex] = 0.02 [tex]m^{3}[/tex]
Therefore, the assumptions will be as follows.
The given system is very well insulated.The work is done on the system because the given process is a compression process.Assume that there is no friction so, work done on the system is equal to the heat energy liberated.[tex]m \times C_{p} \Delta T[/tex] = W
Putting the given values into the above formula as follows.
[tex]m \times C_{p} \Delta T[/tex] = W
W = [tex]100 moles \times 100 J/mol K \times (800 K - 300 K)[/tex]
= [tex]5 \times 10^{6}[/tex] J
= 5 MJ
Hence, this shows that a minimum of 5 MJ work needs to be done.
Since, work is very less. Hence, it will not compress the given system to 800 K and 0.02 [tex]m^{3}[/tex].
Which type of microscope can be used to view cellular organelles such as the endoplasmic reticulum and Golgi?
An electron microscope is used to view cellular organelles such as the endoplasmic reticulum and Golgi because it provides significantly higher resolution and magnification compared to a light microscope.
Explanation:The type of microscope used to view cellular organelles such as the endoplasmic reticulum and Golgi is the electron microscope. This instrument magnifies an object using an electron beam that passes and bends through a lens system, providing much higher resolution and magnification than a light microscope. However, most student microscopes are light microscopes, which use a beam of visible light and are typically used for viewing living organisms, as the staining required to make cellular components visible usually kills the cells. Electron microscopes are commonly used in labs and can give detailed visualizations of organelles and the endomembrane system, which involves a group of organelles and membranes that work together in modifying, packaging, and transporting lipids and proteins.
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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
Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engine. The upper temperature of this combination is 260F, the lower temperature is 40F. If each engine has the same thermal efficiency, determine the exhaust temperature of the first engine (the inlet temperature of the second engine). Ans: T = 140F 3. A nuclear power plant generates 750 MW of power. The heat engine uses a nuclear reactor operating at 315C as the source of heat. A river is available (at 20C) which has a volumetric flow rate of 165 m/s. If you use the river as a heat sink, estimate the temperature rise in the river at the point where the heat is dumped. Assume the actual efficiency of the plant is 60% of the Carnot efficiency.
Answer:
(a) 140 F
(b) The temperature rise at the point where the heat is dumped is 2.51 degC
Explanation:
(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:
[tex]\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}[/tex]
The temperatures have to be expressed in Rankine (or Kelvin) degrees
[tex]1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F[/tex]
(b) The Carnot efficiency of the cycle is
[tex]\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502[/tex]
If the efficiency of the plant is 60% of the Carnot efficiency, we have
[tex]\eta=0.6*\eta_{c}=0.6*0.502=0.302[/tex]
The heat used in the plant can be calculated as
[tex]Q_i=W/\eta=750MW/0.302=2483MW[/tex]
And the heat removed to the heat sink is
[tex]Q_o=Qi-W=2483-750=1733MW[/tex]
If the flow of the river is 165 m3/s, the heat per volume in the sink is
[tex]\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3[/tex]
Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is
[tex]\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C[/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]
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.
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.
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:
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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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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.
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.
Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction? A. carbon + oxygen + carbon dioxide B. carbon + oxygen → carbon dioxide C. carbon dioxide → carbon + oxygen D. carbon dioxide + carbon → oxygen
Answer:
Hello my friend! The correct answer to this quastion is "B. carbon + oxygen → carbon dioxide"
Explanation:
Carbon uses oxygen and heat as fuel for the O2 chemical bond breakdown reaction, and the new reaction between carbon and formed oxygen or carbon dioxide.
C + O2 ----> CO2
Carbon burns in the presence of oxygen to give carbon dioxide. The chemical equation describing this reaction is [tex]\text { carbon }+\text { oxygen } \rightarrow \text { carbon dioxide }[/tex]
Answer: Option B
Explanation:
The chemical equations are the representation of a particular reaction.This equation used to avoid the description of the reaction and narrowing it to precise statement. It consists of two parts namely,
The reactants that initially present undergoes mutual reaction.The products, the aftermath of the reaction.Here Carbon and oxygen are the reactants, the arrow symbol shows the direction of the reaction and the product here is carbon dioxide.