Answer:
2486 mmHg
Explanation:
Gay-Lussac's Law states the pressure varies directly with temperature when volume remains constant:
P₁/T₁ = P₂/T₂
Where P₁ and T₁ are initial pressure and temperature and P₂ and T₂ are final pressure and temperatue.
The problem says initial pressure is 721 mmHg, initial temperature is 25°C and final temperature is 755°C. The question is final pressure.
°C must be converted to absolute temperature (K), thus:
25°C + 273,15 = 298,15 K
755°C + 273,15 = 1028,15 K
Thus, pressure P₂ is:
(T₂·P₁) / T₁ = P₂
1028,15K · 721mmHg / 298,15 K = 2486 mmHg
I hope it helps!
If a solid has a heat of fusion of 17.02 kJ/mol and an entropy of fusion of 38.98 J/mol- K, what is the melting point in °C) of this pure solid? Type your answer rounded to 1 decimal place without units (i.e. NN.N).
Explanation:
Melting point is defined as the point at which a solid substance starts to change into liquid state.
Whereas entropy is the degree of randomness of molecules present in a substance.
Heat of fusion is defined as the amount of heat energy necessary to melt a solid substance at its melting point.
Relation between entropy and heat of fusion is as follows.
[tex]\Delta S = \frac{\Delta H}{T}[/tex]
where, [tex]\Delta S[/tex] = 38.98 J/mol K
[tex]\Delta H[/tex] = 17.02 kJ/mol
= [tex]17.02 kJ/mol \times \frac{1000 J}{1 kJ}[/tex]
= 17020 J/mol
Therefore, calculate the melting point as follows.
[tex]\Delta S = \frac{\Delta H}{T}[/tex]
38.98 J/mol K = [tex]\frac{17020 J/mol}{T}[/tex]
T = 436.63 K
Change the temperature into degree celsius as follows.
[tex](436.63 - 273)^{o}C[/tex]
= [tex]163.63^{o}C[/tex]
Thus, we can conclude that the melting point in [tex]^{o}C[/tex] is [tex]163.63^{o}C[/tex].
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 16.4. 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!
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 of the following statements about resonance are correct?
Which of the following statements about resonance are correct? Atoms can move between different resonance structures. Resonance generally involved lone pairs, pi bonds, and formal charges. Each resonance structure must be a valid Lewis structure. Resonance structures are separated by a double-headed arrow: OElectrons can move between different resonance structures. Resonance often involves sp3 hybridized carbon atoms. The actual molecular structure will alternate between all possible resonance structures.
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.
100.0 kg of liquid methanol and 100.0 kg of liquid water are mixed in a stirred tank.
Assuming volume additivity of methanol and water, determine the moles and volumes of the two substances in the mixture.
M(l) W(l)
MW (kg / kmol) 32.04 18.01
rho (kg / L) .791 1.00
1. kmol of methanol?
2. kmol of water?
3. Liters of methanol?
4. L of water?
Answer:
1. kmol of methanol= 3.12 Kmol
2. kmol of water= 5.55 Kmol
3. Liters of methanol= 126.4 L
4. L of water= 100 L
Explanation:
1. kmol of methanol?
32.04 kg methanol ______________ 1 kmol of methanol
100 kg of methanol_______________ X= 3.12 kmol ofmethanol
2. kmol of water?
18.01 kg water ______________ 1 kmol of wáter
100 kg of wáter_______________ X= 5.55 kmol of water
3. Liters of methanol?
0.791 kg methanol _______________________1.00 L of methanol
100kg methanol _______________________x= 126.4 L of methanol
4. L of water?
1kg water _______________________1.00 L of water
100kg water _______________________x= 100 L of water
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.
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.
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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In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several of the chemical constituents of wood but not cellulose. The slurry of undissolved chips in solution is further processed to recover most of the original solution constituents and dried wood pulp. In one such process, wood chips with a specific gravity of 0.640 containing 45.0 wt% water are treated to produce 2000.0 tons/day of dry wood pulp containing 85.0 wt% cellulose. The wood chips contain 47.0 wt% cellulose on a dry basis. Estimate the feed rate of logs (logs/min), assuming that the logs have an average diameter of 8.00 inches and an average length of 9.00 feet. 21.67 Ulogs/min
Final answer:
To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate using the given information.
Explanation:
To estimate the feed rate of logs, we first need to calculate the dry wood pulp production rate. The dry wood pulp production rate can be calculated using the equation:
Dry wood pulp production rate = Feed rate of logs x (1 - water content of wood chips) x (1 - cellulose content in wood chips on a dry basis)
Plugging in the given values, we have:
2000 tons/day = Feed rate of logs x (1 - 0.45) x (1 - 0.47)
Solving for the feed rate of logs, we find that it is approximately 21.67 logs/min.
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]
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]
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]
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
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.
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.
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 .
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.
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.
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!
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.
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]
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
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.
Learn more about electron microscope here:https://brainly.com/question/507443
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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]
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.
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].
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]