Answer: The volume of solution required is 1.816 L
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]
We are given:
Mass of solute (KCl) = 15.00 g
Molar mass of potassium chloride = 74.6 g/mol
Molarity of solution = 0.1107 M
Putting values in above equation, we get:
[tex]0.1107M=\frac{15.00g}{74.6g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=1.82L[/tex]
Hence, the volume of solution required is 1.816 L
To determine the volume of 0.1107 M KCl solution containing 15.00 g of KCl, we can use the formula Molarity (M) = moles of solute / volume of solution (in liters). By calculating the moles of KCl and using the molarity formula, the volume of the solution is found to be 1.814 L.
Explanation:To calculate the number of liters of 0.1107 M KCl that contain 15.00 g of KCl, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, we need to calculate the moles of KCl using its molecular weight:
Moles of KCl = mass of KCl / molecular weight of KCl
Moles of KCl = 15.00 g / 74.6 g/mol = 0.201 moles of KCl
Now, we can use the formula to calculate the volume of solution:
Volume of solution = moles of solute / molarity
Volume of solution = 0.201 moles / 0.1107 M = 1.814 L
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 .
Define "Anomeric" carbon
Anomeric carbon is a stereocenter present in the cyclic structures of carbohydrates (mono or polysaccharides). Being a stereocenter, more exactly an epimer, two diastereosoimers derive from it, designated by the letters α and β; these are anomers, and are part of the extensive nomenclature in the world of sugars.
During the experiment a student precipitated and digested the BaSO4. After allowing the precipitate to settle, they added a few drops of BaCl2 solution, and the previously clear solution became cloudy. Explain what happened.
Answer:
Incomplete precipitation of barium sulfate
Explanation:
The student has precipitated and digested the barium sulfate on his/her side. But on the addition of [tex]BaCl_2[/tex] in the solution, the solution become cloudy. This happened because incomplete precipitation of barium sulfate by the student. When [tex]BaCl_2[/tex] is added, there are still sulfate ions present in the solution with combines with [tex]BaCl_2[/tex] and forms [tex]BaSO_4[/tex] and the formation of this precipitate makes the solution cloudy.
The solution became cloudy after adding BaCl2 because the additional Ba2+ ions reacted with any remaining [tex]SO4^2-[/tex]orming more BaSO4 precipitate, indicating incomplete digestion of the initial BaSO4.
Explanation:During the experiment, the student added a few drops of BaCl2 solution to the clear solution that contained digested BaSO4. BaSO4 is known for its low solubility in water; however, it's soluble in solutions containing ions that can form more soluble compounds with the constituent ions. In the experiment, adding more BaCl2 likely introduced additional Ba2+ ions into the solution. If any unreacted sulfate ions (SO42-) were present, these extra Ba2+ ions could have reacted with them, forming additional BaSO4 precipitate, thus causing the solution to become cloudy. This suggests that the digestion process was not complete, leaving some sulfate ions in the solution which reacted with the added barium ions to form more BaSO4 precipitate.
a.) How would you prepare .250 L of a 0.300 M phosphate buffer at pH=3 (H3PO4: pka1 = 2.12, pka2 = 7.21, pka = 12.32) using the appropriate weak acid and conjugate salt. b.) How would you prepare the assigned buffer (given in question #1a) if using combining the weak acid and 0.400 M NaOH? c.) How would you prepare the assigned buffer (given in question #1a) if using combining the weak acid and 0.400 M HCl?
Answer:
a) 0,857 g of H₃PO₄ with 9,016 g of KH₂PO₄
b) 166 mL of 0,400M NaOH
c) 22 mL of 0,400M HCl
Explanation:
a) The appropriate weak acid and conjugate salt are:
H₃PO₄ ⇄ H₂PO₄⁻ + H⁺ where pka = 2,12
Henderson–Hasselbalch equation finding pH = 3:
3 = 2,12 + log₁₀ [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex]
7,59 = [tex]\frac{[H2PO4-]}{[H3PO4]}[/tex] (1)
If buffer concentration is 0,300M:
0,300 M = [H₃PO₄] + [H₂PO₄⁻] (2)
Replacing (2) in (1):
[H₃PO₄] = 0,035 M
Thus:
[H₂PO₄⁻] = 0,265 M
Thus, to prepare this buffer you need weight:
0,035 M × 0,250 L = 8,75x10⁻³ moles × [tex]\frac{97,994 g}{1mol}[/tex] = 0,857 g of H₃PO₄
And:
0,265 M × 0,250 L = 6,63x10⁻² moles × [tex]\frac{136,086 g}{1mol}[/tex] = 9,016 g of KH₂PO₄
b) Using 0,400 M NaOH the equilibrium is:
H₃PO₄ + NaOH ⇄ H₂PO₄⁻ + H₂O
Knowing the equilibrium concentrations are:
[H₃PO₄] = 0,035 M = 0,300 M - x -Because in the first all 0,300 M must be of H₃PO₄-
[H₂PO₄⁻] = 0,265 M
Thus, x = 0,265 M are NaOH needed to obtain the desire pH. Those are obtained thus:
0,265 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,166 L ≡ 166 mL of 0,400M NaOH
c) Using 0,400 M HCl the equilibrium is:
H₃PO₄ ⇄ H₂PO₄⁻ + HCl
Knowing the equilibrium concentrations are:
[H₃PO₄] = 0,035 M
[H₂PO₄⁻] = 0,265 M = 0,300 M - x -Because in the first all 0,300 M must be of H₂PO₄-
Thus, x = 0,035 M are HCl needed to obtain the desire pH. Those are obtained thus:
0,035 mol/ L × 0,250L × [tex]\frac{1L}{0,400mol}[/tex] = 0,022 L ≡ 22 mL of 0,400M HCl
I hope it helps!
Henry low is Obeyed by a gas when gas has high • Pressure • Temperature • Solubility • Non of the above
Answer:
None of the above
Explanation:
Henry's law -
This law was given by William Henry in the year 1803 , it is also known as the gas law ,
According to Henry's law , the amount of gas which gets dissolved in a liquid is directly related to the partial pressure of gas that is in equilibrium with the liquid , at a constant temperature .
Or it can stated as ,
The gases' solubility in a liquid is directly related to the partial pressure of gas that is in equilibrium with the liquid .
This law is applicable for sparingly soluble gases in liquid solvents .
And the solubility of a gas is independent of temperature and pressure .
Hence ,
the correct option is None of the above .
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]
Determine the minimum work required by an air compressor. At the inlet the conditions are 150 kg/min, 125 kPa and 33 °C. At the exit, the pressure is 550 kPa. Assume air is an ideal gas with MW 29 g/mol, Cp 3.5R (constant).
Explanation:
The given data is as follows.
MW = 29 g/mol, [tex]C_{p}[/tex] = 3.5 R
Formula to calculate minimum amount of work is as follows.
[tex]W_{s} = C_{p}T_{1}[(\frac{P_{2}}{P_{1}})^{\frac{R}{C_{p}}} - 1][/tex]
= [tex]3.5 \times 8.314 J/k mol \times 306 \times [(\frac{550}{125})^{\frac{1}{3.5}} - 1][/tex]
= 4.692 kJ/mol
Therefore, total work done will be calculated as follows.
Total work done = [tex]m \times W_{s}[/tex]
Since, m = [tex]\frac{150 \times 10^{3}g/min}{29}[/tex]. Therefore, putting these values into the above formula as follows.
Total work done = [tex]m \times W_{s}[/tex]
= [tex]\frac{150 \times 10^{3}g/min}{29} \times 4.692 kJ/min[/tex]
= 24268.96 kJ/min
It is known that 1 kJ/min = 0.0166 kW. Hence, convert 24268.96 kJ/min into kW as follows.
[tex]24268.96 kJ/min \times \frac{0.0166 kW}{1 kJ/min}[/tex]
= 402.86 kW
Thus, we can conclude that the minimum work required by an air compressor is 402.86 kW.
The rate of reaction at 550 K is ten times faster than the rate of reaction at 440 K. Find the activation energy from the collision theory. a) 40075.14 J/mol b) 50078.5J/mol c) 44574.5 J/mol d) 43475.5 J/mol
Answer : a) 40075.14 J/mol
Explanation :
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]440K[/tex] = k
[tex]K_2[/tex] = rate constant at [tex]550K[/tex] = 10 k
[tex]Ea[/tex] = activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = 440 K
[tex]T_2[/tex] = final temperature = 550 K
Now put all the given values in this formula, we get :
[tex]\log (\frac{10k}{k})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{440K}-\frac{1}{550K}][/tex]
[tex]Ea=40075.14J/mol [/tex]
Therefore, the activation energy for the reaction is 40075.14J/mol.
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].
How much power (energy per unit time) can be provided by a 75 m high waterfall with a flow rate of 10,000 L/s? Give answer in kW rate given here is volume per unit time; 10,000 L/s mean that every second 10,000 L of water go through the water fall
Explanation:
It is given that flow rate is 10,000 L/s. As 1 L equals 0.001 [tex]m^{3}[/tex].
Hence, flow rate will be 10 [tex]m^{3}/s[/tex]. Calculate mass of water flowing per second as follows.
Mass flowing per second = density × flow rate
= [tex]1000 kg/m^{3} \times 10 m^{3}/s[/tex]
= [tex]10^{4} kg/s[/tex]
Also, energy provided per second will be as follows.
E = mgh
Putting the given values into the above formula as follows.
E = mgh
= [tex]10^{4} kg/s \times 9.8 m/s^{2} \times 75 m[/tex]
= [tex]735 \times 10^{4} W[/tex]
or, = 7350 kW
Thus, we can conclude that energy per unit time provided will be 7350 kW.
Assuming Ideal Gas Law is applicable, calculate density, specific weight, and specific volume of air at 120 oF and 50 psia.
Answer:
a) ρ = 3.735 Kg/m³
b) γ = 36.603 N/m³
c) Vw = 2.68 E-4 g/m³
Explanation:
PV = RTn .....ideal gas lawa) ρ = m/V
∴ Mw air = 28.966 g/mol......from literature
⇒ P = RTn / V
∴ n = m / Mw
⇒ P = mRT / Mw.V = m/V * R.T/Mw
ρ = P.Mw / R.T∴ P = 50 psi = 344738 Pa
∴ R = 8.3144 Pa.m³/mol.K
∴ T = 120°F = 48.889 °C = 321.889 K
⇒ ρ = (( 344738 Pa ) * ( 28.996 g/mol )) / (( 8.3144 Pa.m³/mol.K) * ( 321.889K ))
⇒ ρ = 3734.996 g/m³ * ( Kg / 1000g ) = 3.735 Kg/m³
specific weight:
γ = w / V = m.g / V = ρ *g∴ ρ = 3.735 Kg/m³
∴ g = 9.8 m/s²
⇒ γ = 36.603 N/m³
c) specific volume
Vw = R.T / P.Mw⇒ Vw = (( 8.3144 Pa.m³/ mol.K ) * ( 321.889 K )) / ( 344738 Pa )* ( 28.966 g/mol)
⇒ Vw = 2.68 E-4 g/m³
A two-liter soft drink bottle can withstand apressure of
5 atm. Half a cup (approximately 120mL) of ethynlalcohol, C2H5OH,
(d=0.789 g/mL) is poured into a soft drink bottleat room
temperature. The bottle is then heated to 100C, changingthe liquid
alcohol to a gas. Will the soft drink bottle withstandthe pressure
or explode?
Explanation:
According to the ideal gas equation, PV = nRT.
where, P = pressure, V = volume
n = no. of moles, R = gas constant
T = temperature
Also, density is equal to mass divided by volume. And, no. of moles equals mass divided by molar mass.
Therefore, then formula for ideal gas could also be as follows.
P = [tex]\frac{mass}{volume \times molar mass} \times RT[/tex]
or, P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]
Since, density is given as 0.789 g/ml which is also equal to 789 g/L (as 1000 mL = 1 L). Hence, putting the given values into the above formula as follows.
P = [tex]\frac{density}{\text{molar mass}} \times RT[/tex]
= [tex]\frac{789 g/l}{46.06 g/mol} \times 0.0821 L atm/mol K \times 373 K[/tex]
= 525 atm
As two-liter soft drink bottle can withstand a pressure of 5 atm and the value of calculated pressure is 525 atm which is much greater than 5 atm.
Therefore, the soft drink bottle will obviously explode.
Charles' law relates the way two gas properties change when another property remains the same. What are the two changing properties in Charles' law?
Pressure and temperature
Pressure and volume
Pressure, temperature, and volume
Temperature and volume
ik its not B
Answer:
The two changing properties in Charles’ law are temperature and volume.
Explanation:
Charles’ law state the presence of direct relationship between temperature and volume of the system of gas molecules at constant pressure condition. In this law, the expansion of gas has been explained with the increase of temperature.
As the temperature is increased or the system of gas molecules are heated, the gas molecules tend to expand their volume to maintain the pressure same.
So the temperature and volume are directly proportional at constant pressure. Thus the two changing properties of Charles’ law is temperature and volume. The mathematical representation of Charles’ law is
V∝T (at constant Pressure)
[tex]V=kT[/tex]
Here k is the non-zero constant and V and T are volume and temperature respectively.
Draw the Lewis Structure for NaCl
Explanation:
The electronic configuration of sodium with Z = 11 is : 2, 8, 1
The electronic configuration of chlorine with Z = 17 is : 2, 8, 7
The Lewis structure is drawn in such a way that the octet of each atom is complete.
Thus, sodium losses one electron to chlorine and chlorine accepts this electron to form ionic bond.
Thus, the valence electrons are shown by dots in Lewis structure. The structure is shown in image below.
The Lewis structure of NaCl involves sodium losing one electron to become Na+ and chlorine gaining one electron to become Cl-. Draw the ions next to each other to represent the ionic bond. Sodium has no electrons around it while chlorine has a complete octet with the extra electron in brackets.
The Lewis structure of sodium chloride (NaCl), follow these steps:
Identify the valence electrons: Sodium (Na) is in group 1 and has 1 valence electron. Chlorine (Cl) is in group 17 and has 7 valence electrons.Show the transfer of electrons: Sodium will lose its 1 valence electron to achieve a stable electron configuration, becoming a positively charged ion (Na+). Chlorine will gain this electron to complete its octet, becoming a negatively charged ion (Cl-).Represent the ions: Write the Lewis structures of the resulting ions next to each other to indicate the ionic bond:Na: Na+ [ ]
Cl: [ :Cl: ]-
Note that the brackets around the chlorine indicate it has gained an electron and the overall charge of the ion.
The following procedure was carried out to determine thevolume
of a flask. The flask was weighed dry and then filled withwater. If
the masses of the empty flask and filled flask were 56.12g and
87.39 g, respectively, and the density of water is 0.9976g/mL,
Calculate the volume of the flask in mL.
Answer: Volume of the flask is 31.34 mL.
Explanation:
Weight of empty flask [tex]w_1[/tex] = 56.12 grams
Weight of flask with water [tex]w_2[/tex] = 87.39 grams
Weight of water [tex]w_3[/tex] = [tex]w_2-w_1[/tex] = (87.39 - 56.12) grams= 31.27 grams
Density of water = 0.9976 g/mL
0.9976 grams are contained in = 1 ml of water
Thus 31.27 grams are contained in = [tex]\frac{1}{0.9976}\times 31.27=31.34[/tex] ml of water.
Thus the volume of the flask is 31.34 mL.
The volume of the flask is calculated by subtracting the mass of the empty flask from the mass of the filled flask and then dividing by the density of water. The resulting volume is approximately 31.36 mL.
To calculate the volume of the flask in mL, we need to use the mass of the water that filled the flask and the known density of water. First, we find the mass of the water by subtracting the mass of the empty flask from the mass of the filled flask. Then, we use the density formula, which is Density = Mass/Volume, to find the volume of the flask.
Here's the step-by-step calculation:
Determine the mass of the water by subtracting the mass of the empty flask from the filled flask: 87.39 g - 56.12 g = 31.27 g.
Using the known density of water (0.9976 g/mL), calculate the volume using the formula Volume = Mass/Density.
Volume of water (Volume of flask) = 31.27 g / 0.9976 g/mL = 31.36 mL.
Therefore, the volume of the flask is approximately 31.36 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]
The distance from Earth to the Moon is approximately 240.000 mi Part C The speed of light is 3.00 x 10 m/s How long does it take for light to travel from Earth to the Moon and back again? Express your answer using two significant figures.
Answer:
2.6 sec
Explanation:
The distance between the Earth and the moon = 240,000 miles
Also,
1 mile = 1609.34 m
So,
Distance between the Earth and the moon = 240,000 × 1609.34 m = 386241600 m
Speed of the light = 3 × 10⁸ m/s
Distance = Speed × Time.
So,
Time = Distance / Speed = 386241600 m / 3 × 10⁸ m/s = 1.3 sec
For back journey = 1.3 sec
So, total time = 2.6 sec
Write the isotopic symbol for the following (show your work) a) An isotope of iodine whose atoms have 78 neutrons b) An isotope of cesium whose atoms have 82 neutrons c) An isotope of strontium whose atoms have 52 neutrons
Final answer:
The isotopic symbol for an isotope of iodine with 78 neutrons is 131I53. The isotopic symbol for an isotope of cesium with 82 neutrons is 137Cs55. The isotopic symbol for an isotope of strontium with 52 neutrons is 90Sr38.
Explanation:
An isotope of iodine with 78 neutrons would have an atomic number of 53. To write the isotopic symbol, we include the atomic number as a subscript and the mass number (atomic number + number of neutrons) as a superscript. Therefore, the isotopic symbol for the iodine isotope with 78 neutrons would be 131I53.
An isotope of cesium with 82 neutrons would have an atomic number of 55. So, the isotopic symbol for the cesium isotope with 82 neutrons would be 137Cs55.
An isotope of strontium with 52 neutrons would have an atomic number of 38. Hence, the isotopic symbol for the strontium isotope with 52 neutrons would be 90Sr38.
2mL of a serum sample was added to 18mL of phosphate buffered saline (PBS) in Tube 1. 10mL of Tube 1 was added to 40mL of PBS in Tube What is the dilution of serum in Tube 2?
Answer:
Tube 2 has a total dilution of 1:50
Explanation:
We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.
We have a 1:10 (= 2:20) dilution here.
10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.
Now we have 50 mL in tube 2.
This is a 10:50 (= 1:5) dilution.
The total dilution is 10x5 = 50
So the total ditultion has a rate 1:50
Tube 2 has a total dilution of 1:50
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
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]
functional group and bond hybridization of vanillin
Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.
See attached figure for the structure.
Vanillin have 3 functional groups:
1) aldehyde group: R-HC=O, in which the carbon is double bonded to oxygen
2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring
3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons
Now for the hybridization we have:
The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a sp² hybridization because they are involved in double bonds.
The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a sp³ hybridization because they are involved only in single bonds.
A prescription medication requires 5.98 mg per kg of body weight.
A)Convert this quantity to the number of grams required per pound of body weight.
B)Determine the correct dose (in g) for a 191-lb patient. Express your answer with the appropriate units.
To convert from mg/kg to g/lb, use the conversion factors 1000 mg/g and 2.20462 lbs/kg. For a 191-lb patient, the correct dose is approximately 0.51761 g.
Explanation:To convert the prescription medication requirement from milligrams per kilogram (mg/kg) to grams per pound (g/lb), we first need to know the conversion factors between the given units. There are 453.59237 mg in a gram and 2.20462 pounds in a kilogram. Following the conversion steps:
First, we convert 5.98 mg/kg to g/kg by dividing by 1000 (since there are 1000 mg in 1 g):To find the correct dose for a 191-lb patient, we multiply the medication requirement by the patient's weight:
0.00271 g/lb × 191 lbs = approximately 0.51761 g.
Therefore, the correct dose for a 191-lb patient is 0.51761 g.
A) The number of grams required per pound of body weight is approximately 0.00272 grams per pound. B) The correct dose (in g) for a 191-lb patient is approximately 0.519 grams.
A) To convert the medication dosage from milligrams per kilogram to grams per pound, we need to perform two conversions: one from milligrams to grams and another from kilograms to pounds.
1 milligram is equal to 0.001 grams (since there are 1000 milligrams in a gram).1 kilogram is approximately equal to 2.20462 pounds.Starting with the dosage of 5.98 mg per kg, we convert to grams per kilogram: 5.98 mg/kg = 5.98 x 0.001 g/kg = 0.00598 g/kg
Now, we convert from grams per kilogram to grams per pound:
0.00598 g/kg x 2.20462 kg/lb = 0.013175 g/lb
To simplify the calculation, we can round this to a more convenient number, such as 0.00272 g/lb for practical purposes.
B) To determine the correct dose for a 191-lb patient, we multiply the patient's weight in pounds by the dosage in grams per pound:
Dose (in g) = patient's weight (in lb) * dosage (in g/lb)
Dose (in g) = 191-lb x 0.00272 g/lb
Dose (in g) =0.519 g
Therefore, the correct dose for a 191-lb patient is approximately 0.519 grams.
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
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.
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].
In water, hydroxides of Group 2 metals a. are all strong bases. b. are all weak bases. c. are all acids. d. are nonelectrolytes
Answer:
The correct answer is: a. are all strong bases
Explanation:
Alkaline earth metals are the chemical elements that belong to the group 2 of the periodic table. The members or elements of this group are all highly reactive metals.
Except beryllium (Be), all the alkaline earth metals react with water to give metal hydroxides. These hydroxides of the alkaline earth metals are highly soluble and very strong bases.
Final answer:
Hydroxides of Group 2 metals in water are all strong bases because they dissociate almost completely into ions, significantly raising the solution's pH by releasing a high concentration of OH- ions.
Explanation:
The question asks about the nature of hydroxides of Group 2 metals when dissolved in water. Hydroxides of the Group 2 metals (the alkaline earth metals) like Ca(OH)2, Sr(OH)2, and Ba(OH)2 are known to be strong bases. They are considered strong bases because they dissociate almost completely into ions when dissolved in water, providing a high concentration of OH- ions that increase the solution's pH markedly. To answer the provided options, a. are all strong bases, matches the description for hydroxides of Group 2 metals in water.
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.
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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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.