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.
Consider water at 1400 kPa and 200 C. What is the specific volume (in m3 /kg)?
Answer:
Vw = 2.80907 E-3 m³/Kg
Explanation:
specific volume is the inverse of density:
⇒ Vw = 1 / ρ
∴ water at:
⇒ P = 1400 KPa * ( 1000 Pa / KPa ) = 1400000 Pa
⇒ T = 200°C = 473 K
Water at these conditions is found as saturated steam, specific volume would be:
⇒ Vw = R*T / P
∴ R = 8.3144 Pa.m³/ Kg.K
⇒ Vw = (( 8.3144 ) * ( 473)) / 1400000
⇒ Vw = 2.80907 E-3 m³/Kg
1826.5g of methanol (CH3OH), molar mass = 32.0 g/mol is added to 735 g of water, what is the molality of the methane 0.0348 m 1.13m 2.03 m 3.61 m 36.1 m Navigator F10 Delete Backspace
Answer:
Molality = 1.13 m
Explanation:
Molality is defined as the moles of the solute present in 1 kilogram of the solvent.
Given that:
Mass of [tex]CH_3OH[/tex] = 26.5 g
Molar mass of [tex]CH_3OH[/tex] = 32.04 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{26.5\ g}{32.04\ g/mol}[/tex]
[tex]Moles\ of\ CH_3OH= 0.8271\ moles[/tex]
Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )
So, molality is:
[tex]m=\frac {0.8271\ moles}{0.735\ kg}[/tex]
Molality = 1.13 m
Friction factor for fluid flow in pipe does not depend upon the A. pipe length. B. pipe roughness. C. fluid density & viscosity. D. mass flow rate of fluid.
Answer:
C. fluid density & viscosity
Explanation:
In 1850, Darcy-Weisbach experimentally deduced an equation to calculate shear losses ("friction"), in a tube with permanent flow and constant diameter:
hf = (f x L x V^2) / (D x 2g)
where:
hf: shear losses
f: shear loss factor (pipe roughness)
g: gravity acceleration
D: tube diameter
L: tube length
V: fluid average speed in the tube
To calculate the loss factor “f” in the Poiseuille laminar region, he proposed in 1846 the following equation:
f = 64 / Re
Where:
Re: Reynolds number
The influence of the parameters on f is quantitatively different according to the characteristics of the current.
In any straight pipeline that transports a liquid at a certain temperature, there is a critical speed below which the regimen is laminar. This critical value that marks the transition between the two regimes, laminar and turbulent, corresponds to a Re = 2300, although in practice, between 2000 and 4000 the situation is quite inaccurate. Thus:
Re <2000: laminar regimen
2000 <Re <4000: critical or transition zone
Re> 4000: turbulent regime
Final answer:
The friction factor for fluid flow in a pipe does not depend upon the pipe length, pipe roughness, fluid density & viscosity, or mass flow rate of fluid.
Explanation:
The friction factor for fluid flow in a pipe does not depend upon the pipe length (A), pipe roughness (B), fluid density & viscosity (C), or mass flow rate of fluid (D).
This is because the friction factor, also known as the Darcy-Weisbach factor, is determined by the characteristics of the flow itself, such as the Reynolds number, which is a dimensionless quantity that relates the inertia of the fluid to the viscous forces acting on it.
The friction factor can be calculated using the Colebrook-White equation or obtained from Moody's diagram based on the relative roughness of the pipe and the Reynolds number.
Draw the Lewis Structure for NI3.
Explanation:
Nitrogen triiodide (NI₃)
Valence electrons of nitrogen = 5
Valence electrons of iodine = 7
The total number of the valence electrons = 5 + 3(7) = 26
The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,
The Lewis structure is shown in image below.
Nitrogen posses one lone pair and thus the geometry is pyramidal.
The Lewis structure of NI3 is drawn by counting the valence electrons, creating single bonds between nitrogen and iodine, completing octets for iodine atoms, and placing remaining electrons on the nitrogen to complete its octet. This results in a structure where nitrogen is single-bonded to three iodine atoms, each surrounded by three lone pairs, and nitrogen has one lone pair.
Explanation:To draw the Lewis structure for NI3 (nitrogen triiodide), follow these steps:
Count the total number of valence electrons. Nitrogen has 5 valence electrons, and each iodine has 7 valence electrons, totaling (5 + 3*7) = 26 valence electrons.
Draw a single bond between the nitrogen atom and each iodine atom. This will use up 6 of the valence electrons (2 for each bond).
Complete the octets for the iodine atoms by adding six more electrons to each iodine in the form of electron pairs, using up 18 of the remaining valence electrons.
Place any remaining electrons (2) on the nitrogen atom to complete its octet.
Examine the structure. Each iodine has 8 electrons, and nitrogen has 8 electrons, making the structure complete.
If you follow these steps, the resulting Lewis structure for NI3 will show a central nitrogen atom single-bonded to three iodine atoms, with each iodine atom surrounded by three lone pairs of electrons, and one lone pair of electrons on the nitrogen atom.
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Attem A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 91.0 mL. The liquid solvent has a mass of 26.6 g and a density of 0.865 g/mL. Determine the mass of the solid given its density is 1.75 g/mL. mass: 38.852
The mass of the insoluble solid is determined by first calculating the volume of the liquid solvent, then using that information to find the volume of the solid, and finally multiplying the volume of the solid by its density.
Explanation:To determine the mass of the solid, we first calculate the volume of the liquid solvent. The volume of a substance can be calculated using the formula volume = mass/density. Thus, the volume of the liquid solvent is 26.6 g / 0.865 g/mL = approximately 30.8 mL.
Now, knowing that the total volume of the solid and liquid together is 91.0 mL, we can find out the volume of the solid as follows: 91.0 mL (total volume) - 30.8 mL (volume of the liquid) = 60.2 mL.
We can then use the given density of the solid (1.75 g/mL) to calculate its mass. According to the formula mass = volume x density, the mass of the solid is therefore 60.2 mL x 1.75 g/mL = 105.35g.
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All final to the right of the decimal place are significant
Water (25◦C) flows through 1-inch Schedule 40 steel pipe at 2.0 gpm. What is the Reynolds number of the flow? What is the friction factor? Is the flow laminar or turbulent?
Answer:
Re=8561.79
Friction factor is 0.014
Flow is turbulent flow.
Explanation:
Given that
Diameter ,d=1 in
d=0.0254 m (1 in =0.0254 m )
Volume flow rate,Q = 2 gpm
We know that
[tex]1\ gpm=7.5\times 10^{-5}\ m^3/s[/tex]
[tex]2\ gpm=2\times 7.5\times 10^{-5}\ m^3/s[/tex]
[tex]2\ gpm=15\times 10^{-5}\ m^3/s[/tex]
We know that
Q= A x V
[tex]A=\dfrac{\pi}{4}\times 0.0254^2\ m^2[/tex]
[tex]A=0.00050\ m^2[/tex]
So
[tex]V=\dfrac{15\times 10^{-5}}{0.00050}[/tex] m/s
V=0.3 m/s
So Reynolds number(Re)
[tex]Re=\dfrac{\rho VD}{\mu }[/tex]
Properties of water at 25 C
[tex]\mu=8.9\times 10^{-4}\ Pa.s[/tex]
[tex]Re=\dfrac{1000\times 0.0254\times 0.3}{8.9\times 10^{-4}}[/tex]
Re=8561.79
Re>4000 ,It means that flow is turbulent flow.
Friction factor
If we assume that pipe is smooth
[tex]f=\dfrac{0.136}{Re^{0.25}}[/tex]
[tex]f=\dfrac{0.136}{8561.79^{0.25}}[/tex]
f=0.014.
Friction factor is 0.014
To identify whether the flow is laminar or turbulent and to determine the friction factor, you should first find the Reynolds number using the given parameters, such as the flow rate of 2.0 gpm and then refer to a Moody chart or similar source.
Explanation:The Reynolds number (NR) and friction factor of the flow can be determined using the given parameters (Flow rate, pipe diameter, and water temperature). The Reynolds number is an indicator that can reveal whether flow is laminar or turbulent. For flow in a tube of uniform diameter, the Reynolds number is defined as an equation related to the properties of the fluid and the characteristics of the flow.
In this case, if the Reynolds number (NR) is below 2000, the flow is considered laminar. If NR is above 3000, the flow becomes turbulent. For values of NR between 2000 and 3000, it may be either or both, depending on factors such as the roughness of the pipe's internal surface and the flow velocity.
The friction factor is a measure of the total resistance created by the force on a fluid as it moves through a pipe. The calculation of this factor also considers the Reynolds number and pipe roughness.
To determine whether your given conditions (2.0 gpm water flow through a 1-inch Schedule 40 steel pipe at 25◦C) will result in laminar or turbulent flow and the corresponding friction factor, please consider calculating the Reynolds number first using the suitable formula and the given conditions and then consulting a Moody chart or a similar source to find the corresponding friction factor.
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how is the use of magnetic fields to plasma related to trying to generate energy using nuclear fusion?
Answer:
See explanation
Explanation:
Fusion is the process of fusing two isotopes of Hydrogen namely Tritium and Deuterium to produce Helium. To achieve this, tremendous heat required (about a million degrees Celsius). The same for the pressure. To achieve this, the hydrogen is made into a plasma through rarefaction so it becomes susceptible to magnetic fields. Magnetic confinement fusion is an approach to generating thermonuclear fusion power that uses magnetic fields to confine the hot fusion fuel in the form of a plasma. Electro-Magnets surround the chamber/reactor and are pulsed adiabatically (as in a bicycle pump) and the gas becomes extremely hot that may melt the surroundings.
As the ions in the plasma are charged (the plasma is so hot all the negatively-charged electrons are stripped off the atoms, leaving them with a positive charge) they respond to magnetic fields. Extra fields help shape the plasma and hold it stable.
Express your answer using two significant figures.
2.7 cm3 = m3
2.0 mm3= m3
To convert cm³ or mm³ to m³, one must divide by 1,000,000 or 1,000,000,000 respectively. Thus, 2.7 cm³ is equal to 2.7 x 10^-6 m³ and 2.0 mm³ is equal to 2.0 x 10^-9 m³.
Explanation:To convert cubic centimeters (cm³) and cubic millimeters (mm³) to cubic meters (m³), you need to know the unit conversions. One square meter is equal to 1,000,000 cubic centimeters and 1,000,000,000 cubic millimeters.
This means you can convert 2.7 cm³ to meters by dividing by 1,000,000, yielding an answer of 0.0000027 m³ (to two significant figures, or 2.7 x 10^-6 m³).
Similarly, 2.0 mm³ can be converted to meters by dividing by 1,000,000,000, resulting in an answer of 0.000000002 m³ or 2.0 x 10^-9 m³.
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The conversion from cm3 and mm3 to m3 is done by multiplying the original number by 1e-6 for cm3 and 1e-9 for mm3. The results for the examples given are 2.7 cm3 equals 2.7e-6 m3, and 2.0 mm3 equals 2.0e-9 m3.
Explanation:The task is to convert measurements from one unit (cubic centimeters or cubic millimeters) to another (cubic meters). It's important to understand the relevant conversion factors:
1 cm3 = 1e-6 m3
1 mm3 = 1e-9 m3
Applying these to your examples we get:
2.7 cm3 = 2.7 * 1e-6 m3 = 2.7e-6 m3
2.0 mm3 = 2.0 * 1e-9 m3 = 2.0e-9 m3
So these are the conversions using two significant figures.
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Convert 6.23 x 10^-3 m to the equivalent length in nanometers. 6.23 x 10^-3 m =
Answer:
Try to remember that 1 nanometer is 1 x 10^9, so calculate 6.23 x 10^-3 m x 1 x 10^9. Your answer is 6.23 x 10^-3 m = 6.23 x 10^6 nm
Explanation:
Answer:
6.23 x 10^-3 m = 6.23 x 10^6 nm
Explanation:
Assume that the complete combustion of one mole of glucose to carbon dioxide and water liberates 2870 kJ/mol2870 kJ/mol ( Δ????°′=−2870 kJ/molΔG°′=−2870 kJ/mol ). If one contraction cycle in muscle requires 67 kJ67 kJ , and the energy from the combustion of glucose is converted with an efficiency of 39%39% to contraction, how many contraction cycles could theoretically be fueled by the complete combustion of one mole of glucose? Round your answer to the nearest whole number.
Answer:
Number of contraction cycles that could theoretically be fueled by the complete combustion of one mole of glucose is around 17
Explanation:
Energy released during the complete combustion of 1 mole glucose = 2870 kJ
Energy required/muscle contraction cycle = 67 kJ/contraction
Energy conversion efficiency = 39%
Actual amount of energy converted to contraction per mole of glucose is:
[tex]=\frac{39}{100} *2870 kJ=1119.3 kJ[/tex]
Total contraction cycles fueled by the above energy is:
[tex]=\frac{1119.3\ kJ}{67\ kJ/contraction} =16.7\ i.e.\ around\ 17\ contractions[/tex]
A solid sample (Sample 1) is analyzed and found to contain 1.47 g carbon and 0.123 g hydrogen. A second sample (Sample 2) is expected to be composed of the same pure compound. If Sample 2 is found to contain 2.17 g hydrogen, how much carbon is expected in the sample ?
Explanation:
Here it is given that carbon is sample 2 = 25.9 g
For sample 1, mass carbon = 1.47 gNo. of moles of carbon will be calculated as follows.
No. of moles of carbon = [tex]\frac{\text{mass carbon}}{\text{molar mass carbon}}[/tex]
= [tex]\frac{1.47 g}{12.01 g/mol}[/tex]
= 0.1224 mol
It is also given that mass of hydrogen = 0.123 g
Hence, calculate number of moles of hydrogen as follows.
No. of moles of hydrogen = [tex]\frac{\text{mass hydrogen}}{\text{molar mass hydrogen}}[/tex]
= [tex]\frac{0.123 g}{1.008 g/mol}[/tex]
= 0.122 mol
Therefore, [tex]\frac{\text{moles of carbon}}{\text{moles of hydrogen}}[/tex]
= [tex]\frac{0.1224 mol}{0.122 mol}[/tex]
= 1.003
For sample 2, mass of hydrogen = 2.17 gTherefore, calculate the number of moles of hydrogen as follows.
No. of moles of hydrogen = [tex]\frac{\text{mass hydrogen}}{\text{molar mass hydrogen}}[/tex]
= [tex]\frac{2.17 g}{1.008 g/mol}[/tex]
= 2.1528 mol
Hence, calculate the moles of carbon as follows.
Moles of carbon = [tex]\text{moles hydrogen} \times \frac{\text{moles of carbon}}{\text{moles hydrogen}} [/tex]
= [tex]2.1528 mol \times 1.003[/tex]
= 2.16 mol
Mass of carbon = moles carbon × molar mass carbon
= (2.16 mol) × (12.01 g/mol)
= 25.9 g
Thus, we can conclude that 25.9 g of carbon is expected in the sample.
Final answer:
The expected amount of carbon in Sample 2 would be 25.93 grams, calculated using the constant carbon-to-hydrogen mass ratio determined from Sample 1.
Explanation:
The question requires determining how much carbon is expected in Sample 2, given the known amounts of carbon and hydrogen in Sample 1 and the amount of hydrogen in Sample 2. We assume that the compound is the same in both samples, meaning the ratio of carbon to hydrogen must be constant. First, we find the ratio of carbon to hydrogen in Sample 1:
Sample 1: 1.47 g C / 0.123 g H = 11.95 g C/g H
Using this ratio, we calculate the expected amount of carbon in Sample 2:
Expected carbon in Sample 2: 2.17 g H x 11.95 g C/g H = 25.93 g C
Therefore, we would expect Sample 2 to contain 25.93 grams of carbon, assuming it is composed of the same pure compound as Sample 1.
Use the given data at 500 K to calculate ΔG°for the reaction
2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g)
Substance H2S(g) O2(g) H2O(g) SO2(g)
ΔH°f(kJ/mol) -21 0 -242 -296.8
S°(J/K·mol) 206 205 189 248
Answer : The value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,
[tex]2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)[/tex]
First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].
[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]
[tex]\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}][/tex]
where,
[tex]\Delta H^o[/tex] = enthalpy of reaction = ?
n = number of moles
[tex]\Delta H_f^0[/tex] = standard enthalpy of formation
Now put all the given values in this expression, we get:
[tex]\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)][/tex]
[tex]\Delta H^o=-1035.6kJ=-1035600J[/tex]
conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].
[tex]\Delta S^o=S_f_{product}-S_f_{reactant}[/tex]
[tex]\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}][/tex]
where,
[tex]\Delta S^o[/tex] = entropy of reaction = ?
n = number of moles
[tex]\Delta S_f^0[/tex] = standard entropy of formation
Now put all the given values in this expression, we get:
[tex]\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)][/tex]
[tex]\Delta S^o=-153J/K[/tex]
Now we have to calculate the Gibbs free energy of reaction [tex](\Delta G^o)[/tex].
As we know that,
[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]
At room temperature, the temperature is 500 K.
[tex]\Delta G^o=(-1035600J)-(500K\times -153J/K)[/tex]
[tex]\Delta G^o=-959100J=-959.1kJ[/tex]
Therefore, the value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ
Magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Write a balanced chemical equation for the reaction. If 0.510 g of magnesium reacts with 0.500 L of hydrochloric acid, determine the molarity and pH of the hydrochloric acid solution. Determine the volume of hydrogen gas that will be produced if the reaction takes place at 25 °C and 1.00 atm.
Answer:
The balanced chemical equation is Mg + 2HCl ⇒ MgCl2 + H2
The molarity of the hydrochloric acid solution is HCl 0.04 M and the pH = 1.4.
The volume of hydrogen gas produced by the reaction of 0.510 g of Mg will be 0.482 L.
Explanation:
First, for the balanced equation you have to consider the oxidation state of the elements to find subscripts. Then you can find the correct coeficients. Mg= +2, Cl = -1.
Mg + HCL ⇒ MgCl2 + H2
For the molarity of the solution you have to notice tha if 0.510 grams of Mg reacts with 0,5 L of hydroclhoric acid, and from the previous equation 1 mol of Mg reacts with 2 mol HCl.
The atomic mass of Mg = 24.31 grs/mol
24.31 grs------------ 1 mol Mg
0.510 grs------------ x=0.02 mol Mg.
If 1 mol of Mg reacts with 2 mol HCl, then 0.02 mol of Mg will react with
0.04 mol HCl. So, the molarity of the solution is 0.04 M HCl.
Then to calculate the pH we use the formula pH = - log [H+]
⇒ pH = -log [0.04]⇒ pH=1.4.
Finally, from the balanced equation and the findings described, and considering that at 25°C and 1.00 atm 1 mol of gas has volume of 24.1 L.
1 mol H2----------- 24.1 L
0.02 mol H2----- x= 0.482L.
A 2.00 m2 tank of a fluid that has a density of 1000 kg/m is draining at a rate of S liters/min. Determine the time for the tank to completely empty out. a.400 seconds b. 10 minutes У С. 400 minutes d. Not enough information
Answer:
the "d" option
Explanation:
Cómo se pronuncia
To calculate the flow I need the volume and time to comply with the formula:
S = V / t
V = volume
t = time
In this problem I have volume and density as data.
There is no way to calculate the time so the correct answer is the "d" option
Modern commercial airliners are largely made of aluminum, a light and strong metal. But the fact that aluminum is cheap enough that airplanes can be made out of it is a bit of historical luck. Before the discovery of the Hall-Héroult process in , aluminum was as rare and expensive as gold. What would happen if airplanes had to be made of steel?
Answer:
Explanation:
The metallic properties of steel would strongly not favor its use in construction of aircraft.
Steel is an alloy of carbon and iron. It is denser and generally weighs more than equivalent amount of aluminum. This would imply that the weight of aircrafts would be much more heavier than usual. Weight is a very key component in construction of aircraft. The lighter the mass, the faster propellers can move the craft and lesser amount of energy used. The weight would be a major problem.
Although alloying iron and carbon improves the resistant of the steel to corrosion, it still cannot be compared to that of aluminium. We would build airplanes that would not be durable for so long and that can readily rust on frequent usage.
If airplanes were made of steel, they would be heavier, requiring more fuel and more powerful engines, reducing efficiency and payload capacity. Aluminum's lightness, toughness, and resistance to corrosion are pivotal for modern airplane performance and cannot be matched by steel without compromising efficiency and increasing operating costs.
If airplanes had to be made of steel instead of aluminum, there would be significant consequences for the aviation industry. Steel is much heavier than aluminum, which means that airplanes would be considerably heavier as well. This increase in weight would lead to a need for more powerful engines to provide the necessary lift, which could result in higher fuel consumption and lower efficiency. Moreover, the added weight would decrease the payload capacity of the airplane, as the plane's own structure would constitute a larger portion of its maximum takeoff weight.
The major advantages of aluminum in airplane construction include its lightness, toughness, and resistance to corrosion. These properties are crucial for ensuring that airplanes are able to carry substantial payloads, achieve adequate fuel efficiency, and maintain structural integrity over time. The use of aluminum, a metal that meets these critical needs, is therefore one of the key reasons why modern airplanes are able to achieve the performance levels that they do today. Utilizing steel, despite its strength, would compromise these benefits and could alter the economics and environmental impact of air travel.
Write 41,405,000 in Engineering Notation with 3 significant figures.
Answer: [tex]41.4\times 10^{6}[/tex]
Explanation:
Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s preceding the first integers are never significant.
All zero’s after the decimal point are always significant.
Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power. It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.
The engineering notation written in the form:
[tex]a\times 10^b[/tex]
where,
a = the number which is greater than 0 and less than 999
b = an integer multiple of 3
Now converting the given value of 41,405,000 into engineering notation, we get [tex]41.4\times 10^{6}[/tex]
Therefore the scientific notation with 3 significant figures is [tex]41.4\times 10^{6}[/tex]
Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) O -7.3J/s O 7.3 kJ/s O -210 kJ/s O 3451 kJ/s
Explanation:
For an isothermal process equation will be as follows.
W = nRT ln[tex]\frac{P_{1}}{P_{2}}[/tex]
It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{10000 g/s}{18 g/mol}[/tex]
= 555.55 mol/s
= 556 mol/s (approx)
As T = [tex]50^{o}C[/tex] or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.
W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]
= [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}[/tex]
= [tex]556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303[/tex]
= -3440193.809 J/s
Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.
[tex]3440193.809 J/s \times \frac{0.001 kJ}{1 J}[/tex]
= 3440.193 kJ/s
= 3451 kJ/s (approx)
Thus, we can conclude that the pump work is 3451 kJ/s.
1 mol of an Ideal Gas expands through a turbine adiabatically to produce work. Assuming steady flow conditions, how much work is lost due to the entropy generated, if the surroundings are at 300 K? Give the value of lost work to the nearest J. Inlet conditions: 494.8 K and 6.3 bar Outlet conditions: 383.5 K and 1.7 bar Cp = (9/2)R where R = 8.314 J/mol-K.
Answer:
W = 1.8 KJ
Explanation:
turbine adiabatically: Q = 0
∴ Lost work:
W = To [ ( S2 - S1 ) + ΔSo ]∴ To = 300 K
ideal gas:
S2 - S1 = Cp Ln (T2/T1) - R Ln (P2/P1)⇒ S2 - S1 = (9/2)*(8.314) Ln (383.5/494.8) - (8.314) Ln (1.7/6.3)
⇒ S2 - S1 = 1.36 J/mol.K
entropy generated in the sourroundings:
ΔSo = Q/To∴ Q = ΔU = Cv*ΔT
∴ Cv = 3/2*R = 12.5 J/mol.K.....ideal gas
∴ ΔT = 494.8 - 383.5 = 111.3 K
⇒ Q = 12.5 * 111.3 = 1391.25 J/mol
⇒ ΔSo = 1391.25/300 = 4.638 J/mol.K
⇒ W = ( 300 K ) * [ 1.36 J/mol.K + 4.638 J/mol.K ]
⇒ W = 1799.25 J/mol * 1 mol * ( KJ/1000J )
⇒ W = 1.799 KJ ≅ 1.8 KJ
Chromium has an atomic mass of 51.9961 u and consists of four isotopes, Cr50, Cr52, Cr53, and Cr54. The Cr52 isotope has a natural abundance of 83.79% and an atomic mass of 51.9405 u. The Cr54 isotope has a natural abundance of 2.37% and an atomic mass of 53.9389 u. The natural abundances of the Cr50 and Cr53 isotopes exist in a ratio of 1:0.4579, and the Cr50 isotope has an atomic mass of 49.9460 u. Determine the atomic mass of the Cr53 isotope.
The atomic mass of Cr53 can be calculated by establishing an equation that takes into account the atomic masses and abundances of all isotopes of Chromium. Solve the equation for 'x' (representing the atomic mass of Cr53) to find your answer.
Explanation:The atomic mass of a given isotope of an element is determined by the weighted average of the masses of its isotopes, each multiplied by the abundance of that isotope. Since the atomic masses and abundances of Cr50, Cr52, and Cr54 are already provided, we only need to account for Cr53's mass. Using Chromium's atomic mass (51.9961 u), we can establish an equation to solve for Cr53's atomic mass.
Here's our equation:
51.9961 u = (49.9460 u * Cr50's abundance) + (51.9405 u * 0.8379) + (x * Cr53's abundance) + (53.9389 u * 0.0237)
First calculate the contribution of Cr50, Cr52, and Cr54 to the atomic weight, then subtract this from the total atomic weight (51.9961 u). Divide this value by the Cr53's abundance to get the atomic weight of Cr53.
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The atomic mass of the Cr53 isotope is approximately 23.83 u, calculated based on its natural abundance and atomic mass in the given chromium isotope mixture.
To determine the atomic mass of the Cr53 isotope, we can use the information provided about the natural abundances and atomic masses of the chromium isotopes.
Let's denote the natural abundance of Cr50 as x. Since the ratio of Cr50 to Cr53 is 1:0.4579, the natural abundance of Cr53 would be 0.4579x. The total natural abundance of Cr50 and Cr53 is 1, so we have the equation:
x + 0.4579x = 1
Solving for x, we find that x is approximately 0.6852.
Now, we can calculate the total contribution of Cr53 to the atomic mass:
Atomic mass of Cr53 = Natural abundance of Cr53 * Atomic mass of Cr53
Atomic mass of Cr53 = 0.4579 * 51.9961
Calculating this gives us the atomic mass of the Cr53 isotope.
Atomic mass of Cr53 = 23.8256 u
In summary, the atomic mass of the Cr53 isotope is approximately 23.83 u.
Consider the formation of nitrogen dioxide from nitric
oxideand oxygen.
2NO(g) + O2(g) -- 2NO2(g)
If 9L of NO are reacted with excess 02 at STP, what is
thevolume in liters of the NO2 produced?
Answer:
9 L
Explanation:
According to the question , the given reaction is -
2NO(g) + O₂(g)------->2NO₂(g)
Since ,
At STP ,
One mole of a gas occupies the volume of 22.4 L.
Hence , as given in the question -
9 L of NO , i.e .
22.4 L = 1 mol
1 L = 1 / 22.4 mol
9 L = 1 / 22.4 * 9 L = 0.40 mol
From the chemical reaction ,
The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .
Hence ,
2 moles of NO will produce 2 moles of NO₂.
Therefore ,
0.40 mol of NO will produce 0.40 mol of NO₂.
Hence , the volume of NO₂ can be calculated as -
1 mol = 22.4 L
0.40 mol = 0.40 * 22.4 L = 9 L
To find the volume of NO2 produced, calculate the number of moles of NO reacted using the ideal gas equation. Convert moles to volume using the molar volume of gases at STP.
Explanation:To find the volume of NO2 produced, we first need to determine the number of moles of NO reacted. Using the ideal gas equation, we can calculate the number of moles:
n = PV/RT = (1 atm)(9 L)/(0.0821 L atm/(mol K))(273 K) = 0.408 mol
Since the reaction has a 2:2 mole ratio, 0.408 mol of NO will produce 0.408 mol of NO2. To convert moles to volume at STP, we can use the molar volume of gases, which is 22.4 L/mol:
Volume of NO2 = 0.408 mol * 22.4 L/mol = 9.1392 L
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Digoxin injection is supplied in ampules of 500 mcg per 2 mL. How many milliliters must a nurse administer to provide a dose of 0.2 mg? img20.img 0.5 mg-2 Sooring= 0r5mg 0 12mg = x 0.001X500mcg = 0.5 mg 0,2 mg x 2mL = 0.8mL 0.5 m
Answer:
0.8 mL.
Explanation:
You need to know that the equivalence in mcg to mg is 1000 mcg are 1 mg; so 500 mcg are 0.5 mg. Now you know that there are 0.5 mg per 2 mL, so if you divide all by 2, you will know that you have 0.25 mg per mL. Now you applied a rule of three: ((0.2 mg)(1 ml))/(.25 mg) = 0.8 mL. So the nurse need to administer 0.8 mL to provide 0.2 mg of dose.
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
reactor. What is the limiting reactant?
Answer:
N₂ is the limiting reactant
Explanation:
The balanced reaction between N₂ gas and H₂ gas is:
N₂ + 3H₂ → 2NH₃
In order to determine the limiting reactant, we have to calculate the number of moles of each rectant, using their molecular weight:
Moles of N₂= 100 kg * [tex]\frac{1kmol}{28kg}[/tex] = 3.57 kmolMoles of H₂= 100 kg * [tex]\frac{1kmol}{2kg}[/tex] = 50.0 kmolLastly, we multiply the number of moles of N₂ by 3, and the number of moles of H₂ by 1; due to the coefficients in the balanced reaction. Whichever number is lower, belongs to the limiting reactant.
N₂ => 10.7
H₂ => 50.0
Thus N₂ is the limiting reactant
Calculate the volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute. Enter your answer in the provided box. IL
Answer:
The volume of a 0.200 M KCl solution containing 5.00 10-2 mol of solute is 0,25 L
Explanation:
Molarity (M) means: moles of solute which are contained in 1 L of solution.
In this case we have 0,2 moles which are in 1 L, so, as we have 5x10*-2 moles we have to apply a rule of three to find out the volume.
0,2 moles ........... 1 L
5x10*-2 moles ........... x
x= (5x10*-2 moles . 1 L) / 0,2 moles = 0.25L
(we can also say 250 mL)
. Given a Fischer Projection, explain how to determine if a carbohydrates is D or L
Answer:
Fischer projection is the method used for representing a three-dimensional organic molecule as a two dimensional molecule.
This method can be used for determining the D- and L- configuration of the organic molecules.
In the Fisher projection of carbohydrate molecule, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the right-hand side, then the carbohydrate is said to have D-configuration.
Whereas, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the left-hand side, then the carbohydrate is said to have L-configuration.
Final answer:
To determine if a carbohydrate is D or L in a Fischer projection, look at the hydroxyl (-OH) group on the penultimate carbon; if it's to the right, it's a D-sugar, and if it's to the left, it's an L-sugar. This classification does not directly relate to the sugar's optical activity but its stereochemistry relative to glyceraldehyde.
Explanation:
How to Determine if a Carbohydrate is D or L Using a Fischer Projection
When examining a Fischer projection of a monosaccharide, you can determine whether it is a D-sugar or an L-sugar by looking at the orientation of the hydroxyl (-OH) group on the penultimate carbon (second-last carbon) in the chain. The rule is straightforward: if the -OH group on this carbon is to the right side of the Fischer projection, the sugar is designated as a D-sugar. Conversely, if the -OH group is to the left side, the sugar is an L-sugar.
This method of classification is based on the relative configuration to glyceraldehyde, where D-glyceraldehyde has the -OH on the right at the chiral center farthest from the carbonyl group, hence all D-sugars follow this pattern. L-sugars are the mirror images (enantiomers) of the D-sugars, with their -OH groups flipped to the opposite side.
It's important to note that the D/L configuration does not directly correlate with the optical activity of the sugar (++/--) but rather describes its stereochemistry relative to glyceraldehyde. The D/L nomenclature is fundamental in distinguishing the stereochemistry of sugars and their derivatives.
This patient has a bone density of 820mg/cm3. What is the volume of a 25g sample?
Answer: The volume of bone for given sample is [tex]30.49cm^3[/tex]
Explanation:
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 bone = [tex]820mg/cm^3=0.82g/cm^3[/tex] (Conversion factor: 1 g = 1000 mg)
Mass of bone = 25 g
Putting values in above equation, we get:
[tex]0.82g/cm^3=\frac{25g}{\text{Volume of bone}}\\\\\text{Volume of bone}=30.49cm^3[/tex]
Hence, the volume of bone for given sample is [tex]30.49cm^3[/tex]
Suppose you dissolve 0.1 moles of 1-aminobutane (NH2-CH2-CH2-CH2-CH3) in 1.0 liter of water. a. What are all of the molecules and ions you would expect to find in the solution? b. Which two of the above will be found in the greatest total amount? Hint: what is the approximate pKa of an amino group?) c. Which one of the above will be found in the least total amount?
Answer:
a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.
b) The greatest total amount is of 1-aminobutane.
c) The least total amount is of the conjugate acid.
Explanation:
The equilibrium in water of 1-aminobutane is:
CH₃(CH₂)₃NH₃⁺ ⇄ CH₃(CH₂)₃NH₂ + H⁺
a) The two compounds you will expect in these solution are 1-aminobutane and its conjugate acid.
b) The equlibrium constant is: K = 1,66x10⁻¹¹.
That means you will have 1-aminobutane:Conjugate acid in a ratio of 6x10¹⁰ : 1 .
The greatest total amount is of 1-aminobutane
c) Thus, The least total amount is of the conjugate acid.
I hope it helps!
Write 10,847,100 in Scientific Notation with 4 significant figures.
Answer:
The number 10,847,100 in Scientific Notation is [tex]1.0847x10^{7}[/tex]
Explanation:
Scientific notation is an easy form to write long numbers and it is commonly used in the scientific field. To write a long number in a shorter way it is necessary to 'move' the decimal point to the left the number of positions that are necessary until you get a unit. Then you write the number and multiplied it by 10 raised to the number of positions you moved the decimal point. In this case, it is necessary to move the decimal point 7 positions so, we multiply the number by 10 raised to 7.
HClO4 acid solution has a concentration of 5 Molarity. Calculate the concentration of this solution in
1. Percentage by weight
2. Molar fractions
Answer:
1. Percentage by weight = 0.5023 = 50.23 %
2. molar fraction =0.153
Explanation:
We know that
Molar mass of HClO4 = 100.46 g/mol
So the mass of 5 Moles= 5 x 100.46
Mass (m)= 5 x 100.46 = 502.3 g
Lets assume that aqueous solution of HClO4 and the density of solution is equal to density of water.
Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.
We know that
Mass = density x volume
Mass of 1000 ml solution = 1 x 1000 =1000 ( density = 1 gm/ml)
m'=1000 g
1.
Percentage by weight = 502.3 /1000
Percentage by weight = 0.5023 = 50.23 %
2.
We know that
molar mass of water = 18 g/mol
mass of water in 1000 ml = 1000 - 502.3 g=497.9 g
So moles of water = 497.7 /18 mole
moles of water = 27.65 moles
So molar fraction = 5/(5+27.65)
molar fraction =0.153
What is the pH of a 0.05 M solution of TRIS
acid(pKa = 8.3)?
Answer:
pH of the solution = 4.80
Explanation:
pKa = 8.3
Concentration of tris acid = 8.3
[tex]pH = \frac{1}{2} \times p_{ka} - \frac{1}{2} logC[/tex]
[tex]pH = \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(0.05)[/tex]
[tex]= \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(5\times 10^{-2})[/tex]
[tex]= 4.15 - \frac{1}{2}\times (-1.30103)[/tex]
[tex]= 4.15 - (-0.650515)[/tex]
[tex]= 4.15 + 0.650515[/tex]
= 4.80
pH of the solution = 4.80