Answer:
690 kg of water.125 kg of KCl crystals are formed after cooling processExplanation:
1) A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent. The solubility of KCl at 370K is 42%mass -42kg of KCl in 100kg of water+KCl , thus, the amount of water added to 500kg of KCl is:
500 kg of KCl × [tex]\frac{100 kg Water+KCl}{42kgKCl}[/tex] = 1190 kg of water +KCl
1190 kg of water +KCl - 500 kg of KCl = 690 kg of water
2) The maximum amount that this solution could solubilize of KCl at 320K is:
1190 kg of water + KCl × [tex]\frac{31,5 kgKCl}{100kgWater+KCl}[/tex] = 375 kg
Thus, the mass of KCl crystals formed are:
500 kg of KCl - 375 Kg of KCl = 125 kg of KCl
I hope it helps!
Which of the following describes carbon dioxide? O Has polar covalent bonds and is a polar molecule O Has non-polar covalent bonds and is a polar molecule O Has ionic bonds and is an polar molecules O Has ionic bonds and is a non-polar molecule O Has polar bonds and is a non-polar molecule O Has non-polar bonds and is a non-polar molecules
Answer: Option "Has polar bonds and is a non-polar molecule" is the correct answer.
Explanation:
A covalent compound is defined as the compound in which there will be sharing of electrons between the combining atoms.
An ionic compound is defined as the compound in which there will be transfer of electrons from one atom to another.
A non-polar covalent compound is defined as the compound in which there will be equal sharing of electrons between the combining atoms.For example, a chlorine atom has 7 valence electrons.
So, when a chlorine atom chemically combines with another chlorine atom then there occurs equal sharing of electrons.
Hence, molecule [tex]Cl_{2}[/tex] is formed which is non-polar covalent compound in nature with non-polar covalent bond.
A polar covalent bond is defined as the bond in which there will be unequal sharing of electrons between the combining atoms.For example, a carbon atom has 4 valence electrons and an oxygen atom has 6 valence electrons.
To complete their octet a carbon atom needs 4 more electrons and an oxygen atom needs 2 more electrons.
And, when one carbon and two oxygen atoms chemically combine together then there occurs unequal sharing of electrons.
Hence, a polar covalent compound [tex]CO_{2}[/tex] is formed.
Since, oxygen is more electronegative than carbon atom. Hence, dipoles in the linear carbon dioxide molecules will cancel out each other.
Therefore, a carbon dioxide molecule is non-polar compound that has polar covalent bonds.
Thus, we can conclude that the statement, has polar bonds and is a non-polar molecule, correctly describes carbon dioxide.
Why does acetic acid lose its buffering ability at very high pH values? Why does acetic acid lose its buffering ability at very low pH values? In other words, what is happening when the buffering range is exceeded? The pKa of formic acid is 3.75.
Answer:
Because the optimal range of buffering for a formic acid potassium formate buffer is 2.74 ≤ pH ≤ 4.74.
Explanation:
Every buffer solution has an optimal effective range due to pH = pKa ± 1. Outside this range, there is not enough acid molecules or conjugate base molecules to sustain the pH without variation. There is a certain amount of both molecules that has to be in the solution to maintain a pH controlled.
Being for the formic acid the pKa 3.74, the optimal effective range is between 2.74 and 4.74. Upper or lower these range a formic acid/potassium formate buffer does not work.
Alcohols like ethanol CH3CH2OH, have a proton attached to the oxygen atom that can be acidic. Can a base like CH3NH2 deprotonate the -OH group? (pka CH3CH2OH=15.7 pKa CH3NH3=10.8) justify your answer
Answer:
No. CH₃NH₂ cannot deprotonate CH₃CH₂OH.
Explanation:
Organic species like ethanol and methylamine can behave as acid or bases. The pKa value is necessary to indentify in a reaction if the species can act as an acid or a base.
In this case, despite alcohols being recognised as acids and amines as bases, the amine CH₃NH₂ has an smaller pKa than the alcohol CH₃CH₂OH , which means that CH₃NH₂ is a stronger acid than CH₃CH₂OH . Therefore, CH₃NH₂ will not deprotonate the -OH group.
The likelihood of methylanime (CH3NH2) deprotonating the -OH group of ethanol (CH3CH2OH) is low as the conjugate acid CH3NH3+ (formed after CH3NH2 accepts a proton) is a stronger acid than ethanol, demonstrated by its lower pKa value (10.8 compared to 15.7).
Explanation:In the realm of chemistry, it's known that the strength of an acid is determined by its pKa value. The lower the pKa, the stronger the acid. As given, ethanol (CH3CH2OH) has a pKa of 15.7 and CH3NH3+ (ammonium ion) has a pKa of 10.8. As such, ethanol is a weaker acid than the ammonium ion.
For a base to deprotonate an acid, the conjugate acid of that base (formed after it accepts a proton) has to be weaker than the initial acid. In your case, the conjugate acid of CH3NH2 (methylamine) would be CH3NH3+. As CH3NH3+ is a significantly stronger acid than ethanol, it is therefore unlikely that methylamine could effectively deprotonate the -OH group in ethanol.
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Spread Spectrum (SS) encoding: a) Name four spread spectrum techniques. b) Which one of them was originally devised against eavesdropping in the military environment?
Explanation:
( a )
The four types of spread spectrum techniques are as follows -
1. Direct sequence spread spectrum .
2. frequency hopping spread spectrum .
3. chirp spread spectrum .
4. time hopping spread spectrum .
( b )
The Direct sequence spread spectrum was devised for eavesdropping in the military .
In the field of telecommunications , the Direct sequence spread spectrum , it is the technique of spread spectrum modulation which is used to reduce the overall inference of the signal .
Assume that you have developed a tracer with first-order kinetics and k = 0.1 s − 1 . For any given mass, how long will it take the tracer concentration to drop by (a) 50% and (b) by 75%?
Answer: It will take 6.93 sec for tracer concentration to drop by 50% and 13.9 sec for tracer concentration to drop by 75%
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant =[tex]0.1s^{-1}[/tex]
t = age of sample
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process
a) for tracer concentration to drop by 50%
a - x = amount left after decay process = 50
[tex]t=\frac{2.303}{k}\log\frac{100}{50}[/tex]
[tex]t=\frac{0.693}{0.1}[/tex]
[tex]t=6.93sec[/tex]
It will take 6.93 sec for tracer concentration to drop by 50%
b) for tracer concentration to drop by 75 %
a - x = amount left after decay process = 25
[tex]t=\frac{2.303}{k}\log\frac{100}{100-75}[/tex]
[tex]t=\frac{2.303}{0.1}\times \log\frac{100}{25}[/tex]
[tex]t=13.9sec[/tex]
It will take 13.9 sec for tracer concentration to drop by 75%.
Several models of atoms were mentioned in the lesson this week. Plum pudding, planetary and quantum mechanical. Which reason below explains why there are different atom models?
Answer:
Explanation:
There are different atomic models due to the fact that with advancement in technology and research, the atom began to open up. More scientists began to formulate models that would completely unify the different observations about the atom.
From the different model of atom, noticeable progressions could be observed in terms of sophistication and what they entail. As more details of the atoms were revealed, the need to modify the model becomes important.
The Prandtl number, Npr, is a dimensionless group important in heat transfer calculations. It defined as Cuk, where C, is the heat capacity of a fluid, u is the fluid viscosity, and k is the thermal conductivity. For a particular fluid, C, = 0.583 J/(g.°C), k = 0.286 W/(m:°C), and 1936 lb./(ft·h). Estimate the value of Npr without using a calculator (remember, it is dimension less), showing your calculations; then determine it with a calculator.
Answer:
The Prandlt number for this fluid is 1630.
Explanation:
The Prandlt number is defined as:
[tex]Pr=\frac{C_p\mu}{k}[/tex]
To compute the Prandlt number for this case, is best if we use the same units in every term of the formula.
[tex]\mu=1936\frac{lb}{ft*h}*\frac{1000g}{2.205lb}*\frac{3.281ft}{1m}*\frac{1h}{3600s} \\ \\\mu=800 \frac{g}{m*s}[/tex]
Now that we have coherent units, we can calculate Pr
[tex]Pr=\frac{C_p\mu}{k}=\frac{0.583*800}{0.286}= 1630[/tex]
The Prandtl number (Npr) is a dimensionless group used in heat transfer calculations. To estimate its value, we can multiply the given heat capacity, viscosity, and thermal conductivity. Without a calculator, an estimated value of Npr is approximately 143.53. With a calculator, the exact value is approximately 144.97.
Explanation:The Prandtl number (Npr) is a dimensionless group used in heat transfer calculations. It is defined as the product of the heat capacity (C) of a fluid, its viscosity (u), and its thermal conductivity (k). To estimate the value of Npr for a particular fluid without a calculator, we can multiply the given values: C = 0.583 J/(g·°C), u = 1936 lb./(ft·h), and k = 0.286 W/(m·°C). Converting units for convenience, we have C = 583 J/(kg·°C) and u = 0.8644 kg/(m·s). Therefore, Npr ≈ C·u·k ≈ (583 J/(kg·°C))·(0.8644 kg/(m·s))·(0.286 W/(m·°C)) ≈ 143.53. Using a calculator, we can perform the exact calculation and obtain Npr ≈ 144.97.
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For this device the liquid to be tested is drawn into the tube to a level above the top etched line. The time is then obtained for the liquid to drain to the bottom etched line. The kinematic viscosity, ν, in m2/s is then obtained from the equation ν = KR4t where K is a constant, R is the radius of the capillary tube in mm, and t is the drain time in seconds. When glycerin at 20 ˚C (υ=1.19·10-3 m2/s) is used as a calibration fluid in a particular viscometer the drain time is 1430 s. When a liquid having a density of 970 kg/m3 is tested in the same viscometer the drain time is 900 s. What is the dynamic viscosity of this liquid?
Answer:
The dynamic viscosity of the liquid is 0.727 kg/m*s
Explanation:
In the equation for that viscosimeter, ν = KR⁴t, the terms K and R are not dependent on the liquid that is being tested, unlike ν and t.
Using that equation and the data given in the problem, we can calculate the product of K and R⁴.
1.19*10⁻³m²/s = (KR⁴)* 1430 s
KR⁴=8,32*10⁻⁷m²/s²
We can now calculate the kinematic viscosity of the unknown liquid.
ν=8,32*10⁻⁷m²/s²*900s
ν=7.49*10⁻⁴m²/s
The relationship between the kinematic viscosity and the dynamic viscosity is given by the equation μ=ν * ρ, where μ is the dynamic viscosity and ρ is the density. Thus:
μ=7.49*10⁻⁴m²/s * 970 kg/m³
μ=0.727 kg/m*s
Answer:
the dynamic viscosity is 0.72653 N s/m²
Explanation:
the solution is in the attached Word file
Find the volume in m3 of 52.6 lbm of iron:
Answer: The volume of iron is [tex]3.031\times 10^{-3}m^3[/tex]
Explanation:
To calculate volume of a substance, we use the equation:
[tex]\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}[/tex]
We are given:
Mass of iron = 52.6 lbm = 23.86 kg (Conversion factor: 1 kg = 2.205 lbm)
Density of iron = [tex]7873kg/m^3[/tex]
Putting values in above equation, we get:
[tex]7873kg/m^3=\frac{23.86kg}{\text{Volume of iron}}\\\\\text{Volume of iron}=3.031\times 10^{-3}m^3[/tex]
Hence, the volume of iron is [tex]3.031\times 10^{-3}m^3[/tex]
Measurement of heat conductivity A metal panel of area A = 100 cm² and thickness Y = 0.5 cm was found to conduct heat at a rate of 3 W at steady state with temperatures To = 25 °C and T, = 30 °C imposed on the two main surfaces. What is the average thermal conductivity of the material at this temperature range?
Answer:
The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.
Explanation:
The heat flow through this area of the metal panel can be expressed as:
[tex]q/A=-k\frac{(T_2-T_1)}{Y}[/tex]
being A the surface area, k the thermal conductivity, T2 and T1 the temperatures in each side and Y the thickness of the metal panel.
We have to rearrange the equation to clear k:
[tex]k=\frac{-(q/A)*Y}{\Delta T} = \frac{-(3 W/100 cm2)*0.5cm}{5 ^\circ K}*(\frac{1m}{100cm} )^{3} \\\\k=\frac{0.015}{5*100^{3} }W/mK= 3*10^{-9}W/mK[/tex]
The average thermal conductivity of the material at this temperature range is 3*10^(-9) W/mK.
A freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient air at 20 °C and a coefficient of h 2 W/(m2 K) characterizes heat transfer by natural convection from the exposed surface of the layer, estimate the time required to completely melt the frost. The frost may be assumed to have a mass density of 700 kg/m3 and a latent heat of 334 kJ/kg.
The time required to completely melt the frost can be calculated by first determining the mass of the frost using its thickness and the assumed surface area of the freezer compartment. Once we know the mass, we can calculate the amount of heat required to melt the frost. We can then find the time by dividing the heat required by the rate at which heat is transferred.
Explanation:To solve this problem, we first need to calculate the volume of frost since we know its thickness (2 mm) and lets assume average surface area of the freezer. Lets say the surface area of the freezer compartment is A. Hence volume of frost would be 0.002m * A. Subsequently, using the volume and the density of the frost, we can find the mass of the frost. Mass = Density * Volume = 700 kg/m3 * 0.002m * A.
Once we have the mass of the frost, we can then calculate the amount of heat needed to completely melt the frost, Q. The formula for this is the mass of the frost multiplied by the latent heat of fusion (Q= mass * Lf), where Lf here is 334 kJ/kg. The time required to complete the process can finally be calculated by dividing the heat required to do so (Q) by the heat transferred per unit time (P = h*A*ΔT). Here, A is the surface area, h is the coefficient of heat transfer and ΔT is the temperature difference.
Note: Here we have used the area A which is lesser when the frost starts to melt and greater when the frost has been almost been melted.
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Define "Dielectric" and list the Dielectric Constant values for vacuum and water
Answer:
The term dielectric makes reference to materials that are insulators (the charges in the material does not flow freely and for this reason an insulator material cannot conduce electricity). BUT if we applied this material in an electric field they can be polarized.
Dielectric Constant: It is also defined as a relative permittivity and it is the amount of charge required to produce one unit of electric flux( electric field in a surface) in a given medium.
it is typically denoted as Epsilon (ε)
and its formula is: ε = ε(ω) /ε(0)
where:
ε(ω) = the frequency-dependent permittivity of the material
ε(0) = the dielectric constant value in vacuum
Dielectric Constant values:
For vacuum is always 1 for water depending on the temperature can be 87.9 at 0º C, 80.2 at 20º C and 55.5 at 100º C.
A man weighs 180lbf at a place where g=32.1 ft/s2. What is his weigh(in lbf) on the moon where g=5.47ft/s2? what is his mass(in lb
Answer:
Mass = 5.6075 lb
Weight on moon = 30.7 lbf
Explanation:
The weight of the man at gravity, g = [tex]32.1\ ft/s^2[/tex] is 180 lbf
Also,
Weight = Mass × Gravitational acceleration
So,
180 lbf = Mass × [tex]32.1\ ft/s^2[/tex]
Mass = 5.6075 lb
Given, g = [tex]5.47\ ft/s^2[/tex] and mass = 5.6075 lb
Thus,
Weight = Mass × Gravitational acceleration = 5.6075 × [tex]5.47\ ft/s^2[/tex] = 30.7 lbf
Weight on the moon = 30.7 lbf
What is the heat of vaporization of ethanol, given that ethanol has a normal boiling point of 63.5°C and the vapor pressure of ethanol is 1.75 atm at 780 g.
Explanation:
According to Clausius-Claperyon equation,
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
The given data is as follows.
[tex]T_{1} = 63.5^{o}C[/tex] = (63.5 + 273) K
= 336.6 K
[tex]T_{2} = 78^{o}C[/tex] = (78 + 273) K
= 351 K
[tex]P_{1}[/tex] = 1 atm, [tex]P_{2}[/tex] = ?
Putting the given values into the above equation as follows.
[tex]ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times [\frac{1}{T_{2}} - \frac{1}{T_{1}}][/tex]
[tex]ln (\frac{1.75 atm}{1 atm}) = \frac{-\text{heat of vaporization}}{8.314 J/mol K} \times [\frac{1}{351 K} - \frac{1}{336.6 K}][/tex]
[tex]\Delta H[/tex] = [tex]\frac{0.559}{1.466 \times 10^{-4}} J/mol[/tex]
= [tex]0.38131 \times 10^{4} J/mol[/tex]
= 3813.1 J/mol
Thus, we can conclude that the heat of vaporization of ethanol is 3813.1 J/mol.
A slab of dry wood of 4-in thickness and sealed edges is exposed to air of 40% relative humidity. Assuming that the two unsealed surfaces of the wood immediately jump to an equilibrium moisture content of 8 lb H2O per 100 lb of dry wood, determine the time for the moisture content at the center of the slab to reach 1% of the equilibrium value using the semi-infinite slab solution. Assume a diffusivity of water of 6.00e-006 cm2/s.
Explanation:
It is given here that a slab is made up of dry wood.
Thickness of slab = 4 inch
It is given condition that the edges of the slab are sealed and exposed to air.
Given the relative humidity of air = 40%
The surface of slab is unsealed and reached to the equilibrium moisture content = [tex]\frac{8 lb H_{2}O}{100 lb dry wood}[/tex].
Given, the diffusivity of water in the slab = [tex]6 \times 10^{-6} cm^{2}/s[/tex].
As per the given condition 1% moisture content will be taken by multiplying the moisture content by 1%, i.e. 0.01.
It is given that the thickness of slab is in inch, so converting the thickness of slab in cm as follows.
Thickness of slab = [tex]4 inch \times 2.54 inch/cm[/tex]
= 10.16 cm (As, 1 inch = 2.54 cm)
Since, the equilibrium moisture content is required to calculate at the center of the slab.
The center of the slab, from the semi-infinite slab solution will be taken from thickness t = 0.
The thickness at center = 10.16 cm
Hence, calculate total thickness at the center of slab as follows.
[tex]\frac{\text{total thickness of slab}}{2}[/tex]
= 5.08 cm
Now, by semi-infinite solution, and by 40% relative humidity of air, time for moisture content to reach at the center of the slab, to 1% of its equilibrium value will be calculated as follows.
Time(Seconds) = [tex]\frac{thickness^{2}(cm^{2}) \times \text{relative humidity} \times 0.01 \times \text{equilibrium content} \times \text{equilibrium moisture}}{\text{diffusivity(cm/s)}}[/tex]
Time (sec) = [tex]\frac{5.08 cm \times 5.08 cm \times 0.4 \times \text{relative humidity} \times 0.01 \text{equilibrium content} \times 8 lb H_{2}O}{100 lb dry.wood \times 6 \times 10{-6} cm^{2}/s \text{diffusivity}}[/tex]
Time (Seconds) = 1376.34 seconds
Thus, we can conclude that the time required for the moisture content at the center of the slab to reach to 1 % of equilibrium value is 1376.34 seconds.
Which of the following reactions is incorrectly labeled?
a. HCl(aq) + NaOH(aq) = NaCl(aq) + H2O(l) – Precipitation reaction.
b. CH4(g) + 3 O2(g) = CO2(g) + 2 H2O(l) – Combustion reaction.
c. No right choice.
Answer:
The option a termed as precipitation reaction is incorrectly labelled.
Explanation:
The chemical reactions are classified based on the reactants used and products formed in a reaction. They are decomposition reaction, single displacement reaction, double displacement reaction, acid-base neutralisation reaction, precipitation reaction, combustion reaction, redox reaction and organic reaction.
Among these, the given options are labelled as precipitation and combustion reaction. The one which is labelled as combustion reaction is correct because combustion reactions occur in the presence of oxygen only and the products of combustion reaction should include water, oxygen or carbon and heat.
The other option which is labelled as precipitation reaction is incorrect because precipitation reaction occurs when an ionic substance will come out of a solution due to heating it or stirring it making the solubility of the ionic substance in a solution zero such that it will come out as solid and form a layer at the bottom of the solution.
But in this case all the products are in aqueous state, there is absence of any ionic substance in solid state, so the option which is labelled as precipitation reaction is incorrectly labelled.
Final answer:
The incorrectly labeled reaction is a. HCl(aq) + NaOH(aq) = NaCl(aq) + H₂O(l) as a Precipitation reaction; it should be labeled as a Neutralization reaction.
Explanation:
The reaction that is incorrectly labeled is a. HCl(aq) + NaOH(aq) = NaCl(aq) + H₂O(l) - Precipitation reaction. This reaction is actually an example of a neutralization reaction, where an acid (HCl) and a base (NaOH) react to form water and a salt (NaCl). A precipitation reaction involves the formation of an insoluble solid, known as a precipitate, from the reaction of two soluble compounds, which does not occur in this case.
Combustion reaction: CH₄(g) + 3O₂(g) = CO₂(g) + 2H₂O(l)
Prepare benzoic acid from from benzyl alcohol. What is the purpose of the addition of the sulfuric acid?Include chemical reaction.
Answer:
Benzoic acid can be prepared from benzyl alcohol using potassium dichromate and sulfuric acid.
Explanation:
Benzyl alcohol is a primary alcohol that can be transformed into benzoic acid by an oxidation reaction.
The reactants needed are potassium dichromate and sulfuric acid.
The sulfuric acid is needed to react with the potassium dichromate and form cromic acid, which is a strong oxidizing agent that will convert the alcohol to its equivalent carboxylic acid.
The chemical reaction is
3ФCH₂OH + 2Cr₂O⁷⁻ + 16 H⁺ 3ФCOOH + 4Cr³⁺ + 11H₂O
Calculate how many g of ammonia will be produced from 9.35 g of nitrogen gas and excess hydrogen using the following equation: 2 NH(g) 3 H2(g) N(g) 4 answer
Answer: The mass of ammonia produced in the reaction is 11.36 g
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of nitrogen gas = 9.35 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of nitrogen gas}=\frac{9.35g}{28g/mol}=0.334mol[/tex]
The chemical reaction for the formation of ammonia from hydrogen and nitrogen follows:
[tex]3H_2+N_2\rightarrow 2NH_3[/tex]
As, hydrogen gas is present in excess. So, it is considered as an excess reagent.
Nitrogen gas is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of nitrogen gas is producing 2 moles of ammonia gas
So, 0.334 moles of nitrogen gas will produce = [tex]\frac{2}{1}\times 0.334=0.668mol[/tex] of ammonia gas.
Now, calculating the mass of ammonia gas by using equation 1, we get:
Moles of ammonia gas = 0.668 mol
Molar mass of ammonia gas = 17 g/mol
Putting values in equation 1, we get:
[tex]0.668mol=\frac{\text{Mass of ammonia gas}}{17g/mol}\\\\\text{Mass of ammonia gas}=11.36g[/tex]
Hence, the mass of ammonia produced in the reaction is 11.36 g
Final answer:
To calculate the mass of ammonia produced from 9.35 g of nitrogen gas, convert the mass of nitrogen to moles, use the stoichiometric relationship from the balanced reaction to find moles of ammonia, and then convert back to grams. The result is 11.37 g of NH3.
Explanation:
To calculate how many grams of ammonia (NH3) will be produced from 9.35 g of nitrogen gas (N2) using the chemical equation N₂(g) + 3H₂(g) → 2NH3(g), we need to perform stoichiometric calculations. First, we determine the molar mass of nitrogen gas (N2) which is 28.02 g/mol. Using this, we find that 9.35 g of N2 is equivalent to 9.35 g / 28.02 g/mol = 0.3338 mol of N2.
From the balanced chemical equation, we know that 1 mole of nitrogen gas produces 2 moles of ammonia. Therefore, 0.3338 mole N2 × (2 moles NH3 / 1 mole N2) = 0.6676 mole NH3. Finally, using the molar mass of ammonia, which is 17.03 g/mol, we can find the mass of ammonia produced: 0.6676 mol × 17.03 g/mol = 11.37 g of NH3.
Which has more kenetic energy: A 150 g bullet fired from a gun achieving a velocity of 440 m/s and a standard baseball weighing 51/4 oz being pitched at he MLB record speed of 105.1 mph. Calculate kinetic energy in Joules.
Answer:
Bullet has more kinetic energy then baseball.
Explanation:
Mass of the bullet , m = 150 g = 0.150 kg
Velocity of the bullet = v = 440 m/s
Kinetic energy =[tex]\frac{1}{2}\times mass\times (velocity)^2[/tex]
Kinetic energy of the bullet = K.E
[tex]K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 0.150 kg\times (440 m/s)^2[/tex]
[tex]=14,520 Joules[/tex]
Mass of the baseball, m' = [tex]\frac{51}{4} oz[/tex]= 0.36146 kg
(1 ounce = 0.0283495 kg)
Velocity of the baseball= v' = 105.1 mph
1 mile = 1609.34 m
1 hour = 3600 seconds
[tex]v'=\frac{ 105.1\times 1609.34 m}{3600 s}=46.98 m/s[/tex]
Kinetic energy of the baseball = K.E'
[tex]K.E'=\frac{1}{2}m'v'^2=\frac{1}{2}\times 0.36146 kg\times (46.98 m/s)^2[/tex]
[tex]=398.83 Joules[/tex]
14,520 Joules > 398.83 Joules
K.E > K.E'
Bullet has more kinetic energy then baseball.
Define "aldose" and "ketone" and recognize them structurally
Explanation:
Aldose is monosaccharide sugar in which the carbon backbone chain has carbonyl group on endmost carbon atom which corresponds to an aldehyde, and the hydroxyl groups are connected to all other carbon atoms.
Example - Glucose
Ketone is functional group with structure RC(=O)R', where the groups, R and R' can be variety of the carbon-containing substituents.
Example, Ethylmethylketone
Structurally, Aldose can be recognized by observing the carbonyl group on endmost carbon atom which corresponds to an aldehyde in a sugar.
Structurally, Ketone can be recognized by observing the functional group with structure RC(=O)R'.
A 32 L samples of xenon gas at 10oC is expanded to 35 L. Calculate the final temperature.
299.3K
325.4K
273.5K
309.5K
ik its not C
Answer:
T = 309.5 K
Explanation:
∴ V1 = 32 L
∴ V2 = 35 L
∴ T1 = 10°C ( 283 K )
the system is considered under constant pressure, so by Charles's law we have:
V1 / T1 = V2 / T2⇒ T2 = V2 / ( V1 / T1 )
⇒ T2 = 35 L / ( 32L / 283 K )
⇒T2 = 309.53 K
The final temperature of the xenon gas after expanding from 32 L to 35 L at a constant pressure can be calculated using Charles's Law. After converting the initial temperature to Kelvin and rearranging Charles's Law, the final temperature is found to be approximately 309.5 K.
To calculate the final temperature of the xenon gas after expansion, we can use Charles's Law, which states that for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its temperature. This can be represented mathematically as V1/T1 = V2/T2, where V represents volume and T represents temperature in kelvins.
First, we need to convert the initial temperature from Celsius to Kelvin by adding 273.15: T1 = 10°C + 273.15 = 283.15 K.
Using Charles's Law, we can then solve for the final temperature (T2) after the expansion to 35 L:
V1 / T1 = V2 / T2
(32 L) / (283.15 K) = (35 L) / T2
Multiplying both sides by T2 and then by 283.15 K, we get:
T2 = (35 L) / (32 L) * 283.15 K
T2 ≈ 309.5 K as the final temperature after expansion.
Compare Newton's law of viscosity and Hooke's law of elasticity. What is the origin of these "laws"?
Answer:
Newton's law of viscosity
It relates shear stress in a fluid flow to velocity gradient in the direction perpendicular to the flow of fluid.
[tex]\tau\ \alpha \ \frac{\mathrm{d} u}{\mathrm{d} y}[/tex]
Or
[tex]\tau\ =\mu \times \frac{\mathrm{d} u}{\mathrm{d} y}[/tex]
[tex]\tau\ =Shear\ stress[/tex]
[tex]\frac{du}{dy} =[/tex] Rate of shear deformation
[tex]\mu\ =Viscosity[/tex]
Hooke's Law
It states that within the limit of elasticity, the stress-induced (σ ) in the solid due to some external force is always in proportion with the strain (ε ). In other words, the force causing stress in a solid is directly proportional to the solid's deformation.
[tex]\sigma\ \alpha\ \epsilon[/tex]
[tex]\sigma=E\ \epsilon[/tex]
where E is constant of proportionality known as Young's Modulus and it represents the stiffness of the material.
The type of matter is different in both Newton's and Hook's laws.
Comparison between Newton's law of viscosity and Hooke's law of elasticityNewton's law of viscous deformation deals with deformation of fluids that is subjected to a load. This law states that shears stress is proportional to shear strain.
While on the other hand, Hooke's law of elasticity deals with deformation in solids which are subjected to a load so we can conclude that the type of matter is different in both Newton's and Hook's laws.
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Radioactive decay can be described by the following equation where is the original amount of the substance, is the amount of the substance remaining after time , and is a constant that is characteristic of the substance. For the radioactive isotope phosphorus-32, is . If the original amount of phosphorus-32 in a sample is 57.9 mg, how much phosphorus-32 remains after 13.7 days have passed
Answer:
29.8 mg
Explanation:
Just slightly less than 1/2 decay
Calculate the number of oxygen atoms in a 120.0 g sample of glucose (C H,206). Be sure your answer has a unit symbol if necessary, and round it to 4 significant digits. Oro X 5 ?
There are approximately 6.011 × 10²⁴ oxygen atoms in the 120.0 g sample of glucose.
The chemical formula for glucose is C₆H₁₂O₆, indicating 6 carbon, 12 hydrogen, and 6 oxygen atoms. The molar mass of glucose may be used to calculate the number of oxygen atoms in 120.0 g of glucose:
Glucose molecule mass (C₆H₁₂O₆) = (6 * carbon) + (12 * hydrogen) + (6 * oxygen)
≈ (6 * 12.01 g/mol) + (12 * 1.008 g/mol) + (6 * 16.00 g/mol)
The molar mass may be used to compute glucose moles in the 120.0 g sample:
To calculate the number of moles of glucose, divide the sample mass by the molar mass: 120.0 g / 72.06 g/mol = 1.664 moles.
Since each glucose molecule has 6 oxygen atoms, we may compute the total:
The total quantity of oxygen atoms is calculated by multiplying the number of glucose moles by the number of oxygen atoms per molecule: 1.664 moles * 6 = 9.984 moles.
Oxygen atoms total roughly 9.984 × 6.022 × 10²³ atoms/mole
= 6.011 × 10²⁴ atoms.
In a 120.0 g glucose sample, there are roughly 6.011 × 10²⁴ oxygen atoms.
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Final answer:
To calculate the number of oxygen atoms in a 120.0 g sample of glucose, use the mole ratio to find the number of moles of oxygen. Then, convert moles of oxygen to atoms using Avogadro's number.
Explanation:
To calculate the number of oxygen atoms in a 120.0 g sample of glucose, we need to determine the number of moles of glucose and use the mole ratio to find the number of moles of oxygen. The molar mass of glucose is 180.16 g/mol, so the number of moles is 120.0 g / 180.16 g/mol = 0.6664 mol. According to the balanced chemical equation, one mole of glucose reacts with 6 moles of oxygen to yield 6 moles of CO2 and 6 moles of H2O. Therefore, the number of moles of oxygen is 0.6664 mol x 6 mol O2 / 1 mol glucose = 3.9984 mol O2. Finally, to convert moles of oxygen to atoms, we use Avogadro's number (6.022 × 10^23 atoms/mol), so the number of oxygen atoms is 3.9984 mol O2 x 6.022 × 10^23 atoms/mol = 2.406 × 10^24 oxygen atoms.
A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express the composition of this as: (a) weight percent HNO3; (b) [lb HNO3/ft3]; HNO3 molarity; [mol HNO3/m3).
Answer:
(a) The weight percent HNO3 is 63%.
(b) Density of HNO3 = 111.2 lb/ft3
(c) Molarity = 13792 mol HNO3/m3
Explanation:
(a) Weight percent HNO3
To calculate a weight percent of a component of a solution we can express:
[tex]wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\ \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63[/tex]
(b) Density of HNO3, in lb/ft3
In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:
[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w[/tex]
The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,
[tex]\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3[/tex]
(c) HNO3 molarity (mol HNO3/m3)
If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are 1704/63.012=27.04 mol HNO3.
When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.
If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is
[tex]Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3[/tex]
We can now calculate the molarity as
[tex]Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3} =13792 \frac{molHNO3}{m3}[/tex]
The composition of an HNO3 solution can be expressed in various ways. The weight percent is calculated by dividing the mass of HNO3 by the total mass of the solution and multiplying by 100. The quantity in [lb HNO3/ft3] can be obtained by unit conversion. The molarity is calculated by dividing the moles of HNO3 by the volume of the solution in liters. Lastly, [mol HNO3/m3] is obtained by multiplying the molarity by 1000.
Explanation:To tackle this problem, we need to make a variety of calculations using the information given and our knowledge of chemical molar masses and concentration calculations.
Firstly, (a) The weight percent HNO3 can be calculated by dividing the mass of HNO3 by the total mass of the solution (HNO3 + water), and then multiplying by 100 to get a percentage. Since the solution has 1.704 kg of HNO3 for every 1 kg of water, the total mass of the solution is 2.704 kg. Thus, the weight percent of HNO3 is (1.704 kg / 2.704 kg) x 100 = ~63%.
Secondly, (b) To express the composition as [lb HNO3/ft3], we have to convert kg of HNO3 to lb, and then use the specific gravity to determine the volume of the solution in ft3.
Next, we calculate the HNO3 molarity by dividing the moles of HNO3 by the liters of solution. The moles of HNO3 can be calculated from the mass using the molar mass of HNO3, which is about 63.01 g/mol.
Finally, [mol HNO3/m3] can be calculated by multiplying the molarity by 1000, since 1 m3 is equal to 1000 L.
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You have a red ballooe that has a volume of 20 liters, a pressure of 1.5 atmospheres and a temperature of 28 C. What is the volume of the balloon if you change the pressure to 5 atmospheres and change the temperature to S0°C Oa.1.07 L Ob3.0 L 0.64 L O d.o.42 L
Explanation:
The given data is as follows.
[tex]P_{1}[/tex] = 1.5 atm, [tex]V_{1}[/tex] = 20 L, [tex]T_{1}[/tex] = (28 + 273) K = 301 K
[tex]P_{2}[/tex] = 5 atm, [tex]V_{2}[/tex] = ?, [tex]T_{2}[/tex] = (50 + 273) K = 323 K
Formula to calculate the volume will be as follows.
[tex]\frac{P_{1} \times V_{1}}{T_{1}}[/tex] = [tex]\frac{P_{2} \times V_{2}}{T_{2}}[/tex]
Putting the given values into the above formula as follows.
[tex]\frac{P_{1} \times V_{1}}{T_{1}}[/tex] = [tex]\frac{P_{2} \times V_{2}}{T_{2}}[/tex]
[tex]\frac{1.5 atm \times 20 L}{301 K}[/tex] = [tex]\frac{5 atm \times V_{2}}{323 K}[/tex]
[tex]V_{2}[/tex] = 0.64 L
Thus, we can conclude that the change in volume of the balloon will be 0.64 L.
Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation
4 P + 5 O2 ----> 2 P2O5
Answer: 14.2 grams
Explanation:-
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
given mass of phosphorous (P) = 3.07 g
Molar mass of phosphorous (P) = 31 g/mol
Putting in the values we get:
[tex]\text{Number of moles of phosphorous}=\frac{3.07g}{31g/mol}=0.1moles[/tex]
given mass of oxygen [tex]O_2[/tex] = 6.09 g
Molar mass of oxygen [tex]O_2[/tex] = 32 g/mol
Putting in the values we get:
[tex]\text{Number of moles of oxygen}=\frac{6.09g}{32g/mol}=0.2moles[/tex]
According to stoichiometry:
[tex]4P+5O_2\rightarrow 2P_2O_5[/tex]
4 moles of phosphorous combine with 5 moles of oxygen
Thus 0.1 moles of phosphorous combine with =[tex]\frac{5}{4}\times 0.1=0.125[/tex] moles of oxygen
Thus phosphorous acts as limiting reagent as it limits the formation of product and oxygen is the excess reagent.
4 moles of phosphorous gives= 2 moles of [tex]P_2O_5[/tex]
Thus 0.1 moles of phosphorous gives =[tex]\frac{2}{4}\times 0.1=0.05[/tex] moles of [tex]P_2O_5[/tex]
mass of [tex]P_2O_5=moles\times {\text {Molar mass}}=0.05\times 284=14.2g[/tex]
Thus the theoretical yield of [tex]P_2O_5[/tex] is 14.2 grams.
Which observation allows you to tell that two cubes are made of different metals? Select the two correct answers.
A. Their volumes are the same, and their masses are the same.
B. Their volumes are the same, and their masses are different.
C. Their volumes are different, and their masses are the same.
D. Their volumes are unknown, and their masses are the same.
E. Their volumes are the same, and their masses are unknown.
Answer:
B. Their volumes are the same, and their masses are different.
C. Their volumes are different, and their masses are the same.
Explanation:
In order to ascertain that two metal cubes are made of different materials, their densities would be very helpful in our experiment.
Every metal has its own unique density and we can use it to differentiate it from other metals.
Density is defined as the mass per unit volume of a substance. Therefore, if the mass and volume are the same, their densities would be the same. Density is expressed a:
Density = [tex]\frac{mass}{volume}[/tex]
If the volumes are the same but given different masses, we can differentiate the metals using their quite different densities.
If the volumes are different but the masses the same, we would still obtain different values of density and the metals can be differentiated.
Calculate the pH of the following simple solutions:
53.1 mM HCl
0.223 M KOH
53.1 mM HCl + 0.223 M KOH
Is the solution "c" a buffer? Why or why not?
Answer :
(1) pH = 1.27
(2) pH = 13.35
(3) The given solution is not a buffer.
Explanation :
(1) 53.1 mM HCl
Concentration of HCl = [tex]53.1mM=53.1\times 10^{-3}M[/tex]
As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.
So, Concentration of hydrogen ion= [tex]53.1\times 10^{-3}M[/tex]
pH : It is defined as the negative logarithm of hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (53.1\times 10^{-3})[/tex]
[tex]pH=1.27[/tex]
(2) 0.223 M KOH
Concentration of KOH = 0.223 M
As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.
So, Concentration of hydroxide ion= 0.223 M
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (0.223)[/tex]
[tex]pOH=0.65[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35[/tex]
(3) 53.1 mM HCl + 0.223 M KOH
Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.
It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.
As we know that the pH of strong acid and strong base solution is always 7.
So, the given solution is not a buffer.
Arrange the following cellular components in order of decreasing size: hexokinase ribosome CO2 glucose
Answer:
Ribosome, Hexokinase, Glucose, CO2.
Ribosomes are proteins that sintetize the proteins in the cell, depending of the organism, the can size up to 30 nm. Hexokinase is an enzyme that measures approximately 50 kDa, and in its spatial conformation it sizes about 5 to 6 nm in diameter. Glucose is a molecule that sizes about 1 nm, and CO2 is another molecule that sizes 0.232 nm.