Explanation:
1) Edge length of the metal in BCC unit cell = [tex]a=0.33226 nm[/tex]
Atomic radius of the metal atom = r
For BCC unit cell, relationship between edge length and radius is given as:
[tex]r=\frac{\sqrt{3}}{4}\times a=0.4330a[/tex]
[tex]r=0.4330\times 0.33226 nm=0.144 nm[/tex]
[tex]1 nm=10^{-7} cm[/tex]
[tex]r=0.144 nm=0.1439\times 10^{-7} cm=1.44\times 10^{-8} cm[\tex]
The atomic radius of the metal atom in BCC unit cell is [tex]1.44 \times 10^{-8} cm[/tex].
2) Edge length of the metal in FCC unit cell = [tex]a=4.3992 \AA[/tex]
Atomic radius of the metal atom = r
For FCC unit cell, relationship between edge length and radius is given as:
[tex]r=\frac{1}{2\sqrt{2}}\times a=0.3535a[/tex]
[tex]r=0.3535\times 4.3992 \AA=1.56 \AA[/tex]
[tex]1 \AA=10^{-8} cm[/tex]
[tex]1.56\AA=1.56 \times 10^{-8} cm[/tex]
The atomic radius of the metal atom in FCC unit cell is [tex]1.56 \times 10^{-8} cm[/tex].
Find the molar concentration of sulfuric acid in a 20.0 wt% solution of sulfuric acid in water (SG=1.13).
Answer: The molarity of sulfuric acid solution is 2.3 M
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 = 1.13
Density of water = 1.00 g/mL
Putting values in above equation we get:
[tex]1.13=\frac{\text{Density of sulfuric acid solution}}{1.00g/mL}\\\\\text{Density of sulfuric acid solution}=(1.13\times 1.00g/mL)=1.13g/mL[/tex]
We are given:
20% (m/m) sulfuric acid solution. This means that 20 g of sulfuric acid is present in 100 g of solution
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 = 1.13 g/mL
Mass of Solution = 100 g
Putting values in above equation, we get:
[tex]1.13g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.13g/mL}=88.5mL[/tex]
To calculate the molarity of solution, we use the equation:[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Mass of solute (sulfuric acid) = 20 g
Molar mass of sulfuric acid = 98 g/mol
Volume of solution = 88.5 mL
Putting values in above equation, we get:
[tex]\text{Molarity of solution}=\frac{20g\times 1000}{98g/mol\times 88.5mL}\\\\\text{Molarity of solution}=2.3M[/tex]
Hence, the molarity of sulfuric acid solution is 2.3 M
The digits that are reported in an answer are called
A sample of gold has a mass of 645.0 lb and a volume of 0.5348 ft. What is its specific gravity? Number SG= What are the units of specific gravity? Ob/t kg/m2 g/cm none
Explanation:
The given data is as follows.
Mass of gold = 645 lb, Volume = 0.5348 ft3
Density of water = 62.4 lbs/ft3
It is known that specific gravity is defined as density of substance divided by the density of standard fluid.
Mathematically, Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water}}[/tex]
Specific gravity = [tex]\frac{\text{density of gold}}{62.4 (lbs.ft^{-3})}[/tex]
Now, calculate the density of gold then from density we will calculate specific gravity as follows
Since, Density = [tex]\frac{mass}{volume}[/tex]
Density = [tex]\frac{645 lbs}{0.5348 ft^{3}}[/tex]
= 1206.06 [tex]lbs/ft^{3}[/tex]
As, Specific gravity = [tex]\frac{\text{density of gold}}{\text{density of water (standard fluid)}}[/tex]
= [tex]\frac{1206.06 (lbs/ft^{3})}{62.4 (lbs/ft^{3})}[/tex]
= 19.32
Therefore, the value of specific gravity is 19.32.
Specific gravity has no units as it is density divided by density. Hence, all the units get canceled out.
The wavelength of the bright red line in the spectrum of atomic
hydrogen is 652 nm. What is the energy of the photon generated in
the transition?
Answer:
The energy of the photon generated in the transition is 3.14*10⁻¹⁹ J
Explanation:
There are two equation that we need to use in order to solve this problem:
The first one is Planck's equation, which describes the relationship between energy and frequency:
E = h*v eq. 1)
Where E is energy, h is Planck's constant (6.626 * 10⁻³⁴ J*s) and v is the radiation frequency.
In order to know the frequency, we use the second equation, which is the wave equation:
c = λ*v eq. 2)
Where c is the speed of light in vacuum (aprx 3 * 10⁸ m/s), and λ is the wavelength. If we solve that equation for v we're left with
v=c/λ eq. 3)
We replace v in eq. 1):
E= c*h/λ
Lastly we put the data we know and solve the equation, keeping in mind the correct use of units (converting 652 nm into m):
[tex]E=\frac{3*10^{8}ms^{-1} *6.626*10^{-34}Js}{633*10^{-9}m }=3.14*10^{-19}J[/tex]
Final answer:
The energy of the photon generated in the transition using the given formula. Calculating gives approximately 3.04 x 10^-19 joules.
Explanation:
The energy of the photon generated in the transition can be calculated using the formula:
E = hc / λ
Where E is the energy of the photon, h is the Planck's constant, c is the speed of light, and λ is the wavelength of the light. Substituting the given values:
E = (6.63 x 10^-34 J s x 3 x 10^8 m/s) / (652 x 10^-9 m)
Calculating this gives the energy of the photon as approximately 3.04 x 10^-19 joules.
If a weak acid, HA, is 3% dissociated in a 0.25 M
solution,calculate the Ka and the pH of the solution.
Answer:
Ka = [tex]2.32 \times 10^{-4}[/tex]
pH = 2.12
Explanation:
Calculation of Ka:
% Dissociation = 3% = 0.03
Concentration of solution = 0.25 M
HA dissociated as:
[tex]HA \rightarrow H^+ + A^{-}[/tex]
C(1 - 0.03) C×0.03 C×0.03
HA] after dissociation = 0.25×0.97 = 0.2425 M
[tex][H^+]= 0.25\times0.03 = 0.0075 M[/tex]
[tex][A^{-}]= 0.25 \times 0.03 = 0.0075 M[/tex]
[tex]Ka= \frac{[H^+][A^{-}]}{[HA]}[/tex]
[tex]Ka = \frac{(0.0075)^2}{0.2475} =2.32 \times 10^{-4}[/tex]
Calculation of pH of the solution
[tex]pH = -log [H^+][/tex]
[tex]H^+[\tex] = 0.0075 M[/tex]
[tex]pH = -log 0.0075 = 2.12[/tex]
The Ka of the weak acid is calculated using the concentrations of the dissociated and non-dissociated parts of the weak acid and the pH is derived from the concentration of H+ ions.
Explanation:The subject in concern relates to the dissociation of a weak acid and the calculation of the Acid dissociation constant (Ka) and the pH of the solution.
We are given that 3% of the weak acid, HA, is dissociated. Meaning, 3% of 0.25 M HA dissociates into H+ and A-. This will be 0.03 * 0.25 M = 0.0075 M.
The equilibrium for the reaction is HA <--> H+ + A-. Due to water autoionization, the concentration of water is approximately taken as constant in the denominator of Ka calculation. Hence, Ka for HA can be approximated as [H+][A-] / [HA]. We know that [H+] = [A-] = x, and [HA] = 0.25 - x. We approximate x as 0.0075 since only 3% of HA dissociates. Hence Ka=~(0.0075)^2/(0.25-0.0075).
The pH of the solution is -log[H+] = -log(0.0075).
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For the balanced equationLaTeX: 2Li\left(s\right)\:+\:2H_2O\left(l\right)\:\longrightarrow\:2LiOH\left(aq\right)\:+\:H_2\left(g\right)identify what is:
oxidized
reduced
oxidizing agent
reducing agent
Answer: Lithium is getting oxidized and is a reducing agent. Hydrogen is getting reduced and is oxidizing agent.
Explanation:
Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.
[tex]X^{n+}+ne^-\rightarrow X[/tex]
Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.
Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.
For the given chemical reaction:
[tex]2Li(s)+2H_2O(l)\rightarrow 2LiOH(aq.)+H_2(g)[/tex]
The half reactions for the above reaction are:
Oxidation half reaction: [tex]2Li(s)\rightarrow 2Li^{+}(aq.)+2e^-[/tex]
Reduction half reaction: [tex]2H^(aq.)+2e^-\rightarrow H_2(g)[/tex]
From the above reactions, lithium is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.
Hydrogen is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.
Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl, according to the reaction in the following: Mg(OH)2(aq)+2HCl(aq)→2H2O(l)+MgCl2(aq) What mass of HCl, in grams, is neutralized by a dose of milk of magnesia containing 3.26 g of Mg(OH)2?
Answer:
4.076g of HCl is neutralized by a dose of milke of magnesia containing 3.26g of Mg(OH)2
Explanation:
Step 1: The balanced equation
Mg(OH)₂(aq) + 2 HCl(aq) ⇔ 2 H₂O(l) + MgCl₂(aq)
This means that for 1 mole of Magnesium hydroxide consumed, there i s consumed 2 moles of HCl and there is produced 1 mole of MgCl2 ( and 2 moles of H2O)
Step 2: Calculating moles of Mg(OH)2
moles of Mg(OH)2 = mass of Mg(OH)2 / Molar mass of Mg(OH)2
moles Mg(OH)2 = 3.26g / 58.32g/mole = 0.0559 mole Mg(OH)2
Step 3: Calculating moles of HCl
Since there is for 1 mole Mg(OH)2 consumed, there is consumed 2 moles of HCl
There is for each 0.0559 moles of Mg(OH)2, 2*0.0559 = 0.1118 moles of HCl consumed.
Step 4: Calculating mass of HCl
mass of HCl = moles of HCl x Molar mass of HCl
mass of HCl = 0.1118 moles * 36.46 g/mole
mass of HCl = 4.076 g
4.076g of HCl is neutralized by a dose of milke of magnesia containing 3.26g of Mg(OH)2
To find the mass of HCl neutralized by Mg(OH)2, calculate the number of moles in 3.26 g of Mg(OH)2 and multiply by the molar mass of HCl, taking account that each mole of Mg(OH)2 neutralizes two moles of HCl.
Explanation:The subject of this question is stoichiometry, which is a part of Chemistry. The equation provided showcases the reaction between magnesium hydroxide (Mg(OH)2, the active ingredient in milk of magnesia) and stomach acid (HCl). To find the mass of HCl that is neutralized by 3.26 g of Mg(OH)2, we first need to determine the molar mass of Mg(OH)2, which is 58.3197 g/mol. Then, calculate the number of moles in 3.26 g of Mg(OH)2. Given the balanced formula, 1 mole of Mg(OH)2 neutralizes 2 moles of HCl. Therefore, the moles of HCl that can be neutralized is twice the moles of Mg(OH)2. The molar mass of HCl is 36.461 g/mol. Multiply the moles of HCl by the molar mass to acquire the mass of the HCl that can be neutralized.
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A KNO3 solution containing 45 g of KNO3 per 100.0 g of water is cooled from 40 ∘C to 0 ∘C. What will happen during cooling? Match the words in the left column to the appropriate blanks in the sentences on the right.
(A) 58
(B) 14
(C) 85
(D) 55
(E) 35
(F) 63
(G) 41
(H)10
At 40 ∘C the solution has __ g of KNO3 per 100 g of water and it can contain up to __ g of KNO3 per 100 g of water. At 0 ∘C the solubility is ~ __ gKNO3per 100 g of water, so __ gKNO3 per 100 g of water will precipitate
Answer:
Hello my friend! The amount of 31g of KNO3 will precipitate!
Explanation:
At 40 ∘C the solution has 45 g of KNO3 per 100 g of water and it can contain up to 63 g of KNO3 per 100 g of water. At 0 ∘C the solubility is ~ 14 gKNO3per 100 g of water, so 31 gKNO3 per 100 g of water will precipitate
The cooling of a solution with 45g of KNO3 in 100g of water from 40 degrees Celsius to 0 degrees Celsius results in precipitation of excess KNO3 because of decreased solubility.
Explanation:This question is related to the solubility of Potassium Nitrate (KNO3) in water at different temperatures. At 40 degrees Celcius, the solution already has 45g of KNO3 per 100g of water. The solubility of KNO3 at this temperature is approximately 60g per 100g of water. However, upon cooling to 0 degrees Celcius, the solubility of KNO3 drops to around 13g per 100g of water. As a result, the excess KNO3 which is around 32g per 100g of water, will precipitate out of the solution.
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What is the minimum number of nucleobase "digits" required to code for the 20 different amino acids? (Groups of nucleotides, each coding for a single amino acid in a protein chain, are called "codons" by biologists.) (Hint: How many four-state "digits" are needed to represent 20 unique things?)
Answer:
3 "digits" are required to code for the 20 different amino acids.
This means that in order to code for one amino acid, you require a group of 3 nucleotides, which is called a 'codon'
Explanation:
If each nucleotide determined one amino acid, we could only code for four different amino acids, since DNA contains only four kinds of nucleotides.
If an amino acid were to be coded by a group of two nucleotides, the total number of diniclueotides we could get would be 4^2 = 16. This means that we could only code for 16 amino acids, which is an inferior amount than the number of amino acids required for protein synthesis (20).
If an amino acid were to be coded by a group of three nucleotides, the total number of trinucleotides we could get would be 4^3 = 64. This means that the total amount of triplets we could get is 64, which is more than enough to be able to code for the 20 different amino acids.
In natural convection heat transfer, the parameter is the
Choose the right answer
a) Nusselt number
b) Rayleigh number
c) Grashof number
d) Lewis number
Answer:
correct option is c
Explanation:
The Grashof number is a dimensionless number, which is named after renowned scientist Franz Grashof. The Grashof quantity is defined as the proportion of the buoyant force to viscous force performing on a fluid in a pace boundary layer.
Its function in natural convection is more or less the same as that of Reynolds's number in compelled convection.
Calculate the fraction of atom sites that are vacant for silver at 552°C. Assume an energy for vacancy formation of 0.63 eV/atom.
Answer:
[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]
Explanation:
given data:
temperature = 552 degree celcius = 825 Kelvin
energy for vacancy formation is given as 0.63 eV/atom
fraction of atom can be obtained from following formula
[tex]N_C = N e^{\frac{-Q_V}{kT}}[/tex]
Where, K IS BOLTZMAN CONSTANT [tex]= 8.62\times 10^{-5}eV/K[/tex]
[tex]\frac{N_C}{N} =e^{\frac{-0.63}{8.62\times 10^{-5} \times 825}[/tex]
[tex]\frac{N_C}{N} = e^{-8.8858}[/tex]
[tex]\frac{N_C}{N} = 1.42\times 10^{-4}[/tex]
The fraction of atom sites that are vacant for silver at 552°C is approximately [tex]\( 7.347 \times 10^{-5} \).[/tex]
The fraction of atom sites that are vacant for silver at a given temperature can be calculated using the formula derived from the statistics of a canonical ensemble:
[tex]\[ f = \exp\left(-\frac{\Delta G_{\text{form}}}{k_B T}\right) \][/tex]
Now, we can plug in the values:
[tex]\[ f = \exp\left(-\frac{0.63 \text{ eV/atom}}{(8.617 \times 10^{-5} \text{ eV/K}) \times (825.15 \text{ K})}\right) \] \[ f = \exp\left(-\frac{0.63}{8.617 \times 10^{-5} \times 825.15}\right) \] \[ f = \exp\left(-\frac{0.63}{0.0658}\right) \] \[ f = \exp\left(-9.574\right) \] \[ f \approx \exp\left(-9.574\right) \] \[ f \approx 7.347 \times 10^{-5} \][/tex]
One mole of pure N2 and 1 mole of pure O2 are contained in separate compartments of a rigid and insulated container at 1 bar and 298 K. The separator between the compartments are later removed to allow the mixing of the gases. Assume that both gases are in the ideal gas state.
a) What is the final T and P of the mixture?
b) For a mixture of ideal gases, each component carries a partial pressure that is proportional to its mole fraction: i.e., p????2 = y????2 P and p????2 = y????2 P, P being the total pressure of the mixture. Calculate the entropy change of the mixing process and draw a schematic showing the hypothetical path you used for the calculation.
I am unclear why you need enthalpy information when the question asks for entropy. Can someone please solve part A?
Answer:
a. T and P remain the same (T=298 K and P=1 bar)
b. 11.23J/K
Explanation:
a. Since the mixing process of an idea gas doesn't present a change in the enthalpy, we could state that no change in neither temperature and pressure are given.
b. It is not necessary to know enthalpy data, the following formula is enough to compute the entropy change:
Δ[tex]S_{mix}=-n_{N_2}R ln(x_{N_2})-n_{O_2}R ln(x_{O_2})[/tex]
Thus, the molar fractions are equal to 0.5, and the result yields:
Δ[tex]S_{mix}=-(1mol)[(8.314J/(mol*K)]ln(0.5)-(1mol)[(8.314J/(mol*K)]ln(0.5)[/tex]
Δ[tex]S_{mix}=11.23J/K[/tex]
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Calculate the pressure exerted by Ar for a molar volume 0.45 L at 200 K using the van der Waals equation of state. The van der Waals parameters a and b for Ar are 1.355 bar dm mol-2 and 0.0320 dm3mol?, respectively. Please write your answer (unit: bar) with 2 decimals, as 12.23. Please do not add unit to your answer.
Explanation:
It is known that the Van der Waals equation is a description of real gases, as in this equation there are all those interactions which we previously ignore in the ideal gas law.
In Vander Waals equation, the repulsion and collision, between molecules of gases are being considered. They are no longer ignored and they also are not considered a "point" particle.
According to the ideal gas law, PV = nRT
or, [tex]P(\frac{V}{n})[/tex] = RT
and, let [tex]\frac{V}{n}[/tex] = v; which is called molar volume
Hence, P × v = RT
As, the van der Waals equation corrects pressure and volume as follows.
[tex](P+ \frac{a}{v^{2}}) \times (v - b)[/tex] = RT
where, R = idel gas law; recommended to use the units of a and b; typically bar/atm and dm/L
T = absolute temperature, in K
v = molar volume, v = [tex]\frac{\text{Volume of gas}}{\text{moles of gas}}[/tex]
P = pressure of gas
Now, substitute the data in Vander Waal,s equation as follows.
For argon, [tex](P+ \frac{a}{v^{2}}) \times (v - b)[/tex] = RT
[tex](P+ \frac{1.355}{(0.45)^{2}}) \times (0.45- 0.0320)[/tex] = 0.08314 [tex]bar dm^{3}/molK \times (200)K[/tex]
(P+ 6.691) = [tex]0.08314 \times \frac{200}{(0.45- 0.0320)}[/tex]
P = (39.7799 - 6.691) bar
P = 33.0889 bar
or, P = 33.09 bar (approx)
Thus, we can conclude that the pressure exerted by Ar in the given situation is 33.09 bar.
Consider water at 500 kPa and a specific volume of 0.2 m3/kg, what is the temperature (in oC)?
Answer:
[tex]T=-272.9^{o}C[/tex]
Explanation:
We have the ideal gasses equation [tex]PV=nRT[/tex] and the expression for the specific volume [tex]v=\frac{V}{m}[/tex], that is the inverse of the density, and for definition the number of moles is equal to the mass over the molar mass, that is [tex]n=\frac{m}{M}[/tex]
And we can relate the three equations as follows:
[tex]PV=nRT[/tex]
Replacing the expression for n, we have:
[tex]PV=\frac{m}{M}RT[/tex]
[tex]P\frac{V}{m}=\frac{RT}{M}[/tex]
Replacing the expression for v, we have:
[tex]Pv=\frac{RT}{M}[/tex]
Now resolving for T, we have:
[tex]T=\frac{PvM}{R}[/tex]
Now, we should convert all the quantities to the same units:
-Convert 500kPa to atm
[tex]500kPa*\frac{0.00986923}{1kPa}=4.93atm[/tex]
-Convert 0.2[tex]\frac{m^{3}}{kg}[/tex] to [tex]\frac{L}{kg}[/tex]
[tex]0.2\frac{m^{3} }{kg}*\frac{1L}{1m^{3}}=0.2\frac{L}{kg}[/tex]
- Convert the molar mass M of the water from [tex]\frac{g}{mol}[/tex] to [tex]\frac{kg}{mol}[/tex]
[tex]18\frac{g}{mol}=\frac{1kg}{1000g}=0.018\frac{kg}{mol}[/tex]
Finally we can replace the values:
[tex]T=\frac{(4.93atm)(0.2\frac{L}{kg})(0.018\frac{kg}{mol})}{0.082\frac{atm.L}{mol.K}}[/tex]
[tex]T=0.216K[/tex]
[tex]T=0.216K-273.15\\T=-272.9^{o}C[/tex]
Use Lewis symbols to represent the reaction that occursbetween
Mg and Br atoms.
Explanation:
The electronic configuration of magnesium with Z = 12 is : 2, 8, 12
The electronic configuration of bromine with Z = 35 is : 2, 8, 18, 7
The Lewis structure is drawn in such a way that the octet of each atom is complete.
Thus, magnesium losses two electrons to bromine and 2 atoms of bromine accepts the electron.
Thus, the valence electrons are shown by dots in Lewis structure. The reaction is shown in image below.
Lewis symbols represent electron transfer between Mg and Br atoms in the formation of an ionic compound.
Explanation:Lewis symbols, also known as Lewis dot diagrams or electron dot diagrams, depict the valence electrons of atoms using dots around the symbol of the element. In the formation of an ionic compound between Mg and Br atoms, magnesium (Mg) donates two electrons to each bromine (Br) atom, resulting in the transfer of electrons and the creation of Mg2+ and Br- ions.
According to the following reaction, how many grams of water are produced in the complete reaction of 29.7 grams of ammonia? 4 NH3(g) + 5 O2(g) + 4 NO(g) + 6 H2O(g) grams H2O
Answer: The mass of water produced in the reaction is 47.25 grams.
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 ammonia = 29.7 g
Molar mass of ammonia = 17 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ammonia}=\frac{29.7g}{17g/mol}=1.75mol[/tex]
The given chemical reaction follows:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
By stoichiometry of the reaction:
4 moles of ammonia produces 6 moles of water.
So, 1.75 moles of ammonia will produce = [tex]\frac{6}{4}\times 1.75=2.625mol[/tex] of water.
Now, calculating the mass of water by using equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 2.625 moles
Putting values in equation 1, we get:
[tex]2.625mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=47.25g[/tex]
Hence, the mass of water produced in the reaction is 47.25 grams.
The complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water by converting the given mass of NH3 to moles, using the balanced chemical equation to find the ratio of NH3 to H2O, and then converting the moles of H2O to grams.
Explanation:To find out how many grams of water are produced in the complete reaction of 29.7 grams of ammonia (NH3), we need to use stoichiometry, which involves several steps:
First, calculate the molar mass of NH3 which is 14.01 (for N) + 3×1.01 (for H) = 17.03 g/mol.Next, convert the given mass of NH3 to moles by dividing by the molar mass: 29.7 g ÷ 17.03 g/mol = 1.743 mol of NH3.Using the balanced chemical equation 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l), for every 4 mol of NH3, 6 mol of H2O are produced. This means for 1.743 mol NH3, we'd produce (1.743 mol NH3 × 6 mol H2O) ÷ 4 mol NH3 = 2.6145 mol H2O.Finally, convert the moles of H2O to grams. The molar mass of H2O is 18.02 g/mol, so 2.6145 mol × 18.02 g/mol = 47.099 grams of H2O.Therefore, the complete reaction of 29.7 grams of NH3 would produce 47.099 grams of water (H2O).
Enter your answer in the provided box. an aqueous solution containing 10 g of an optically pure substance was diluted to 500 ml with water and placed in a polarimeter tube 20 cm long. the measured rotation was −3.10°. calculate the specific rotation of this compound. [α] = °
Answer:
[α] = -77.5° / [tex]\frac{\textup{dm-g}}{\textup{mL}}[/tex]
Explanation:
Given;
Mass of optically pure substance in the solution = 10 g
Volume of water = 500 mL
Length of the polarimeter, l = 20 cm = 20 × 0.1 dm = 2 dm
measured rotation = - 3.10°
Now,
The specific rotation ( [α] ) is given as:
[α] = [tex]\frac{\alpha}{c\times l}[/tex]
here,
α is the measured rotation = -3.10°
c is the concentration
or
c = [tex]\frac{\textup{Mass of optically pure substance in the solution}}{\textup{Volume of water}}[/tex]
or
c = [tex]\frac{10}{500}[/tex]
or
c = 0.02 g/mL
on substituting the values, we get
[α] = [tex]\frac{-3.10^o}{0.02\times2}[/tex]
or
[α] = -77.5° / [tex]\frac{\textup{dm-g}}{\textup{mL}}[/tex]
How many carbon atoms are present in 3.900 x 10-4 mol of carbon? Give your answer in scientific notation. x 10 (select) A
Answer: The number of atoms of carbon present in given number of moles are [tex]2.350\times 10^{20}[/tex]
Explanation:
We are given:
Number of moles of carbon = [tex]3.900\times 10^{-4}mol[/tex]
According to mole concept:
1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.
So, [tex]3.900\times 10^{-4}mol[/tex] of carbon will contain = [tex]3.900\times 10^{-4}\times 6.022\times 10^{23}=2.350\times 10^{20}[/tex] number of atoms.
Hence, the number of atoms of carbon present in given number of moles are [tex]2.350\times 10^{20}[/tex]
How many moles of disulfur decafluoride are present in 3.99 x 104 molecules of this compound? moles
Answer: The number of moles of disulfur decafluoride is [tex]6.62\times 10^{-20}[/tex]
Explanation:
We are given:
Number of disulfur decafluoride molecules = [tex]3.99\times 10^4[/tex]
According to mole concept:
[tex]6.022\times 10^{23}[/tex] number of molecules are contained in 1 mole of a compound.
So, [tex]3.99\times 10^4[/tex] number of molecules will be contained in = [tex]\frac{1mol}{6.022\times 10^{23}}\times 3.99\times 10^{4}=6.62\times 10^{-20}mol[/tex] of disulfur decafluoride.
Hence, the number of moles of disulfur decafluoride is [tex]6.62\times 10^{-20}[/tex]
5.45 grams of magnesium sulfate heptahydrate is how many moles?
Answer:
5.45 g is 0.0221 moles of magnesium sulfate heptahydrate.
Explanation:
The molecular formula of magnesium sulfate heptahydrate is Mg SO₄·7H₂O.
The molar mass of this compound is calculated adding the molar mass of each element:
Mg: 24. 3 g
S: 32. 1 g
O: 16 g
H: 1 g
Then the mass of a mole of Mg SO₄·7H₂O is:
24.3 g + 32.1 g +4(16 g) + 7(2(1) + 16) = 246.4 g.
if 246.4 g is 1 mole of Mg SO₄·7H₂O, then 5.45 g will be:
5.45 g *(1 mol / 246.4 g) = 0.0221 mol.
Given a chemical equation, showing the dissociation of an acid in solution, be able to identify the "acid", the "base", the "conjugate acid", and the "conjugate base"
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory:
An acid is defined as a substance which looses donates protons and thus forming conjugate base A base is defined as a substance which accepts protons and thus forming conjugate acid.[tex]HA+H_2O\rightarrow A^-+H_3O^+[/tex]
Suppose acid Ha is getting dissociated in its solution and after dissociation it donates its proton to water molecule and forms conjugate base. Where as water (acting as a base) accepts protons and forms conjugate acid.
HA = Acid
[tex]H_2O[/tex] = Base
[tex]A^-[/tex] = Conjugate base
[tex]H_3O^+[/tex] = Conjugate acid
For example:
[tex]H_2SO_4+2H_2O\rightarrow SO_4^{2-}+2H_3O^+[/tex]
Sulfuric acid dissociating in its solution to form conjugate base and conjugate acid.
Sulfuric acid = Acid
[tex]H_2O[/tex] = Base
[tex]SO_^{2-}[/tex] = Conjugate base
[tex]H_3O^+[/tex] = Conjugate acid
In a capillary tube, how do we tell which side of the meniscus has higher or lower pressure
Answer:
A liquid with a sharp contact angle (e.g., water on glass) will form a concave meniscus, and the liquid pressure under the meniscus will be smaller than the atmospheric pressure
Explanation:
The phenomenon of capillarity is produced by the action of the surface tension of the fluids and is observed when a small diameter tube is immersed within the fluid. If we pay attention to the result, we can see that, depending on the fluid, two different things can happen, that the liquid rises through the tube and that the level inside the tube is greater than that of the liquid or that the opposite happens.
The case in which the liquid rises above the tube occurs when the liquid "wets". This occurs when the adhesion forces with the walls exceed those of cohesion between the fluid molecules. In this case, the concave side is out of the fluid.
The case where the level of the liquid inside the tube is lower than the level of the liquid occurs when the liquid does not get wet. We remember that the liquid does not get wet when the cohesion forces are greater than those of adhesion. This phenomenon is called capillary depression and the concave angle is for the liquid side and is said to be convex.
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.) -7.3 kJ/s O 7.3 kJ/s -210 kJ/s e345 kJ/s 3451 kJ/s
Answer:
The pump work is 3451 kJ/s
Explanation:
Pump work (W) is calculated as
[tex] W = (h_f - h_i) \times \dot{m}[/tex]
where
[tex] h_f [/tex] is the enthalpy of water at its final state
[tex] h_i [/tex] is the enthalpy of water at its initial state
[tex]\dot{m} = 10 kg/s [/tex] is water mass flow
For liquids, properties are evaluated as saturated liquid. From the figure attached, it can be seen that
[tex] h_f = 762.81 kJ/kg [/tex]
[tex] h_i = 417.46 kJ/kg [/tex]
Replacing
[tex] W = (762.81 kJ/kg - 417.46 kJ/kg) \times 10 kg/s[/tex]
[tex] W = 3451 kJ/s [/tex]
When 1.04 g of an unknown non-electrolyte is dissolved in 50.0 g of cyclohexane, the freezing point decreased by 3.91 degrees C. If the Kfp of the solvent is 20.1 K/m, calculate the molar mass of the unknown solute.
Answer:
The molar mass of the unknown solute is 106,9 g/m
Explanation:
Cryoscopic descent formula to solve this
ΔT = Kf . m
Be careful because units in Kfp are K/m, so let's get the ΔT degrees °C in K
3,91°C = 3,91 K
It's a difference, in the end it does not matter
For example you can have 5° C as the final temperature and as initial, 1,09 °C -- ΔT is 5 - 1.09 = 3.91
What happens in Kelvin?
5°C + 273 = 278 K
1,09° C + 273 = 274,09 K
ΔT = 278 K - 274,09 K = 3,91 K
3,91 K = 20,1 K/m * m
3,91 K / 20,1 m/K = m
0,194 = m (molality)
Molality means moles from solute in 1 kg of solvent.
1kg = 1000 g
1000 g ________ 0,194 moles
50 g _________ x
x = (50 g * 13,77 moles) / 1000 g = 9,72 *10-3 moles
Moles = mass / molar mass
Molar mass = mass / moles
Molar mass = 1,04 g / 9,72 *10-3 moles
Molar mass = 106,9 g/m
65 m^3 are fed / h of benzene to a reactor, it is requested: to. What is the mass flow fed in kg / h? b. And the molar flow in mol / s? NOTE: Report the value and source of the density of benzene used in its calculations.
Answer:
a) mass flow = 56940 Kg/h
b) mass flow = 202.5 mol/s
Explanation:
∴ δ C6H6 = 876 Kg/m³,,,,,wwwcarlroth.com
⇒ 65m³/h * 876 Kg/m³ = 56940 Kg/h
⇒ 56940 Kg/h * ( 1000 g/Kg ) * ( mol/ 78.11 g) * ( h/3600s )= 202.5 mol/s
The mass flow rate of benzene is calculated using its density of 0.876 kg/m³, resulting in 56.94 kg/h. The molar flow rate is determined using the molar mass of benzene, 78.11 g/mol, which yields 20.34 mol/s.
Mass Flow and Molar Flow Calculations
To calculate the mass flow fed in kg/h for benzene, the density of benzene is required. Benzene has a density of approximately 0.876 kg/m³ at room temperature. Using this density, the mass flow can be calculated as follows:
Mass flow (kg/h) = Volumetric flow (m³/h) × Density (kg/m³)
Mass flow = 65 m³/h × 0.876 kg/m³ = 56.94 kg/h
To calculate the molar flow in mol/s, we use the molar mass of benzene which is approximately 78.11 g/mol:
Molar flow (mol/s) = Mass flow (kg/h) × (1000 g/kg) / (Molar mass (g/mol) × (3600 s/h))
Molar flow = (56.94 kg/h × 1000 g/kg) / (78.11 g/mol × 3600 s/h) = 20.34 mol/s
Note: The molecular formula of benzene is C₆H₆, not CH as mentioned in the information provided. Therefore, the combustion analysis would yield different amounts of CO₂ and H₂O than suggested by the empirical formula CH. It's important to use the correct molecular formula for accurate calculations.
Which one of the following pairs of substances illustrate the Law of Multiple Proportions? O A. 03, O2 OB.D20, H20 (D = deuterium) O C. SiO2, Sb203 OD.KCI, MgCl2 O E. PC13, PCI
Answer:
The correct answer is: PCl₃ and PCl₅
Explanation:
Law of multiple proportions, also known as the Dalton's Law, states that the when the two different chemical elements combine in order to form two or more than two chemical compounds, then the ratio of mass of the chemical element that combines with the fixed mass of other chemical element is the ratio of small whole numbers.
Example: PCl₃ and PCl₅
Calculate the volume of 48.3 g of carbon monoxide at STP. Enter your answer in the box provided. L
Explanation:
STP means standard temperature and pressure where values are as follows.
T = 273.15 K, P = 1 atm
According to ideal gas equation, PV = nRT. Since, it is given that mass is 48.3 g and we have to find the volume as follows.
n = [tex]\frac{mass}{\text{molar mass}}[/tex]
So, PV = nRT
V = [tex]\frac{mass}{\text{molar mass}} \times \frac{RT}{P}[/tex]
= [tex]\frac{48.3 g}{28 g/mol} \times \frac{0.0821 L atm/mol K \times 273.15 K}{1 atm}[/tex] (molar mass of CO = 28 g/mol)
= 38.68 L
Thus, we can conclude that volume of 48.3 g of carbon monoxide at STP is 38.68 L.
Chlorine oxide (ClO), which plays an important role in the depletion of ozone, decays rapidly according to the equation 2ClO(g) → Cl2(g) + O2(g) From the following data, determine the reaction order and calculate the rate constant of the reaction. Time (s) [ClO] (M) 4.26 × 10−3 7.73 × 10−6 4.88 × 10−3 6.27 × 10−6 5.50 × 10−3 5.28 × 10−6 6.12 × 10−3 4.56 × 10−6 6.74 × 10−3 4.01 × 10−6
The given data suggests that the decay of Chlorine oxide (ClO) is a first-order reaction. The rate constant can be calculated using the first-order rate law equation, which in this case gives a value of approximately 70000 s⁻¹.
Explanation:The reaction order is determined by the relationship between the rate of reaction and the concentration of the reactants. By observing the given data, it appears that the decay of Chlorine oxide (ClO) is halving approximately. This suggests that it could be a first-order reaction, where the rate of the reaction is directly proportional to the concentration of one reactant.
To calculate the rate constant of the reaction, we can use the first-order rate law equation: k = -1/[t]*ln([A]t/[A]0), where 'k' is the rate constant, '[t]' is the elapsed time, '[A]t' is the concentration at time 't' and '[A]0' is the initial concentration.
Using the initial and final concentrations given (at 4.26 × 10−3 s and 6.74 × 10−3 s), the equation for the rate constant becomes: k = -1/(6.74 × 10⁻³ - 4.26 × 10⁻³)*ln((4.01 × 10⁻⁶)/(7.73 × 10⁻⁶)). Calculating gives a rate constant value of approximately 70000 s-1. Remember, these values may vary depending on specific calculation approaches.
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The decomposition of ClO is determined to be first-order with a rate constant of approximately 0.263 s⁻¹. We used the method of initial rates and plotting ln[ClO] versus time to ascertain the reaction order and calculate the rate constant.
To determine the reaction order and rate constant for the decomposition of chlorine oxide (ClO) according to the equation 2ClO(g) → Cl₂(g) + O₂(g), we need to analyze the given concentration data over time.
Reaction Order Determination:We use the method of initial rates to determine the reaction order. For this we need to plot ln[ClO] versus time and [ClO]⁻¹ versus time and determine which gives the best straight line.Data Analysis:Time(s): 2.44 × 10³, [ClO] (M): 8.82 × 10⁻⁶Time(s): 2.97 × 10³, [ClO] (M): 6.97 × 10⁻⁶Time(s): 3.50 × 10³, [ClO] (M): 5.77 × 10⁻⁶Time(s): 4.03 × 10³, [ClO] (M): 4.92 × 10⁻⁶Time(s): 4.56 × 10³, [ClO] (M): 4.28 × 10⁻⁶From our calculations for ln[ClO] versus time, we get a straight line, indicating a first-order reaction:ln(8.82 × 10⁻⁶), ln(6.97 × 10⁻⁶), ln(5.77 × 10⁻⁶), ln(4.92 × 10⁻⁶), ln(4.28 × 10⁻⁶)Reaction Rate Constant (k):For a first-order reaction, the rate constant k can be determined using the equation: k = -slope of ln[ClO] vs. time.Plot the data to obtain the slope, which represents -k. Once we have plotted the data and calculated the slope, we find that k ≈ 0.263 s⁻¹.The reaction is **first-order** with respect to ClO, and the rate constant, **k**, is approximately 0.263 s⁻¹.
What is the boiling point of an aqueuous solution of
anonelectrolyte that has an osmotic pressure of 10.50 atm at 25C?
Kbof water is 0.52C/m. Assume its density is the same as
purewater.
Answer:
100.223°C is the boiling point of an aqueous solution.
Explanation:
Osmotic pressure of the solution = π = 10.50 atm
Temperature of the solution =T= 25 °C = 298 .15 K
Concentration of the solution = c
van'y Hoff factor = i = 1 (non electrolyte)
[tex]\pi =icRT[/tex]
[tex]c=\frac{\pi }{RT}=\frac{10.50 atm}{0.0821 atm L/mol K\times 298.15 K}[/tex]
c = 0.429 mol/L = 0.429 mol/kg = m
(density of solution is the same as pure water)
m = molality of the solution
Elevation in boiling point = [tex]\Delta T_b[/tex]
[tex]\Delta T_b=iK_b\times m[/tex]
[tex]\Delta T_b=T_b-T[/tex]
T = Boiling point of the pure solvent
[tex]T_b[/tex] = boiling point of the solution
[tex]K_b[/tex] = Molal elevation constant
We have :
[tex]K_b=0.52^oC/m[/tex] (given)
m = 0.429 mol/kg
T = 100° C (water)
[tex]\Delta T_b=1\times 0.52^oC/m\times 0.429 mol/kg[/tex]
[tex]\Delta T_b=0.223^oC[/tex]
[tex]\Delta T_b=T_b-T[/tex]
[tex]T_b=\Delta T_b+T=0.223^oC+100^oC=100.223^oC[/tex]
100.223°C is the boiling point of an aqueous solution.
The numeral 4.21 has three significant fire two known figures d e figure numbers are always significant
The significant figures are always:
Different from zero except there are only zeros before the point.
You can round them to the previous significant.
In scientific notation, you have one figure point two more figures.
Examples:
You have 4.21
All different from zero and only two decimals.
Those are all significant figures.
if you have 000.231555
You will shorten this to two significant figures.
Before the point, you only have zeros, so you need to keep only one of them to say its less than one.
After the point, you have a lot of figures, but you need to round this to two.
Because you have a one before the three, you'll keep the three. If you have a five or bigger number, you round it.
In this case, you'll have 0.23 with two significant figures.