Answer : The mass of helium gas in the balloon is, 1.23 grams.
Explanation : Given,
Volume of helium gas = 6.9 L
First we have to calculate moles of helium gas at STP.
As we know that, 1 mole of substance occupy 22.4 L volume of gas.
As, 22.4 L volume of helium gas present in 1 mole of helium
So, 6.9 L volume of helium gas present in [tex]\frac{6.9L}{22.4L}\times 1mole=0.308mole[/tex] of helium
Now we have to calculate the mass of helium gas.
[tex]\text{Mass of He gas}=\text{Moles of He gas}\times \text{Molar mass of He gas}[/tex]
Molar mass of He gas = 4 g/mol
[tex]\text{Mass of He gas}=0.308mol\times 4g/mol=1.23g[/tex]
Thus, the mass of helium gas in the balloon is, 1.23 grams.
To find the mass of helium in the balloon, we can use the ideal gas law equation to calculate the number of moles of helium, and then convert it to grams using the molar mass. The mass of helium in the balloon is 27.14 grams.
To find the mass of helium in the balloon, we need to know the density of helium. The ideal gas law can be used to calculate the density of a gas. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation gives us n = PV/RT. We can then use the molar mass of helium (4 g/mol) to convert the number of moles to grams. Since the volume of the balloon and the pressure are given, we can plug in the values and solve for the mass of helium.
Given:
Volume (V) = 6.9 L
Molar mass of helium = 4 g/mol
R = 8.31 J/mol · K
After performing the calculations, the mass of helium in the balloon is 27.14 grams.
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Two moles of nitrogen are initially at 10 bar and 600 K (state 1) in a horizontal piston/cylinder device. They are expanded adiabatically to 1 bar (state 2). They are then heated at constant volume to 600 K (state 3). Finally, they are isothermally returned to state 1. Assume that N 2 is an ideal gas with a constant heat capacity as given on the back flap of the book. Neglect the heat capacity of the piston/cylinder device. Suppose that heat can be supplied or rejected as illustrated below. Assume each step of the process is reversible. Calculate the net work done overall.
Answer:
Net work done overall = sum of work done for all the processes = 16,995.84 J
Explanation:
From the start, P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = ?
We can obtain V from PV = nRT; n = 2, R = 8.314 J/mol.K
V = 2 × 8.314 × 600/(1000000) = 0.009977 m³
P₁ = 10bar = 1 × 10⁶ Pa, T₁ = 600K, V₁ = 0.009977 m³
For an adiabatic process for an ideal gas,
P(V^γ) = constant
γ = ratio of specific heats = Cp/CV = 1.4,
P₂ = 1 bar = 10⁵ Pa
P₁ (V₁^1.4) = P₂ (V₂^1.4) = k
10⁶ (0.009977^1.4) = 10⁵(V₂^1.4) = 1579.75 = k
V₂ = 0.0517 m³
Work done for an adiabatic process
W = k((V₂^(1-γ)) - (V₁^(1-γ))/(1-γ)
W = 1579.75 ((0.0517^0.4) - (0.009977^0.4))/0.4
W = 582.25 J
We still need T₂
PV = nRT
T₂ = P₂V₂/nR = 100000×0.0517/(2×8.314) = 310.92K
Step 2, constant volume heating,
Work done at constant volume is 0 J.
T₂ = 310.92K, T₃ = 600K
V₂ = 0.0517 m³, V₃ = V₂ = 0.0517 m³ (Constant volume)
P₂ = 1bar, P₃ = ?
PV = nRT
P₃ = nRT₃/V₃ = 2 × 8.314 × 600/0.0517 = 192974.85 Pa = 1.93bar
Step 3, isothermally returned to the initial state.
P₃ = 1.93bar, P₄ = P₁ = 10bar
T₃ = 600K, T₄ = T₁ = 600K (Isothermal process)
V₃ = 0.0517 m³, V₄ = V₁ = 0.009977 m³
Work done = nRT In (V₃/V₁) = 2 × 8.314 × 600 In (0.0517/0.009977) = 16413.59 J
Net work done = W₁₂ + W₂₃ + W₃₁ = 582.25 + 0 + 16413.59 = 16995.84 J
Hope this helps!!
The Henry’s law constant for CO2 is 3.4 × 10−2 M/atm at 25 °C. Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a CO2 concentration of 0.10 M in a can of lemon-lime soda?
Answer: The pressure of carbon dioxide needed is 2.94 atm
Explanation:
To calculate the partial pressure of carbon dioxide, we use the equation given by Henry's law, which is:
[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]
where,
[tex]K_H[/tex] = Henry's constant = [tex]3.4\times 10^{-2}mol/L.atm[/tex]
[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas = [tex]0.10mol/L[/tex]
[tex]p_{CO_2}[/tex] = pressure of carbon dioxide = ?
Putting values in above equation, we get:
[tex]0.10mol/L=3.4\times 10^{-2}mol/L.atm\times p_{CO_2}\\\\p_{CO_2}=\frac{0.10mol/L}{3.4\times 10^{-2}mol/L.atm}=2.94atm[/tex]
Hence, the pressure of carbon dioxide needed is 2.94 atm
Write the rate law for the reaction 2A + B → C if the reaction
(1) is second order in B and overall third order, –rA = ______
(2) is zero order in A and first order in B, –rA = ______
(3) is zero order in both A and B, –rA = ______
(4) is first order in A and overall zero order. –rA = ______
Answer:
1. [tex]R=k[A]^1[B]^2[/tex]
2. [tex]R=k[B]^1[/tex]
3. [tex]R=k[A]^0[B]^0=k[/tex]
4. [tex]R=k[A]^1[B]^{-1}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
(1) is second order in B and overall third order.
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= x + 2 = 3
x = 1
Rate of the reaction =R
[tex]R=k[A]^1[B]^2[/tex]
(2) is zero order in A and first order in B.
2A + B → C
Rate of the reaction =R
[tex]R=k[A]^0[B]^1=k[B]^1[/tex]
Order of the reaction = sum of stoichiometric coefficient
= 0 + 1 = 1
(3) is zero order in both A and B .
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= 0 + 0 = 0
Rate of the reaction =R
[tex]R=k[A]^0[B]^0=k[/tex]
(4) is first order in A and overall zero order.
2A + B → C
Order of the reaction = sum of stoichiometric coefficient
= 1 + x = 0
x = -1
Rate of the reaction = R
[tex]R=k[A]^1[B]^{-1}[/tex]
A 1.00-kg block of copper at 100°C is placed in an in- sulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0°C. Find the entropy change of (a) the cop- per block, (b) the water, and (c) the universe.
Answer:
the entropy change of the copper block = - 117.29 J/K
the entropy change of the water = 138.01 J/K
the entropy change of the universe = 20.72 J/K
Explanation:
For Copper block:
the mass of copper block [tex](m_c)[/tex] = 1.00 kg
Temperature of block of copper [tex](T_c)[/tex] = 100°C
= (100+273)K
= 373K
Standard Heat capacity for copper [tex](C_c)[/tex] = 386 J/kg.K
For water:
We know our volume of liquid water to be = 4.00 L
At 0.0°C Density of liquid water = 999.9 kg/m³
As such; we can determine the mass since : [tex]density = \frac{mass}{volume}[/tex]
∴ the mass of 4.00 L of liquid water at 0.0°C will be its density × volume.
= 999.9 kg/m³ × [tex]\frac{4}{1000}m^3[/tex]
= 3.9996 kg
so, mass of liquid water [tex](m_w)[/tex] = 3.9996 kg
Temperature of liquid water [tex](T_w)[/tex] at 0.0°C = 273 K
Standard Heat Capacity of liquid water [tex](C_w)[/tex] = 4185.5 J/kg.K
Let's determine the equilibruium temperature between the copper and the liquid water. In order to do that; we have:
[tex]m_cC_c \delta T_c =m_wC_w \delta T_w[/tex]
[tex]1.00*386*(373-T_\theta)=3.996*4185.5*(T _\theta-273)[/tex]
[tex]386(373-T_\theta)=16725.26(T_\theta-273)[/tex]
[tex](373-T_\theta)=\frac{16725.26}{386} (T_\theta-273)[/tex]
[tex](373-T_\theta)=43.33 (T_\theta-273)[/tex]
[tex](373-T_\theta)=43.33 T_\theta-11829.09[/tex]
[tex]373+11829.09=43.33 T_\theta+T_\theta[/tex]
[tex]12202.09 =43.33T_\theta[/tex]
[tex]T_\theta= 275.26 K[/tex]
∴ the equilibrium temperature = 275.26 K
NOW, to determine the Entropy change of the copper block; we have:
[tex](\delta S)_{copper}=m_cC_cIn(\frac{T_\theta}{T_c} )[/tex]
[tex](\delta S)_{copper}=1.0*386In(\frac{275.26}{373} )[/tex]
[tex](\delta S)_{copper}=-117.29 J/K[/tex]
The entropy change of the water can also be calculated as:
[tex](\delta S)_{water}=m_wC_wIn(\frac{T_\theta}{T_w} )[/tex]
[tex](\delta S)_{water}=3.9996*4185.5In(\frac{275.26}{373} )[/tex]
[tex](\delta S)_{water}=138.01J/K[/tex]
The entropy change of the universe is the combination of both the entropy change of copper and water.
[tex](\delta S)_{universe}=(\delta S)_{copper}+(\delta S)_{water}[/tex]
[tex](\delta S)_{universe}=(-117.29+138.01)J/K[/tex]
[tex](\delta S)_{universe}=20.72J/K[/tex]
One reason carbon monoxide (CO) is toxic is that it binds to the blood protein hemoglobin more strongly than oxygen does. The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹ . (The units are the reciprocal of the wavelength in centimeters.) Calculate the wavelength (in nm and Å) and the frequency (in Hz) of the absorbed radiation.
Answer:
The wavelength in nm = 5120
The wavelength in A°= 51,200
The frequency of the absorbed radiation is [tex]5.859\times 10^{13} Hertz[/tex]
Explanation:
1) The wave number of the CO bond = [tex]\bar v=1953 cm^{-1}[/tex]
The wavelength corresponding to this wave number =[tex]\lambda [/tex]
[tex]\lambda =\frac{1}{\bar v}=\frac{1}{1953 cm^{-1}}=5.12\times 10^{-4} cm[/tex]
[tex]1 cm = 10^7 nm[/tex]
[tex]5.12\times 10^{-4}\times 10^7 nm=5120 nm[/tex]
[tex]1 cm = 10^8 \AA[/tex]
[tex]5.12\times 10^{-4}\times 10^8 \AA=51,200 \AA[/tex]
The wavelength in nm = 5120
The wavelength in A°= 51,200
2)
Wavelength of the wave = [tex]\lambda =5120 nm = 5120\times 10^{-9} nm[/tex]
[tex]1 nm = 10^{-9} m[/tex]
Frequecy of the wave = [tex]\nu [/tex]
[tex]\nu=\frac{c}{\lambda }[/tex]
c = Speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\nu=\frac{3\times 10^8 m/s}{5120\times 10^{-9} m}=5.859\times 10^{13} s^{-1}[/tex]
The frequency of the absorbed radiation is [tex]5.859\times 10^{13} Hertz[/tex]
The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹. The corresponding wavelength is 5120 nm and 51200 Å. The corresponding frequency is 5.855 × 10¹³ Hz.
The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹, that is, the wavenumber (ν) is 1953 cm⁻¹.
We can calculate the wavelength (λ) using the following expression.
[tex]\lambda = \frac{1}{\nu } = \frac{1}{1953cm^{-1} } = 5.120 \times 10^{-4} cm[/tex]
We will convert 5.120 × 10⁻⁴ cm to nm using the following conversion factors.
1 m = 100 cm.1 m = 10⁹ nm.[tex]5.120 \times 10^{-4} cm \times \frac{1m}{100cm} \times \frac{10^{9}nm }{1m} = 5120 nm[/tex]
We will convert 5.120 × 10⁻⁴ cm to Å using the following conversion factors.
1 m = 100 cm.1 m = 10¹⁰ Å.[tex]5.120 \times 10^{-4} cm \times \frac{1m}{100cm} \times \frac{10^{10}A }{1m} = 51200 nm[/tex]
We can calculate the frequency (f) of the absorbed radiation using the following expression.
[tex]f = \frac{c}{\lambda } = \frac{2.998 \times 10^{8}m/s }{5.120\times 10^{-6}m } = 5.855 \times 10^{13} Hz[/tex]
where,
c is the speed of lightThe bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹. The corresponding wavelength is 5120 nm and 51200 Å. The corresponding frequency is 5.855 × 10¹³ Hz.
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Name the element described in each of the following:
(a) Smallest atomic radius in Group 6A(16)
(b) Largest atomic radius in Period 6
(c) Smallest metal in Period 3
(d) Highest IE₁ in Group 4A(14)
(e) Lowest IE₁ in Period 5
(f) Most metallic in Group 5A(15)
(g) Group 3A(13) element that forms the most basic oxide
(h) Period 4 element with filled outer level
(i) Condensed ground-state electron configuration of [Ne] 3s²3p²
(j) Condensed ground-state electron configuration of [Kr] 5s²4d⁶
(k) Forms 2+ ion with electron configuration [Ar] 3d³
(l) Period 5 element that forms 3+ ion with pseudo–noble gas configuration
(m) Period 4 transition element that forms 3+ diamagnetic ion
(n) Period 4 transition element that forms 2+ ion with a halffilled d sublevel
(o) Heaviest lanthanide
(p) Period 3 element whose 2- ion is isoelectronic with Ar
(q) Alkaline earth metal whose cation is isoelectronic with Kr
(r) Group 5A(15) metalloid with the most acidic oxide
Answer:
a. Smallest atomic radius in 6A – Oxygen (O)
b. Largest atomic radius in Period 6 – Cesium (Cs)
c. Smallest metal in period 3 – Aluminum (Al)
d. Highest IE1 in Group 4A –Carbon (C)
e. Lowest IE1 in period 5 – Rubidium (Rb)
f. Most metallic in Group 5A – Bismuth (Bi) or element 115
g. Group 3A element that forms the most basic oxide – Thallium
(Tl) or element 113
h. Period 4 element with filled outer level – Krypton (Kr)
i. Condensed gound state configuration is [Ne]3s23p2 –
Germanium (Ge)
j. Condensed ground state configuration is [Kr]5s24d6 –
Ruthenium (Ru)
k. Forms 2+ ion with electron configuration of [Ar]3d3 – Vanadium
(V)
l. Period 5 element that forms 3+ ion with pseudo‐noble gas
configuration – Indium (In)
m. Period 4 transition element that forms 3+ diamagnetic ion –
Scandium (Sc)
n. Period 4 transition element that forms 2+ ion with half‐filled d
sublevel – Manganese (Mn)
o. Heaviest Lanthanide – Lutetium (Lu)
p. Period 3 element whose 2‐ ion is isoelectronic with Ar – Sulfur
(S)
q. Alkali earth metal whose cation is isoelectronic with Kr –
Strontium (Sr)
r. Group 5 A metalloid with the most acidic oxide – Arsenic (As) or
Antimony (Sb)
Elements in a group are chemically similar to each other.
The periodic table is an arrangement of elements in groups and periods. The elements in the same group share a lot of chemical similarity with each other. The elements that are in the same period only have the same number of valence shells.
The elements described by each statement is;
(a) Smallest atomic radius in Group 6A(16) - oxygen
(b) Largest atomic radius in Period 6 - cesium
(c) Smallest metal in Period 3 - Aluminium
(d) Highest IE₁ in Group 4A(14) - carbon
(e) Lowest IE₁ in Period 5 - Bismuth
(g) Group 3A(13) element that forms the most basic oxide - Thallium
(h) Period 4 element with filled outer level - krypton
(i) Condensed ground-state electron configuration of [Ne] 3s²3p² - silicon
(j) Condensed ground-state electron configuration of [Kr] 5s²4d⁶- xenon
(k) Forms 2+ ion with electron configuration [Ar] 3d³ - vanadium
(l) Period 5 element that forms 3+ ion with pseudo–noble gas configuration - Indium
(m) Period 4 transition element that forms 3+ diamagnetic ion - Scandium
(n) Period 4 transition element that forms 2+ ion with a halffilled d sublevel - Manganese
(o) Heaviest lanthanide - Lutetium
(p) Period 3 element whose 2- ion is isoelectronic with Ar - sulfur
(q) Alkaline earth metal whose cation is isoelectronic with Kr - Strontium
(r) Group 5A(15) metalloid with the most acidic oxide - nitrogen
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Determine the point groups for a. Naphthalene b. 1,8-Dichloronaphthalene c. 1,5-Dichloronaphthalene d. 1,2-Dichloronaphthalene
Answer:
a). Nepthalene is a D2h molecule because it has 3 C2 axes which are perpendicular and it has a mirror plane which is horizontal. Moreover C2 is taken as principal axis.
b). 1,8-dichloronaphthalene is a C2v molecule beccause it has 1 C2 axis, two rings are joined by the C-C bond and it also has two mirro planes.
c). 1,5-dichloronaphthalene is a C2h molecule because it has only 1 C2 axis which is pependicular to the plane, it has an inversion center and also a mirror plane which is horizontal in position.
d). 1,2-dichloronaphthalene is a Cs molecule because it has only a mirror plane.
The percent copper in an unknown copper compound is 21.90 %. What is the formula weight of the unknown compound? You may assume that one mole of the compound contains one mole of copper. Enter your answer in units of amu.
The formula weight of the unknown copper compound is approximately 13.90145 amu.
Explanation:To find the formula weight (F.W.) of the unknown copper compound, we need to consider the percent copper composition and the atomic weight of copper. The atomic weight of copper is approximately 63.55 amu.
We can use the percent composition to calculate the mass of copper in one mole of the compound:
(0.2190) * F.W. = (0.2190) * (63.55 amu) = 13.90145 amu
Since one mole of the compound contains one mole of copper, we can equate the mass of copper to the formula weight:
13.90145 amu = F.W.
Therefore, the formula weight of the unknown copper compound is approximately 13.90145 amu.
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Potassium (K) has a work function value of 2.29 eV. What is the wavelength of light required to begin to generate a current in an apparatus like the one Hertz used with potassium metal?
Answer: The wavelength of light required is [tex]5.43\times 10^{-7}m[/tex]
Explanation:
To calculate the threshold wavelength for a given work function, we use the equation:
[tex]\phi =h\nu_o[/tex]
where,
[tex]\phi[/tex] = work function of the potassium metal = [tex]2.29eV=3.66\times 10^{-19}J[/tex] (Conversion factor: [tex]1eV=1.6\times 10^{-19}[/tex] )
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\nu_o=\frac{c}{\lambda _o}[/tex]
c = speed of light = [tex]3\times 10^9m/s[/tex]
[tex]\lambda_o[/tex] = wavelength of light
Putting values in above equation:
[tex]3.66\times 10^{-19}J=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{\lambda_o}\\\\\lambda_o=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.66\times 10^{-19}J}=5.43\times 10^{-7}m[/tex]
Hence, the wavelength of light required is [tex]5.43\times 10^{-7}m[/tex]
g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial concentration of A is 0.163 M, how many minutes will it take for the concentration of A to be 66.8% of the initial concentration?
Answer:
In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.
Explanation:
The equation used to calculate the constant for first order kinetics:
[tex]t_{1/2}=\frac{0.693}{k}}[/tex] .....(1)
Rate law expression for first order kinetics is given by the equation:
[tex]t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}[/tex] ......(2)
where,
k = rate constant
[tex]t_{1/2}[/tex] =Half life of the reaction = [tex]2.42\times 10^3 s[/tex]
t = time taken for decay process = ?
[tex][A_o][/tex] = initial amount of the reactant = 0.163 M
[A] = amount left after time t = 66.8% of [tex][A_o][/tex]
[A]=[tex]\frac{66.8}{100}\times 0.163 M=0.108884 M[/tex]
[tex]k=\frac{0.693}{2.42\times 10^3 s}[/tex]
[tex]t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}[/tex]
t = 1,409.19 s
1 minute = 60 sec
[tex]t=\frac{1,409.19 }{60} min=23.49 min[/tex]
In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.
Consider a gas mixture in a 2.00-dm3 flask at 27.0 ºC. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole percent. a) 1.00 g H2 and 1.00 g O2 b) 1.00 g N2 and 1.00 g O2 c) 1.00 g CH4 and 1.00 g NH3
Explanation:
a)
Moles of hydrogen gas = [tex]n_1=\frac{1.00 g}{2 g/mol}=0.5 mol[/tex]
Moles of oxygen gas = [tex]n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol[/tex]
Total moles in container = [tex]n=n_1+n_2=0.5 mol+0.03125 mol=0.53125 mol[/tex]
Total pressure of mixture = P
Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]
Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]
[tex]1 dm^3=1 L[/tex]
[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )
[tex]P=\frac{0.53125 mol\times 0.0821 atm L/mol K\times 300 K}{2.00 L}=6.54 atm[/tex]
Partial pressure of the hydrogen gas :
= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]
[tex]=6.54 atm\times \frac{0.5 mol}{0.53125 mol}=6.16 atm[/tex]
Partial pressure of the oxygen gas :
= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]
[tex]=6.54 atm\times \frac{0.03125 mol}{0.53125 mol}=0.38 atm[/tex]
hydrogen
= [tex]\frac{n_1}{n}\times 100=\frac{0.5 mol}{0.53125 }\times 100[/tex]
= 94.12%
oxygen :
= [tex]\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.53125 mol}\times 100[/tex]
= 5.88%
b)
Moles of nitrogen gas = [tex]n_1=\frac{1.00 g}{28 g/mol}=0.03571 mol[/tex]
Moles of oxygen gas = [tex]n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol[/tex]
Total moles in container = [tex]n=n_1+n_2=0.03571 mol+0.03125 mol=0.06696 mol[/tex]
Total pressure of mixture = P
Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]
Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]
[tex]1 dm^3=1 L[/tex]
[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )
[tex]P=\frac{0.06696 mol\times 0.0821 atm L/mol K\times 300 K}{2.00 L}=0.82 atm[/tex]
Partial pressure of the nitrogen gas :
= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]
[tex]=0.82 atm\times \frac{0.03571 mol}{0.06696 mol}=0.44 atm[/tex]
Partial pressure of the oxygen gas :
= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]
[tex]=0.82 atm\times \frac{0.03125 mol}{0.06696 mol}=0.38 atm[/tex]
Composition of each in mole percent :
nitrogen
= [tex]\frac{n_1}{n}\times 100=\frac{0.03571 mol}{0.06696 }\times 100[/tex]
= 53.33%
oxygen :
= [tex]\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.06696 mol}\times 100[/tex]
= 46.67%
c)
Moles of methane gas = [tex]n_1=\frac{1.00 g}{16 g/mol}=0.0625 mol[/tex]
Moles of ammonia gas = [tex]n_2=\frac{1.00 g}{17g/mol}=0.0588 mol[/tex]
Total moles in container = [tex]n=n_1+n_2=0.0625 mol+0.0588 mol=0.1213 mol[/tex]
Total pressure of mixture = P
Temperature of the mixture = [tex]T = 27^oC =27+273K= 300 K[/tex]
Volume of the container in which mixture is kept = [tex]2.00 dm^3 =2.00 L[/tex]
[tex]1 dm^3=1 L[/tex]
[tex]P=\frac{nRT}{V}[/tex] (from ideal gas equation )
[tex]P=\frac{0.1213 mol\times 0.0821 atm L/mol K\times 300 K}{2.00 L}=1.49 atm[/tex]
Partial pressure of the methane gas :
= [tex]p_1=P\times \chi_1=P\times \frac{n_1}{n}[/tex]
[tex]=1.49 atm\times \frac{0.0625 mol}{0.1213mol}=0.77 atm[/tex]
Partial pressure of the ammonia gas :
= [tex]p_2=P\times \chi_2=P\times \frac{n_2}{n}[/tex]
[tex]=1.49 atm\times \frac{0.0588 mol}{0.1213mol}=0.72 atm[/tex]
Composition of each in mole percent :
Methane :
= [tex]\frac{n_1}{n}\times 100=\frac{0.0625 mol}{0.1213mol}\times 100[/tex]
= 51.52%
Ammonia
= [tex]\frac{n_2}{n}\times 100=\frac{0.0588 mol}{0.1213mol}\times 100[/tex]
= 48.47%
To calculate the partial pressure of each gas and the composition of the mixture, we can use the ideal gas law. By dividing the mass of each gas by its molar mass, we can calculate the number of moles. The total pressure can be obtained by summing up the partial pressures, and the mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.
Explanation:To calculate the partial pressure of each gas, we need to use the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. To calculate the total pressure, we can sum up the partial pressures of each gas. The mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.
a) 1.00 g H2 and 1.00 g O2:
To calculate the number of moles, we divide the mass of each gas by its molar mass. The molar mass of H2 is 2 g/mol and the molar mass of O2 is 32 g/mol. So, the number of moles of H2 is 1 g / 2 g/mol = 0.5 mol, and the number of moles of O2 is 1 g / 32 g/mol = 0.03125 mol. The total number of moles is 0.5 mol + 0.03125 mol = 0.53125 mol.
The partial pressure of H2 can be calculated by multiplying the number of moles of H2 by the gas constant R and the temperature in Kelvin, and then dividing by the volume. The same process can be applied to calculate the partial pressure of O2. Finally, the total pressure can be calculated by summing up the partial pressures. The mole percent composition can be calculated by dividing the number of moles of each gas by the total number of moles and multiplying by 100.
b) 1.00 g N2 and 1.00 g O2:
Using the same process as before, we can calculate the number of moles of N2 and O2. The molar mass of N2 is 28 g/mol, so 1 g of N2 is equal to 1 g / 28 g/mol = 0.03571 mol of N2. The molar mass of O2 is 32 g/mol, so 1 g of O2 is equal to 1 g / 32 g/mol = 0.03125 mol of O2. The total number of moles is 0.03571 mol + 0.03125 mol = 0.06696 mol.
Using the ideal gas law, we can calculate the partial pressures of N2 and O2, the total pressure, and the mole percent composition.
c) 1.00 g CH4 and 1.00 g NH3:
Following the same calculations as above, the molar mass of CH4 is 16 g/mol, so 1 g of CH4 is equal to 1 g / 16 g/mol = 0.0625 mol of CH4. The molar mass of NH3 is 17 g/mol, so 1 g of NH3 is equal to 1 g / 17 g/mol = 0.05882 mol of NH3. The total number of moles is 0.0625 mol + 0.05882 mol = 0.12132 mol.
By using the ideal gas law, we can calculate the partial pressures of CH4 and NH3, the total pressure, and the mole percent composition.
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Consider the second-order reaction: 2HI(g)→H2(g)+I2(g). Use the simulation to find the initial concentration [HI]0and the rate constant k for the reaction. What will be the concentration of HI after t = 1.01×10^10 s ([HI]t) for a reaction starting under the condition in the simulation?
Explanation:
The given reaction equation is as follows.
[tex]2HI(g) \rightarrow H_{2}(g) + I_{2}(g)[/tex]
[tex]\frac{-d[HI]}{dt} = k[HI]^{2}[/tex]
[tex]-\int_{[HI]_{o}}^{[HI]_{t}} \frac{d[HI]}{[HI]^{2}} = k \int_{o}^{t} dt[/tex]
[tex]-[\frac{-1}{[HI]}]^{[HI]_{t}}_{[HI]_{o}} = kt[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{[HI]_{o}} = kt[/tex] .......... (1)
where, [tex][HI_{o}][/tex] = Initial concentration
[tex][HI]_{t}[/tex] = concentration at time t
k = rate constant
t = time
Now, we will calculate the initial concentration of HI as follows.
Initial rate = [tex]1.6 \times 10^{-7} mol/sec[/tex]
k = [tex]6.4 \times 10^{-9}[/tex]
R = [tex]k[HI]^{2}_{o}[/tex]
[tex][HI]^{2}_{o} = \frac{R}{k}[/tex]
= [tex]\frac{1.6 \times 10^{-7}}{6.4 \times 10^{-9}}[/tex]
[tex][HI]_{o}[/tex] = 5 M
Now, we will calculate the concentration of [tex][HI]_{t}[/tex] at t = [tex]1.01 \times 10^{10}[/tex] sec as follows.
Using equation (1) as follows.
k = [tex]6.4 \times 10^{-9}[/tex]
[tex]\frac{1}{[HI]_{t}} - \frac{1}{5}[/tex] = [tex](6.4 \times 10^{-9}) \times 1.01 \times 10^{10}[/tex]
[tex]\frac{1}{[HI]_{t}}[/tex] = 64.44
[tex][HI]_{t}[/tex] = 0.0155 M
Thus, we can conclude that concentration of HI at t = [tex]1.01 \times 10^{10}[/tex] sec is 0.0155 M.
A reception subservient on the second and first-order reactants is called a second-order reaction.
The correct answer is:
The concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
The equation according to the question is:
2 HI (g) ⇒ H₂ (g) + I₂ (g)
[tex]\dfrac{\text{-d}\left[\begin{array}{ccc}\text{HI}\end{array}\right] }{\text{dt}} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2}[/tex]
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{t}} } - \dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] _{\text{o}} } &= \text{kt}[/tex] .......equation (1)
Where, the initial concentration can be represented as: [tex]\left[\begin{array}{ccc}\text{HI}_{o} \end{array}\right][/tex]
The concentration at time t = [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex]
Rate constant will be = k
The time will be = t
The initial concentration of HI can be calculated as:
The initial rate = 1.6 × 10⁻⁷ mol/sec
k = 6.4 × 10⁻⁹
[tex]\text{R} & = \text{k} \left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
[tex]\dfrac{\text{R}}{\text{k}} & =\left[\begin{array}{ccc}\text{HI}\end{array}\right] ^{2} _{0}[/tex]
= [tex]\dfrac{1.6 \times 10^{-7} }{6.4 \times 10^{-9} }[/tex]
[tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right]\end{array}\right] _{0} &= 5 \;\text{M}[/tex]
To calculate the concentration [tex]\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}[/tex] at time (t) = 1.01 × 10 ¹⁰ sec.
Now, using the above equation: (1)
k = 6.4 × 10⁻⁹
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} - \dfrac{1}{5}[/tex]
= (6.4 × 10⁻⁹) × 1.01 × 10¹⁰
[tex]\dfrac{1}{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 64.44[/tex]
[tex]{\left[\begin{array}{ccc}\text{HI}\end{array}\right] \text{t}} = 0.0155 \;\text{M}[/tex]
Therefore, concentration of HI at t = 1.01 × 10¹⁰ sec is 0.0155 M.
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Draw the partial (valence-level) orbital diagram, and write the symbol, group number, and period number of the element:
(a) [Ne] 3s²3p⁵
(b) [Ar] 4s²3d¹⁰4p³
Answer:
a) Element = Chlorine
b) Element = Arsenic
Explanation:
The knowledge of Orbitals and Quantum number and the electronic configuration is applied as shown in the analysis in the attached file.
Naturally occurring iodine has an atomic mass of 126.9045 amu. A 12.3849-g sample of iodine is accidentally contaminated with 1.00070 g of 129I, a synthetic radioisotope of iodine used in the treatment of certain diseases of the thyroid gland. The mass of 129I is 128.9050 amu. Find the apparent "atomic mass" of the contaminated iodine.
The apparent "atomic mass" of the contaminated iodine is approximately 1111.9285 amu.
To determine the apparent "atomic mass" of contaminant iodine, we need to take into account the contribution of both the naturally occurring iodine ([tex]\rm ^1^2^7I[/tex]) and the man-made radioisotope ([tex]\rm ^1^2^9I[/tex]).
The formula for calculating the atomic mass is:
Atomic mass = (mass of isotope 1 * abundance of isotope 1) + (mass of isotope 2 * abundance of isotope 2) + ...
Lets calculating the abundances of [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex]:
Abundance of [tex]\rm ^1^2^7I[/tex] = 1 - Abundance of [tex]\rm ^1^2^9I[/tex]
Given:
Mass of [tex]\rm ^1^2^7I[/tex] = 126.9045 amu
Mass of [tex]\rm ^1^2^9I[/tex] = 128.9050 amu
We have two isotopes: [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex]. The masses of these isotopes and their abundances need to be considered:
Atomic mass = (mass of [tex]\rm ^1^2^7I[/tex] * abundance of [tex]\rm ^1^2^7I[/tex]) + (mass of [tex]\rm ^1^2^9I[/tex]* abundance of [tex]\rm ^1^2^9I[/tex])
We need to calculate the abundances of [tex]\rm ^1^2^7I[/tex] and [tex]\rm ^1^2^9I[/tex] before we can calculate the atomic mass.
Mass of contaminated iodine sample = 12.3849 g + 1.00070 g = 13.3856 g
Now we can calculate the abundances:
Abundance of [tex]\rm ^1^2^7I[/tex] = (mass of [tex]I-127[/tex] in sample) / (total mass of sample)
Abundance of [tex]\rm ^1^2^7I[/tex] = (12.3849 g * 126.9045 amu) / (13.3856 g) = 117.0997 amu / 13.3856 g ≈ 8.7411
Abundance of [tex]\rm ^1^2^9I[/tex] = (mass of I-129 in sample) / (total mass of sample)
Abundance of [tex]\rm ^1^2^9I[/tex]= (1.00070 g * 128.9050 amu) / (13.3856 g) = 128.9050 amu / 13.3856 g ≈ 9.6175
We can calculate the atomic mass:
Atomic mass = (126.9045 amu * 8.7411) + (128.9050 amu * 9.6175) ≈ 1111.9285
Therefore, the apparent "atomic mass" of the contaminated iodine is approximately 1111.9285 amu.
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The apparent atomic mass of the contaminated iodine is 126.9045 amu.
Explanation:
To calculate the apparent atomic mass of the contaminated iodine, we can use the formula:
Apparent atomic mass = (mass of naturally occurring iodine * atomic mass of naturally occurring iodine + mass of synthetic radioisotope * atomic mass of synthetic radioisotope) / total mass of contaminated iodine
Given:
Mass of naturally occurring iodine = 12.3849 gAtomic mass of naturally occurring iodine = 126.9045 amuMass of synthetic radioisotope = 1.00070 gAtomic mass of synthetic radioisotope = 128.9050 amuSubstituting the given values into the formula:
Apparent atomic mass = (12.3849 g * 126.9045 amu + 1.00070 g * 128.9050 amu) / (12.3849 g + 1.00070 g)
Simplifying the expression:
Apparent atomic mass = (1570.6078055 + 128.905070) / 13.3856
Apparent atomic mass = 1699.5128755 / 13.3856
Apparent atomic mass = 126.9045 amu
Therefore, the apparent atomic mass of the contaminated iodine is 126.9045 amu.
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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq), as described by the chemical equationMnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 115 mL Cl2(g)115 mL Cl2(g) at 25 °C and 805 Torr805 Torr?
Answer:
4.98 × 10⁻³ mol
Explanation:
Given data for Cl₂
Volume (V): 115 mL = 0.115 LPressure (P): 1.06 atm[tex]805torr.\frac{1atm}{760torr} =1.06atm[/tex]
Temperature (T): 25°C + 273.15 = 298 KFirst, we will calculate the moles of Cl₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.06 atm × 0.115 L / (0.0821 atm.L/mol.K) × 298 K
n = 4.98 × 10⁻³ mol
Let's consider the balanced equation.
MnO₂(s) + 4 HCl(aq) ⟶ MnCl₂(aq) + 2 H₂O(l) + Cl₂(g)
The molar ratio of MnO₂ to Cl₂ is 1:1. The required moles of MnO₂ are 4.98 × 10⁻³ moles.
The volume of 1 kg of helium in a piston-cylinder device is initially 5 m3 . Now helium is compressed to 2 m3 while its pressure is maintained constant at 180 kPa. Determine the work required to compress helium, in kJ.
Answer: The work required to compress helium gas is 540 kJ
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]
W = amount of work done = ?
P = pressure = 180 kPa
[tex]V_1[/tex] = initial volume = [tex]5m^3[/tex]
[tex]V_2[/tex] = final volume = [tex]2m^3[/tex]
Putting values in above equation, we get:
[tex]W=-180kPa\times (2-5)m^3=540kPa.m^3[/tex]
To convert this into joules, we use the conversion factor:
[tex]1kPa.m^3=1kJ[/tex]
So, [tex]540kPa.m^3=540kJ[/tex]
The positive sign indicates that work is done by the system.
Hence, the work required to compress helium gas is 540 kJ
The work required to compress 1 kg of helium from 5 m³ to 2 m³ at a constant pressure of 180 [tex]K_p_a[/tex] is 540 kJ.
This problem involves the calculation of work done during the compression of helium gas in a piston-cylinder device at a constant pressure. The work required to compress the gas can be determined using the formula for work done by a gas at constant pressure:
Work (W) = P x ΔV
Where:
P is the pressure, which is 180 [tex]K_p_a[/tex] (or 180 kJ/m³).ΔV (change in volume) is the difference in volume, which is the initial volume minus the final volume.Thus, ΔV = [tex]V_i_n_i_t_i_a_l[/tex] - [tex]V_f_i_n_a_l[/tex]
Given:
Initial volume, [tex]V_i_n_i_t_i_a_l[/tex] = 5 m³Final volume, [tex]V_f_i_n_a_l[/tex] = 2 m³Therefore:
ΔV = 5 m³ - 2 m³ = 3 m³
Substituting the values into the work formula:
Work (W) = 180 [tex]K_p_a[/tex] x 3 m³ = 540 kJ
Therefore, the work required to compress the helium is 540 kJ.
Write the appropriate symbol for each of the following isotopes: (a) Z = 19, A = 41 (b) Z = 46, A = 106 (c) Z = 52, A = 125 (d) Z = 38, A = 88
Answer:
(a) ⁴¹₁₉K
(b) ¹⁰⁶₄₆Pd
(c) ¹²⁵₅₂Te
(d) ⁸⁸₃₈Sr
Explanation:
The identity of an element is its atomic number, by convention we write the atomic mass as superscript and the and the atomic number as subscript to the left of the element .
(a) Z = 19 A = 41 symbol: ⁴¹₁₉K
(b) Z = 46 A = 106 symbol: ¹⁰⁶₄₆Pd
(c) Z = 52 A = 125 symbol: ¹²⁵₅₂Te
(d) Z = 38 A = 88 symbol: ⁸⁸₃₈Sr
The isotopes correspond to Potassium-41 (41K), Palladium-106 (106Pd), Tellurium-125 (125Te), and Strontium-88 (88Sr). Z and A denote atomic and mass numbers, which help in identifying particular isotopes.
Explanation:The appropriate symbol for each of these isotopes would be:
(a) Z = 19, A = 41, which corresponds to Potassium-41 (41K), (b) Z = 46, A = 106, this stands for Palladium-106 (106Pd), (c) Z = 52, A = 125, this refers to Tellurium-125 (125Te), (d) Z = 38, A = 88, the symbol for this is Strontium-88 (88Sr).
Z and A represent atomic and mass numbers respectively. These isotopes are written with a mass number (A) preceding the symbol and atomic number (Z) is typically understood from the symbol itself.
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A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What is the final pressure in atm?
Answer:
0.681 atm
Explanation:
To solve this problem, we make use of the General gas equation.
Given:
P1 = 785 torr
V1 = 2L
T1 = 37= 37 + 273.15 = 310.15K
P2 = ?
V2 = 3.24L
T2 = 58 = 58+273.15 = 331.15K
P1V1/T1 = P2V2/T2
Now, making P2 the subject of the formula,
P2 = P1V1T2/T1V2
P2 = [785 * 2 * 331.15]/[310.15 * 3.24]
P2 = 515.715 Torr
We convert this to atm: 1 torr = 0.00132 atm
515.715 Torr = 515.715 * 0.00132 = 0.681 atm
A 20.3 mass % aqueous solution of iron(III) chloride has a density of 1.280 g/mL. Calculate the molality of the solution. Give your answer to 2 decimal places.
Answer:
Molality for the solution is 1.57 m
Explanation:
Molality is mol of solute in 1kg of solvent.
20.3 % by mass means that 20.3 g of solute (FeCl₃) are contained in 100 g of solution..
Let's determine the mass of solvent.
Mass of solution = Mass of solvent + Mass of solute
100 g = Mass of solvent + 20.3 g
100 g - 20.3 g = Mass of solvent → 79.7 g
Let's convert the mass in g to kg
79.7 g . 1kg / 1000 g = 0.0797 kg
Let's determine the moles of solute (mass / molar mass)
20.3 g / 162.2 g/mol = 0.125 mol
Molality = 0.125 mol / 0.0797 kg → 1.57 m
Final answer:
The molality of the 20.3% aqueous solution of iron(III) chloride is 1.60 m.
Explanation:
This question demands basic understanding of molatility.
To calculate the molality of the solution, we need to determine the moles of solute (iron(III) chloride) and the mass of the solvent (water).
First, convert the mass percent to grams of solute:
Mass of solute = (20.3%)(1.280 g/mL)(1000 mL) = 260.48 g
Next, calculate the moles of solute:
Moles of solute = (260.48 g)/(162.2 g/mol) = 1.603 mol
Finally, calculate the molality:
Molality = (1.603 mol)/(1 kg) = 1.60 m
Therefore, The molality of the 20.3% aqueous solution of iron(III) chloride is 1.60 m.
The observed rotation of 2.0 g of a compound in 50 mL of solution in a polarimeter tube 20 cm long is +138°. What is the specific rotation of the compound?
Answer:
+1725
Explanation:
The specific rotation can be calculated using the formula:
Specific Rotation = Observed Rotation / (l * C)Where l is the path length in decimeters (20 cm / 10), 2 dm; and C is the concentration of the compound in mg/mL (2.0g / 50mL), 0.04 g/mL.
Putting the data we're left with:
Specific Rotation = +138 / (2 * 0.04) = +1725The specific rotation of the compound is [tex]\( +1725\° \text{ dm}^{-1} \text{ (g/mL)}^{-1} \).[/tex]
The specific rotation of the compound is given by the formula:
[tex]\[ [\alpha] = \frac{\alpha}{l \cdot c} \][/tex]
where:
- [tex]\( [\alpha] \)[/tex] is the specific rotation,
- [tex]\( \alpha \)[/tex] is the observed rotation in degrees,
- l is the length of the polarimeter tube in decimeters,
- c is the concentration of the solution in grams per milliliter.
Given:
- The observed rotation [tex]\( \alpha = +138\° \)[/tex],
- The length of the polarimeter tube l = 20cm= 2 dm (since 1 dm = 10 cm),
- The mass of the compound m = 2.0 g,
- The volume of the solution V = 50mL.
First, we need to calculate the concentration c of the solution:
[tex]\[ c = \frac{\text{mass of compound}}{\text{volume of solution}} = \frac{2.0 \text{ g}}{50 \text{ mL}} = 0.04 \text{ g/mL} \][/tex]
Now we can calculate the specific rotation:
[tex]\[ [\alpha] = \frac{\alpha}{l \cdot c} = \frac{+138\°}{2 \text{ dm} \cdot 0.04 \text{ g/mL}} = \frac{+138\°}{0.08 \text{ dm} \cdot \text{g/mL}} \]\\[/tex]
[tex]\[ [\alpha] = \frac{+138\°}{0.08} \] \[ [\alpha] = +1725\° \text{ dm}^{-1} \text{ (g/mL)}^{-1} \][/tex]
Therefore,
The answer is: [tex]+1725\° \text{ dm}^{-1} \text{ (g/mL)}^{-1}.[/tex]
A liquid is added to a buret, and the initial measurement is found to be 0.75 mL. After liquid has been added to the flask from the buret, the buret reads 15.20 mL. How much liquid (in mL) was dispensed from the buret, with the correct number of significant figures?
Answer:
14.45 mL
Explanation:
The rule apply for the addition and subtraction for significant digits is :
The least precise number present after the decimal point determines the number of significant figures in the answer.
Initial Burette reading = 0.75 mL ( 2 significant digits)
Final Burette reading = 15.20 mL ( 4 significant digits)
Liquid dispensed = 15.20 mL - 0.75 mL = 14.45 mL ( Answer to two decimal places )
To find the amount of liquid dispensed from the buret, subtract the initial volume (0.75 mL) from the final volume (15.20 mL), resulting in 14.45 mL of liquid dispensed. Ensure that the answer is reported with the same level of precision as the measurements from the buret.
Explanation:To calculate the amount of liquid dispensed from the buret, you subtract the initial measurement from the final measurement. In this case, the initial measurement is 0.75 mL and the final measurement after the liquid is added to the flask is 15.20 mL. Therefore, the amount of liquid dispensed is 15.20 mL - 0.75 mL = 14.45 mL.
Burets are commonly used in titration analyses and allow volume measurements to the nearest 0.01 mL, which is why our final answer should be reported with the same level of precision as the buret readings. According to the proper significant figures rules, the result should be reported with the same number of decimal places as the measurement with the fewest decimal places. Since the readings from the buret are given to two decimal places (0.75 mL and 15.20 mL), our calculated volume of dispensed liquid also should be reported to two decimal places as 14.45 mL.
The half-life of a radioactive isotope is the amount of time it takes for a quantity of radioactive material to decay to one-half of its original amount.
Answer: The statement is true
Explanation:
The half-life of a radioactive isotope is the time taken for half of the total number of atoms in a given sample of the isotope to decay.
For instance
The half-life of radium is 1622 years. This means that if we have 1000 radium atoms at the beginning, then at the end of 1622 years, 500 atoms would have disintegrated, leaving 500 undecayed radium atoms
Thus, the statement is true
A sample of uranium ore contains 6.73 mg of 238U and 3.22 mg of 206Pb. Assuming all of the lead arose from the decay of the uranium and that the half-life of 238U is 4.51 x 109years, determine the age of the ore
Answer:
The age of the ore is 4.796*10^9 years.
Explanation:
To solve this question, we use the formula;
A(t) =A(o)(1/2)^t/t1/2
where;
A(t) =3.22mg
A(o) = 6.731mg
t1/2 = 4.51*70^9 years
t = age of the ore
So,
A(t) =A(o)(1/2)^t/t1/2
3.22 = 6.73 (1/2)^t/4.51*10^9
Divide both sides by 6.73
3.22/6.73= (1/2)^t/4.51*10^9
0.47825= (0.5)^t/4.51*10^9
Log 0.4785 = t/4.51*10^9 • log 0.5
Log 0.4785/log 0.5 • 4.51*10^9 = t
t = 1.0634 * 4.51*10^9
t = 4.796*10^9
So therefore, the age of the ore is approximately 4.796*10^9 years.
(a) Find the concentration of electrons and holes in a sample of germanium that has a concentration of donor atoms equal to 1015 cm−3 . Is the semiconductor n-type or p-type? (b) Repeat part (a) for silicon.
Answer:
a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.
b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.
For clear view of the answers: Please refer to calculation 5 in the attachments section.
Explanation:
So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).
Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:
So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.
So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o). (p subscript o)
Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.
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(a) The germanium with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type. (b) For silicon with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type.
(a) In n-type semiconductors like germanium, donor atoms introduce excess electrons into the crystal lattice. To find the concentration of electrons and holes, we can assume that all the donor atoms ionize and release one extra electron each. Therefore, the concentration of electrons will be equal to the donor atom concentration, which is 10^15 cm^-3. In this case, there are very few holes since most of the electrons are engaged in conduction. So, the semiconductor is n-type because it has an excess of electrons.(b) Silicon behaves similarly to germanium but has a larger energy gap. In silicon with a donor atom concentration of 10^15 cm^-3, the concentration of electrons will also be 10^15 cm^-3 because each donor atom donates one electron. As in germanium, there are very few holes compared to electrons, making it an n-type semiconductor due to the excess of electrons generated by the donor atoms.For more such questions on atoms
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Write an equation that represents the action in water of formic acid (HCOOH)(HCOOH) as a Brønsted–Lowry acid. Express your answer as a chemical equation.
Answer:
HCOOH + H2O <===> HCOO- + H3O+
Explanation:
On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (u) 46Ti 70.600% 45.95263 48Ti 11.900% 47.94795 50Ti 17.500% 49.94479 What is the average atomic mass of titanium on that planet? average atomic mass = 47.867
Final answer:
The average atomic mass of titanium on the hypothetical planet is calculated by using the weighted average of the abundances and masses of its isotopes, resulting in 46.8989 amu.
Explanation:
The average atomic mass of titanium on that hypothetical planet can be calculated by multiplying the abundance of each isotope by its mass (in atomic mass units, amu), then summing these products. The calculation will look as follows:
(0.70600 × 45.95263 amu) for 46Ti(0.11900 × 47.94795 amu) for 48Ti(0.17500 × 49.94479 amu) for 50TiTo find the average atomic mass, we add these values together to get the sum which should give us the average atomic mass of titanium on the planet.
The correct calculation would be:
(0.70600 × 45.95263 amu) + (0.11900 × 47.94795 amu) + (0.17500 × 49.94479 amu)= (32.45785 amu) + (5.70081 amu) + (8.74024 amu)= 46.8989 amuHence, the average atomic mass of titanium for this planet is 46.8989 amu. It's important to note that the average atomic mass of an element is the weighted average of all the isotopes of that element.
Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis.
True or false?
Answer: The given statement is true.
Explanation:
When we increase the amount of solvent which is water in this case then it means there will occur an increase in the molecules. Hence, there will be more number of collisions to take place with increase in number of molecules.
Therefore, more is the amount of interaction taking place between the molecules of a solution more will be its rate of hydrolysis.
Thus, we can conclude that the statement increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis, is true.
Draw a structure containing only carbon and hydrogen that is a stable alkyne of five carbons containing a ring.
Answer:
Ethynylcyclopropane is the stable isomer for given alkyne.
Explanation:
In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,
CₙH₂ₙ₋₂
Putting value on n = 5,
C₅H₂.₅₋₂
C₅H₈
Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.
C₅H₈ ------------(-2 H) ----------> C₅H₆
Hence, the molecular formula for given compound is C₅H₆
Below, 4 different isomers with molecular formula C₅H₆ are attached.
The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.
Compounds 3-ethylcycloprop-1-yne, cyclopentyne and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.
What is the molarity of a 10.5 %% by mass glucose (C6H12O6C6H12O6) solution? (The density of the solution is 1.03 g/mLg/mL .) Express your answer to three significant figures.
The molarity of the 10.5% by mass glucose solution is 0.600 M.
Explanation:To find the molarity of the glucose solution, we need to determine the number of moles of glucose present in the solution first. We can use the percent by mass to calculate this.
Given that the solution is 10.5% by mass, we know that 10.5 grams of glucose is present in a 100 gram solution. We can convert this to moles by dividing the mass of glucose by its molar mass, which is 180.16 g/mol.
So, the number of moles of glucose is 10.5 g / 180.16 g/mol = 0.0583 mol. To find the molarity, we divide the number of moles by the volume of the solution in liters. The volume of the solution can be determined by multiplying the density of the solution by its mass: 100 g / (1.03 g/mL) = 97.09 mL = 0.0971 L.
Therefore, the molarity of the glucose solution is 0.0583 mol / 0.0971 L = 0.600 M.
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The molarity of a 10.5% by mass glucose solution with a density of 1.03 g/mL can be calculated by assuming a 100 g sample of the solution. The mass of glucose per liter is found to be 108.1395 g, and with the molar mass of glucose, the molarity is determined to be 0.600 M.
The question pertains to determining the molarity of a 10.5% by mass glucose solution with a given density of 1.03 g/mL. To calculate the molarity, we need to use the given mass percentage and density to find out how many moles of glucose are present in a liter of solution.
First, assume you have 100 g of this solution. Because it's a 10.5% by mass solution, this means there are 10.5 g of glucose (C6H12O6) and 89.5 g of water in the mixture.
Using the density, we find the volume of 100 g of solution:
100g / 1.03g/mL = 97.09 mL
Because we want to know the molarity per liter, it's important to work with a liter of the solution:
(1000 mL/L) / (97.09 mL) = 10.299 L^-1 multiplication factor
Now, we will use the multiplication factor to scale up the mass of glucose to what would be in one liter:
10.5 g * 10.299 = 108.1395 g glucose per liter
The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol, so the number of moles in one liter would be:
108.1395 g / 180.16 g/mol = 0.600 mol/L
Therefore, the molarity of the glucose solution is 0.600 M.
Citrus fruits are rich in citric acid, a compound with three COOH groups. Explain the following: a. The first pKa (for the COOH group in the center of the molecule) is lower than the pKa of acetic acid. b. The third pKa is greater than the pKa of acetic acid.
Answer:
a. the first pKa is lower because of the presence of an electron withdrawing 'OH' group attached to the carbon that is directly attached to it.
b. the third pKa is greater than that of acetic acid because of the presence of an electron donating methyl group which is directly attached to it.
Explanation:
Induction or Inductive effect is the electronic effects an atom or a group of atoms exert on a compound or a portion of a compound, which could either be electron donating or electron withdrawing, thereby affecting its acidity or basicity.
Electronegativity confers acidity on a compound. In a, the OH group withdraws electrons from the COOH group, conferring more electronegativity on the middle COOH group, thereby reducing the pKa and thus increasing acidity.
In b, the electron donating effect of the methyl group, decreases the electronegativity of the COOH group,p thereby increasing the pKa which also means decreased acidity.
The first pKa of citric acid is lower than acetic acid due to the presence of three COOH groups. The third pKa of citric acid is greater than acetic acid due to steric hindrance caused by the presence of two COOH groups at either end of the molecule.
Explanation:a. The first pKa of citric acid is lower than the pKa of acetic acid due to the presence of three COOH groups. The proximity of the COOH groups in citric acid allows for easier liberation of the H+ ions, resulting in a lower pKa value. This is supported by the concept of resonance stabilization. In contrast, acetic acid has only one COOH group, resulting in a higher pKa value.
b. The third pKa of citric acid is greater than the pKa of acetic acid due to steric hindrance caused by the presence of two COOH groups at either end of the molecule. The close proximity of these two groups makes it difficult for the liberation of the H+ ion, resulting in a higher pKa value compared to acetic acid.
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