Answer:
Mole fraction of alcohols in liquid phase [tex]x_1=0.2727\& x_2=0.7273[/tex].
Mole fraction of alcohols in vapor phase [tex]y_1=0.1468\& y_2=0.8516[/tex].
Explanation:
The total vapor pressure of the solution = p =38.6 Torr
Partial vapor pressure of the n-propyl alcohol =[tex]p^{o}_1=21.0 Torr[/tex]
Partial vapor pressure of the isopropyl alcohol =[tex]p^{o}_2=45.2 Torr[/tex]
[tex]p=x_1\times p^{o}_1+x_2\times p^{o}_2[/tex] (Raoult's Law)
[tex]p=x_1\times p^{o}_1+(1-x_1)\times p^{o}_2[/tex]
[tex]38.6 Torr=x_1\times 21.0 Torr+(1-x_1)\times 45.2 Torr[/tex]
[tex]x_1=0.2727[/tex]
[tex]x_2=1-0.2727=0.7273[/tex]
[tex]x_1\& x_2[/tex] is mole fraction in liquid phase.
Mole fraction of components in vapor phase [tex]y_1\& y_2[/tex]
[tex]p_1=y_1\times p[/tex] (Dalton's law of partial pressure)
[tex]y_1=\frac{p_1}{38.6 Torr}=\frac{p^{o}_1\times x_1}{38.6 Torr}[/tex]
[tex]y_1=\frac{21.0 Torr\times 0.2727}{38.6 Torr}=0.1468[/tex]
[tex]y_1=\frac{p_2}{38.6 Torr}=\frac{p^{o}_2\times x_2}{38.6 Torr}[/tex]
[tex]y_2=\frac{45.2 Torr\times 0.7273}{38.6 Torr}=0.8516[/tex]
Mole fraction of alcohols in vapor phase [tex]y_1=0.1468\& y_2=0.8516[/tex]
What condition leads to the production of ketone bodies from acetyl CoA? a. low blood pH b. low supply of oxaloacetate in the citric acid cycle c. high levels of ATP d. high blood pH e. low supply of carnitine in β-oxidation
Answer:
low supply of oxaloacetate in the citric acid cycle
Explanation:
When there is low supply of oxalo acetate the acetyl CoA gets converted to ketone bodies to enter the TCA cycle.
In the condition of low level of glucose, the supply of oxaloacetate decreases. Thus making it unavailable to react with acetyl CoA, in this condition ketogenesis occur i.e. acetyl CoA gets converted to ketone bodies.
For an ideal gas mixture of A and B, if the total pressure is 100 kPa and component B has a mole fraction of 0.25, calculate the partial pressure of component
Answer:
The partial pressure of component A and B is 75 kPa and 25kPa respectively.
Explanation:
Total pressure of ideal gas mixture = 100kPa
Mole fraction of component B, [tex]\chi_B= 0.25[/tex]
Mole fraction of component A ,[tex]\chi_A[/tex]
As we know sum of all mole fraction in a mixture is equal to zero.
[tex]\chi_A+\chi+B=1[/tex]
[tex]\chi_A=1-\chi_B=1-0.25=0.75[/tex]
For partial pressure of each component we will apply Dalton's law of partial pressure:
[tex]p^{o}_i=p_{total}\times \chi_i[/tex]
Partial pressure of component a in a mixture:
[tex]p^{o}_A=p\times \chi_A=100kPa\times 0.75=75 kPa[/tex]
Partial pressure of component a in a mixture:
[tex]p^{o}_B=p\times \chi_B=100kPa\times 0.25=25 kPa[/tex]
When 2-methyl-2,5-pentanediol is treated with sulfuric acid, dehydration occurs and 2,2-dimethyltetrahydrofuran is formed. Suggest a mechanism for this reaction. Which of the two oxygen atoms is most likely to be eliminated, and why?
Answer:The oxygen present at the tertiary carbon would be eliminated.The suggested mechanism of the reaction can be found in attachment
Explanation:
The Oxygen atom at the tertiary carbon atom would be eliminated because the removal of this oxygen in form of water after the protonation by sulphuric acid would lead to the formation of a stable tertiary carbocation which is vary stable.
The tertiary carbocation is stable on account of inductive effect of the methy groups.
The oxygen atom at the primary carbon would not be eliminated as its elimination would result in a primary carbocation which is unstable in nature,.
The mechanism of the overall reaction is following:
1. In the first step the OH group present at the tertiary carbocation is protonated by sulphuric acid and on account of this protonation the OH group turns into a good leaving group and leaves as (water) H₂O.
2. Once the H₂O molecule is eliminated it leads to the formation of a stable tertiary carbocation.
3. The tertiary carbocation so formed is electrophilic in nature and as there is one more OH group present at the primary carbon which is 3 carbons away . The OH group is weakly nucleophilic in nature and can appreciably attack the carbocation . The attack of OH at the carbocation leads to the formation of a 5-membered ring containing oxygen as heteroatom.
4.The 5-membered ring so formed has Oxygen as hetero atom which is protonated so the protonated oxygen atom is deprotonated using H₂O.
This further leads to the product formation.
Kindly refer the attachment for the complete reaction mechanism:
The most useful ore of aluminum is bauxite, in which Al is present as hydrated oxides, Al2O3⋅xH2O The number of kilowatt-hours of electricity required to produce 3.00kg of aluminum from electrolysis of compounds from bauxite is ________ when the applied emf is 4.50V.
Explanation:
According to Faraday's law, the amount of a substance deposited or liberated in electrolysis process is proportional to the quantity of electric charge passed and to the equivalent weight of the substance.
Formula to calculate the mass of substance liberated according to Faraday's law is as follows.
m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]
where, m = mass of substance liberated at electrode
Q = electric charge passing through the substance
F = Faraday constant = 96,487 C [tex]mol^{-1}[/tex]
M = molar mass of the substance
Z = valency number of ions of the substance
Since, it is given that mass is 3 kg or 3000 g (as 1 kg = 1000 g), molar mass of Al is 27, Z is 3.
Therefore, putting the values in the above formula as follows.
m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]
3000 g = [tex](\frac{Q}{96,487 C mol^{-1}})(\frac{27}{3})[/tex]
Q = 32162333.33 C
As it is given that V = 4.50 Volt. Also, it is known that
Energy = [tex]V \times Q[/tex]
Therefore, calculate the energy as follows.
Energy = [tex]V \times Q[/tex]
= [tex]4.50 V \times 32162333.33 C[/tex]
= 144730500 J
As it is known that [tex]3.6 \times 10^{6}[/tex] J = 1 KW Hr
So, convert 144730500 J into KW Hr as follows.
[tex]\frac{144730500 J \times 1 KW Hr}{3600000 J}[/tex]
= 40.202 KW Hr
Thus, we can conclude that the number of kilowatt-hours of electricity required to produce 3.00 kg of aluminum from electrolysis of compounds from bauxite is 40.202 KW Hr when the applied emf is 4.50V.
The production of aluminum from bauxite involves several chemical reactions and an electrolysis process in a Hall-Héroult cell. Aluminum oxide is reduced to aluminum metal during electrolysis. The exact amount of electricity required cannot be stated without additional industrial data.
Explanation:The process of making aluminum from bauxite involves several stages of chemical reactions. Initially, bauxite, AlO(OH), reacts with hot sodium hydroxide to form soluble sodium aluminate, leaving behind impurities. Aluminum hydroxide is then precipitated out and heated to form aluminum oxide, Al2O3, which is dissolved in a molten mixture of cryolite and calcium fluoride. An electrolytic cell, specifically called the Hall-Héroult cell, is used for the electrolysis process. During electrolysis, the reduction of aluminum ions to aluminum metal takes place at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode. The amount of electricity required to produce 3 kg of aluminum using a 4.5V emf cannot be directly calculated without additional data such as the exact chemical conversion efficiencies and energy losses in the process, which are typically proprietary information for aluminum production companies.
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The boiling point of water, H2O, is 100.000 °C at 1 atmosphere. Kb(water) = 0.512 °C/m In a laboratory experiment, students synthesized a new compound and found that when 13.62 grams of the compound were dissolved in 217.5 grams of water, the solution began to boil at 100.094 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?
Answer:
341.1 g/molExplanation:
1) Data:
a) Tb₁ = 100.000°C
b) Kb = 0.512 °C/m
c) mass of solute = 13.62 g
d) mass of solvent = 217.5 g
e) Tb₂ = 100.094°C
f) Solute: nonvolatile and nonelectrolyte
g) MM = ?
2) Chemical principles and formulae:
a) The boiling point elevation of a non-volatile solute is a colligative property, which follows this equation:
ΔTb = Kb × m × iWhere:
ΔTb is the elevation of the boiling point = Tb₂ - Tb₁,Kb is the molal boiling constant of the solvent,m is the molality of the solution,i is the Van't Hoof constant, and is equal to 1 for non-electrolyte solutes.b) Molality, m:
m = number of moles of solute / kg of solventc) Molar mass, MM:
MM = mass in grams / number of moles3) Solution:
i) ΔTb = Tb₂ - Tb₁ = 100.094°C - 100.000°C = 0.094°C
ii) ΔTb = Kb × m ⇒ m = ΔTb / Kb = 0.094°C / (0.512°C/m) = 0.1836 m
iii) m = number of moles of solute / kg of solvent ⇒
number of moles of solute = m × kg of solvent = 0.1836 m × 0.2175 kg
number of moles of solute = 0.03993 mol
iv) MM = mass in grams / number of moles = 13.62 g / 0.03993 mol = 341.1 g/mol
Final answer:
The molecular weight of the synthesized compound is calculated using the change in boiling point and the molal boiling point elevation constant to first determine molality and then the number of moles of the compound. The molecular weight is then found by dividing the mass of the compound by the moles of the compound, yielding a value of 341.1 g/mol.
Explanation:
To calculate the molecular weight of the synthesized compound, we apply the boiling point elevation formula ΔT = Kb × m, where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant of water, and m is the molality of the solution. Given Kb for water is 0.512°C/m and ΔT is 0.094°C (100.094°C - 100.000°C), we can calculate the molality:
m = ΔT / Kb = 0.094°C / 0.512°C/m = 0.1836 m
Next, we calculate the moles of solute using molality and the mass of the solvent (water):
moles of solute = molality × mass of solvent in kg = 0.1836 m × 0.2175 kg = 0.0399 mol
The molecular weight (MW) is then found by dividing the mass of the compound by the moles of the compound:
MW = mass of compound / moles of compound = 13.62 g / 0.0399 mol = 341.1 g/mol
Consider the two reactions. 2NH3(g)+3N2O(g)4NH3(g)+3O2(g)⟶4N2(g)+3H2O(l)⟶2N2(g)+6H2O(l) Δ????∘=−1010 kJΔ????∘=1531 kJ Using these two reactions, calculate and enter the enthalpy change for the reaction below. N2(g)+12O2(g)⟶N2O(g)
Answer: The [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical reaction for the formation reaction of [tex]N_2O[/tex] is:
[tex]N_2(g)+\frac{1}{2}O_2(g)\rightarrow N_2O(g)[/tex] [tex]\Delta H^o_{formation}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(l)[/tex] [tex]\Delta H_1=-1010kJ[/tex] ( ÷ 3)
(2) [tex]4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l) [tex]\Delta H_2=1531kJ[/tex] ( ÷ 6)
Reversing Equation 1 and then adding both the equations, we get the enthalpy change for the chemical reaction.
[tex]\Delta H^o_{formation}=[\frac{\Delta H_1}{3}]+[\frac{\Delta H_2}{6}][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{formation}=[\frac{1010}{3}]+[\frac{1531}{6}]\\\\\Delta H^o_{formation}=591.9kJ[/tex]
Hence, the [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.
Hess's law is defined as the sum of amount of heat absorbed or released in the given chemical equation remains constant, irrespective of the steps involved in the reaction. The [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex] is for the given reaction.
Given that,
[tex]\Delta \text H^0_{1}&=-1010 \text{kJ}[/tex][tex]\Delta \text H^0_{2}&=-1531 \text{kJ}[/tex][tex]\Delta \text H^0_{\text{formation}}&=? \text{kJ}[/tex]Now, the given chemical equations are:
[tex]\text{N}_2_{\text (g)} + \dfrac{1}{2}\text O_2_{(\text g)} \rightarrow \text N_2\text O\;\;\;\;\Delta \text H^0_{\text{formation}}&=? \text{kJ}[/tex]The intermediate reactions between the above equation are:
[tex]\text {2 NH}_3_\text{(g)} + 3\text N_2\text O \rightarrow 4\text N_2 +3\text H_2\text O\;\;\;\;\;\Delta \text H^0_{1}&=-1010 \text{kJ}[/tex][tex]\text {4 NH}_3_\text{(g)} + \text O_2 \rightarrow 2\text N_2 +6\text H_2\text O\;\;\;\;\;\Delta \text H^0_{1}&=-1531\text{kJ}[/tex]Reversing the equation and then adding both the values, the enthalpy change becomes:
[tex]\Delta \text H^0_{\text{formation}}&=\dfrac{\Delta \text {H}_1}{3}+\dfrac{\Delta \text {H}_2}{6}\\\\\Delta \text H^0_{\text{formation}}&=\dfrac{1010}{3}+\dfrac{1531}{6}\\\\\Delta \text H^0_{\text{formation}}&= 591.9 \text{kJ}[/tex]Therefore, the enthalpy change of the reaction is [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex].
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Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be produced from 5.0 kg of limestone.
Answer: 2800 g
Explanation:
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass = 5 kg = 5000 g
[tex]\text{Number of moles}=\frac{5000g}{100g/mol}=50moles[/tex]
1 mole of [tex]CaCO_3[/tex] produces = 1 mole of [tex]CaO[/tex]
50 moles of [tex]CaCO_3[/tex] produces =[tex]\frac{1}{1}\times 50=50moles[/tex] of [tex]CaO[/tex]
Mass of [tex]CaO=moles\times {\text{Molar mass}}=50moles\times 56g/mole=2800g[/tex]
2800 g of [tex]CaO[/tex] is produced from 5.0 kg of limestone.
A decomposition reaction splits the reactants into two or more products. The mass of the quicklime produced from 5 kg of limestone is 2800 gm.
What is mass?Mass is the amount or the weight of the substance occupied in the system. It can be calculated in grams or kilograms.
The decomposition reaction of the Limestone can be shown as:
[tex]\rm CaCO_{3} \rightarrow CaO + CO_{2}[/tex]
The number of the mole of limestone is given as:
[tex]\rm Moles = \rm \dfrac {Mass}{Molar \;mass}[/tex]
Here, mass is 5000 gm and the molar mass is 100 g/mol
Substituting values in the equation above:
[tex]\begin{aligned}\rm n &= \dfrac{5000}{100}\\\\&= 50\;\rm mol\end{aligned}[/tex]
The stoichiometry coefficient of the reaction gives:
1 mole of limestone = 1 mole quicklime
So, 50 moles of limestone = x moles of quicklime
Solving for x:
[tex]\begin{aligned}\rm x &= \dfrac{50 \times 1}{1}\\\\&= 50 \;\rm mole\end{aligned}[/tex]
Mass of quicklime is calculated as:
[tex]\begin{aligned}\rm mass &= \rm moles \times \rm molar \; mass\\\\&= 50 \times 56\\\\&= 2800\;\rm gm\end{aligned}[/tex]
Therefore, 2800 gm of quicklime is produced.
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The tiara worn by Kate Middleton for her wedding to Prince William of England contains 888 diamonds and belongs to the British monarchy. If each diamond in the tiara is 1.0 carat, and given that diamond is a form of carbon and that 1 carat is defined as 0.200 g, calculate the number of atoms in the gemstones of that tiara.
Answer:
The number of atoms in the gemstones of that tiara is [tex]8.9125\times 10^{24} atoms[/tex].
Explanation:
Number of diamonds ion tiara = 888
Mass of each diamond = 1.0 carat = 0.200 g (given)
Mass of 888 diamonds in tiara:
[tex]888\times 0.200 g=177.600 g[/tex]
Given that diamond is a form of carbon.
Atomic mass of carbon atom = 12 g/mol
Moles of 77.600 g of carbon =[tex]\frac{177.600 g}{12 g/mol}=14.800 mol[/tex]
Number of atoms of carbon 14.800 moles:
[tex]14.800 mol\times 6.022\times 10^{23} atoms =8.9125\times 10^{24} atoms[/tex]
The number of atoms in the gemstones of that tiara is [tex]8.9125\times 10^{24} atoms[/tex].
Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 5.00 atm . Calculate the partial pressure of each gas in the container.
Answer:
Partial pressure of methane: 1.18 atm
Partial pressure of ethane: 1.45 atm
Partial pressure of propane: 2.35 atm
Explanation:
Let the total moles of gases in a container be n.
Total pressure of the gases in a container =P = 5.0 atm
Temperature of the gases in a container =T = 23°C = 296.15 K
Volume of the container = V = 10.0 L
[tex]PV=nRT[/tex] (Ideal gas equation)
[tex]n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol[/tex]
Moles of methane gas =[tex]n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol[/tex]
Moles of ethane gas =[tex]n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol[/tex]
Moles of propane gas =[tex]n_3=?[/tex]
[tex]n=n_1+n_2+n_3[/tex]
[tex]n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol[/tex]
Partial pressure of all the gases can be calculated by using Raoult's law:
[tex]p_i=P\times \chi_i[/tex]
[tex]p_i[/tex] = partial pressure of 'i' component.
[tex]\chi_1[/tex] = mole fraction of 'i' component in mixture
P = total pressure of the mixture
Partial pressure of methane:
[tex]p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}[/tex]
[tex]p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm[/tex]
Partial pressure of ethane:
[tex]p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}[/tex]
[tex]p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm[/tex]
Partial pressure of propane:
[tex]p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}[/tex]
[tex]p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm[/tex]
After calculating moles for each gas, the partial pressure for methane, ethane, and propane was calculated as 1.21 atm, 1.45 atm, and 2.34 atm respectively.
Explanation:To calculate the partial pressure of each gas, we will first need to calculate the moles of each gas. For methane, the molar mass is 16.04 g/mol, so 8.00 g / 16.04 g/mol = 0.499 mol. The molar mass of ethane is 30.07 g/mol, so 18.0 g / 30.07 g/mol = 0.598 mol. The total moles of gas can be calculated by the total pressure and volume using the Ideal gas law, PV=nRT. The total moles of gases are 5.0 atm *10.0L/(0.0821*296.15K) = 2.06 mol. Hence, the moles of propane are 2.06 - 0.499 - 0.598 = 0.963 mol.
Using Dalton's law of partial pressures, the partial pressure of each gas can be calculated by (moles of gas/ total moles) * total pressure. Hence, the partial pressures of methane, ethane, and propane are 0.499 / 2.06 * 5.00 atm = 1.21 atm, 0.598 / 2.06 * 5.00 atm = 1.45 atm, and 0.963 / 2.06 * 5.00 atm = 2.34 atm, respectively.
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Which of the following has the greatest electronegativity difference between the bonded atoms? View Available Hint(s) Which of the following has the greatest electronegativity difference between the bonded atoms? A strong acid made of hydrogen and a halogen, such as HCl A group 1 alkali metal bonded to fluoride, such as LiF. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO A diatomic gas, such as nitrogen (N2).
Answer: A group 1 alkali metal bonded to fluoride, such as LiF.
Explanation:
Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself. The size of an atom increases as we move down the group because a new shell is added and electron gets added up.
1. A strong acid made of hydrogen and a halogen, such as HCl : A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms. Electronegativity difference = electronegativity of chlorine - electronegativity of hydrogen = 3-2.1= 0.9
2. A group 1 alkali metal bonded to fluoride, such as LiF: Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal.
Electronegativity difference = electronegativity of fluorine - electronegativity of lithium= 4-1= 3
3. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO: A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.
Electronegativity difference = electronegativity of oxygen - electronegativity of carbon= 3.5-2.5= 1.0
4. A diatomic gas, such as nitrogen [tex](N_2)[/tex]: Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms.
Electronegativity difference = 0
Thus the greatest electronegativity difference between the bonded atoms is in LiF.
Among the given compounds, a group 1 alkali metal bonded to fluoride, such as LiF, has the greatest electronegativity difference because fluoride is highly electronegative while group 1 alkali metals like lithium have low electronegativity.
To determine which of the given compounds has the greatest electronegativity difference between the bonded atoms, we need to consider the electronegativities of the individual atoms involved. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.
A strong acid made of hydrogen and a halogen, such as HCl, has a significant electronegativity difference due to chlorine being much more electronegative than hydrogen. However, a group 1 alkali metal bonded to fluoride, such as LiF, will typically have an even greater electronegativity difference. This is because fluoride is one of the most electronegative elements, and lithium is a metal with a much lower electronegativity. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO, will have a moderate electronegativity difference, but less than that between lithium and fluoride. A diatomic gas, such as nitrogen (N₂), will have no electronegativity difference because both atoms are the same and thus have the same electronegativity.
Therefore, a group 1 alkali metal bonded to fluoride, such as LiF, has the greatest electronegativity difference of the options provided.
A liquid in the lab has a density of 1.17 g/cm3. What is the volume in liters of 3.02 kg of the liquid?
Answer: The volume of liquid in liters is 2.5812 L.
Explanation:
Density of an object is defined as the ratio of its mass and volume. The chemical equation representing density of an object is:
[tex]\text{Density of an object}=\frac{\text{Mass of an object}}{\text{Volume of an object}}[/tex]
We are given:
Mass of liquid = 3.02 kg = 3020 g (Conversion factor: 1kg = 1000 g)
Density of liquid = [tex]1.17 g/cm^3[/tex]
Putting values in above equation, we get:
[tex]1.17g/cm^3=\frac{3020g}{\text{Volume of liquid}}\\\\\text{Volume of liquid}=2581.2cm^3[/tex]
Converting this into liters, we use the conversion factor:
[tex]1L=1000cm^3[/tex]
So, [tex]\Rightarrow \frac{1L}{1000cm^3}\times 2581.2cm^3[/tex]
[tex]\Rightarrow 2.5812L[/tex]
Hence, the volume of the liquid is 2.5812 L.
The volume in liters of 3.02 kg of the liquid is 2.58 Liters
Density is the ratio of mass to volume of a substance. The density of a substance is given by:
Density = mass / volume
Given that a liquid has a density of 1.17 g/cm³ and a mass of 3.02 kg, the volume is:
Density = 1.17 g/cm³ = 1.17 kg/L
Density = mass/volume
1.17 = 3.02/volume
Volume = 3.02/1.17
Volume = 2.58 Liters
Hence the volume in liters of 3.02 kg of the liquid is 2.58 Liters
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What is a hydrocarbon? What is a hydrocarbon? It is a molecule derived from hydrogen synthesis. It is a wet carbon atom It is any organic molecule. It is a molecule composed of carbon and hydrogen only. none of the above
Answer:
The correct answer is :It is a molecule composed of carbon and hydrogen only.
Explanation:
Hydrocarbons are defined as compounds which majorly made up carbon and hydrogen.There three types of hydrocarbons:
Alkanes : These are saturated hydrocarbons with single bond present in between two carbon atoms.The general structure of an alkane is:[tex]C_nH_{2n+2}[/tex]Alkene :These are unsaturated hydrocarbons with double bond present in between two carbon atoms.The general structure of an alkene is:[tex]C_nH_{2n}[/tex]Alkyne :These are unsaturated hydrocarbons with triple bond present in between two carbon atoms.The general structure of an alkyne is:[tex]C_nH_{2n-1}[/tex]At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s
2HIg=H2g+I2g Suppose a vessel contains HI at a concentration of 1.28M . Calculate how long it takes for the concentration of HI to decrease to 17.0% of its initial value. You may assume no other reaction is important. Round your answer to 2 significant digits.
Answer : The time taken for the reaction is, 28 s.
Explanation :
Expression for rate law for first order kinetics is given by :
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = 0.0632
t = time taken for the process = ?
[tex][A_o][/tex] = initial amount or concentration of the reactant = 1.28 M
[tex][A][/tex] = amount or concentration left time 't' = [tex]1.28\times \frac{17}{100}=0.2176M[/tex]
Now put all the given values in above equation, we get:
[tex]0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}[/tex]
[tex]t=28s[/tex]
Therefore, the time taken for the reaction is, 28 s.
Ammonia, NH3, is used as a refrigerant. At its boiling point of –33 oC, the standard enthalpy of vaporization of ammonia is 23.3 kJ/mol. How much heat is released when 50.0 g of ammonia is condensed at –33 oC?–0.466 kJ–7.94 kJ–36.6 kJ–68.4 kJ–1.17 x 103 kJ
Answer:
-68.4 kJ
Explanation:
The standard enthalpy of vaporization = 23.3 kJ/mol
which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
Thus, Q = -23.3 kJ/mol
Where negative sign signifies release of heat
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.
The heat released when 50.0 g of ammonia condenses at its boiling point is -68.4 kJ. This is calculated by multiplying the moles of ammonia by the enthalpy of vaporization and recognizing that heat is released in condensation.
Explanation:To solve this problem, we need to understand the concept of enthalpy of vaporization, which is the heat needed to convert 1 mole of a substance from a liquid to a gas at constant pressure and temperature. For ammonia (NH3), which boils at -33 °C, the enthalpy of vaporization is 23.3 kJ/mol. However, we want the heat released when 50.0 g (around 2.94 moles) of ammonia condenses, which is the reverse process of vaporization. Thus, the energy would be released rather than absorbed.
Now, let's calculate this value. We multiply the number of moles of ammonia by the enthalpy of vaporization:
2.94 moles x 23.3 kJ/mol = 68.4 kJ
Since this is the reverse of the process of vaporization, heat is released, so the enthalpy change is negative (-68.4 kJ). Therefore, the correct answer is -68.4 kJ.
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You carefully weigh 5.724 g of potato flour into a Kjeldahl flask and add an appropriate reagent to digest the flour. You then carefully distill the material by adding a reagent to the digestion flask and stirring vigorously, then heating while under a closed system. You collect distillate in a collection flask containing a liquid, then titrate with 0.1 N HCl. It takes 11.95 ml to titrate the sample. The volume of titrate for blank titration is 0.1 ml. What was the percent protein in the potato flour if the protein in potato flour is 16% nitrogen?
Hey there!:
In Kjeldahl's method estimation of amount of nitrogen in a sample, the sample is first digested by using sulphuric acid , which converts the nitrogen in the sample to ammonium sulphate.
The ammonium ions in the digested product are converted to ammonia gas by adding excess of base like NaOH. The ammonia gas thus produced is collected by condensation.
The amount of ammonia produced is estimated by titration with standard solution of acid. We are using 0.1 N HCl in this case.
Volume of 0.1 N HCl used = 11.95 mL
Volume of of 0.1 N HCl used in blank titration =0.1 mL
Therefore,actual volume us of 0.1 N HCl
used by analyte( ammonia solution) = 11.95 mL - 0.1 mL
= 11.85 mL
0.1 N HCl = 0.1 M HCl , as HCl is a monobasic acid and therefore its molar mass will be equal to its equivalent mass.
Therefore, concentration of HCl solution used = 0.1 M = 0.1 mol L-1
Volume of HCl solution used =11.85 mL =0.01185 L
Nº.of moles of HCl used = 0.1 mol L-1 * 0.01185 L
= 0.001185 moles
The equation for reaction between NH3 (aq) and HCl during titration is :
NH3 (aq) + HCl (aq) <= > NH4Cl
From the above equation, we see that each mole of HCl reacts completely with 1 mole of NH3 during titration.
Therefore, no. of moles of NH3 that would have been neutralized by 0.001185 moles of HCl = 0.001185 moles
From the formula of NH3 , 1 mole of NH3 contains 1 mole of N atoms.
Therefore, nº of moles of :
N - atoms present = 0.001185 moles
Moar mass of N= 14 g/mol
Mass of 0.001185 moles of N =0.001185 moles * 14 g/ mol
= 0.01659 g
% of N in protein is given as 16%.
16%.of mass of protein = mass of nitrogen in protein sample = 0.01659 g
(16/100) *mass of protein = 0.01659 g
mass of protein = 0.1036875 g
% of protein in potato flour = [mass of protein in potato flour / mass of potato flour] * 100%
( 0.1036875 g / 5.724 ) * 100%
= 1.812 %
Hope this helps!
Final answer:
The percent protein was found to be 1.8113 %.
Explanation:
To calculate the percent protein in the potato flour, we first need to correct the titration volume for the blank, then figure out the amount of Nitrogen (N) in the potato flour and finally convert this to percent protein using the factor that proteins in potato flour contain 16% nitrogen. Since each gram of nitrogen corresponds to 6.25 grams of protein (since 100/16 = 6.25), we can use this conversion factor to get the final percent protein.
Volume of HCl used for the sample is 11.95 mL
Volume of HCl used for the blank is 0.1 mL
Net volume of HCl used for the sample is 11.85 mL (11.95 mL - 0.1 mL)
Since the concentration of HCl is 0.1 N, we multiply the net volume by the normality to find the milliequivalents (meq) of nitrogen:
11.85 mL * 0.1 N = 1.185 meq
To find the grams of nitrogen, we use the fact that 1 meq of N equals 0.014 g:
1.185 meq * 0.014 g/meq = 0.01659 g of N
To convert grams of N to percent nitrogen in the sample, we divide by the sample mass and multiply by 100:
(0.01659 g / 5.724 g) * 100 = 0.2898 % N
Multiply the percent N by the conversion factor to get percent protein:
0.2898 % N * 6.25 = 1.8113 % Protein
A syringe containing 1.75 mL of oxygen gas is cooled from 97.8 ∘C to 0.7 ∘C. What is the final volume Vf of oxygen gas? (Assume that the pressure is constant.)
Answer : The final volume of oxygen gas will be, 1.292 ml
Explanation :
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of oxygen gas = 1.75 ml
[tex]V_2[/tex] = final volume of oxygen gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]97.8^oC=273+97.8=370.8K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]0.7^oC=273+0.7=273.7K[/tex]
Now put all the given values in the above formula, we get the final volume of the oxygen gas.
[tex]\frac{1.75ml}{V_2}=\frac{370.8K}{273.7K}[/tex]
[tex]V_2=1.292ml[/tex]
Therefore, the final volume of oxygen gas will be, 1.292 ml
Arrange the following aqueous solutions in order of decreasing freezing point. (The last three are all assumed to dissociate completely into ions in water.) (a) 0.20 m ethylene glycol (nonvolatile, nonelectrolyte) (b) 0.12 m K2SO4 (c) 0.10 m MgCl2 (d) 0.12 m KBr
The answer would be Lead.
The heat capacity of solid sodiu capacity of solid sodium carbonate (Na2CO3) as estimated by Kopp's rule is A. 2.7 kJ/mol°C B. 0.121 KJ/mol°C C. 1.9 kJ/mol°C D. 0.11 KJ/mol°C
Final answer:
Without specific numeric values for the individual atomic heat capacities of the elements in sodium carbonate, directly answering the heat capacity multiple-choice question using Kopp's rule is not feasible. Additional data is required to estimate the heat capacity of solid sodium carbonate.
Explanation:
The question regarding the heat capacity of solid sodium carbonate (Na2CO3) can be answered by applying Kopp's rule, which is a method used to estimate the heat capacities of solids. According to Kopp's rule, the heat capacity of a compound in the solid state is the sum of the atomic heat capacities of the individual elements that compose the compound. Atomic heat capacities can be estimated using the Law of Dulong and Petit, which suggests that at room temperature, the molar heat capacity of many solid elements roughly equates to 3R, where R is the gas constant with a value approximately equal to 8.314 J/(mol·K).
Applying Kopp's rule would involve calculating the total molar heat capacity for sodium, carbon, and oxygen in sodium carbonate, based on their atomic weights and expected contributions. However, since specific numeric values are not provided in the question or reference information to apply Kopp's rule directly, directly answering the multiple-choice question is not feasible without additional data on the individual atomic heat capacities of the elements involved in sodium carbonate.
if the rate of a particular reaction is 4 ties faster at 373K that it was at 323K what is the activation energy for the reaction
Answer:
27.77 kJ/mol is the activation energy for the reaction.
Explanation:
Rate of the reaction at 323 K =[tex]k_1=k[/tex]
Rate of the reaction at 373 K =[tex]k_2=4k[/tex]
Activation energy for the reaction is calculated by formula:
[tex]\log \frac{k_2}{k_1}=\frac{E_a}{2.303\times R}[\frac{T_2-T_1}{T_2\times T_1}][/tex]
[tex]E_a[/tex] = Activation energy
[tex]T_1[/tex] = Temperature when rate of the reaction was [tex]k_1[/tex]
[tex]T_2[/tex] = Temperature when rate of the reaction was [tex]k_2[/tex]
Substituting the values:
[tex]\log \frac{4k}{k}=\frac{E_a}{2.303\times 8.314 J /mol K}[\frac{373 K-323K}{373 K\times 323 K}][/tex]
[tex]E_a=27,776.98 J/mol=27.77 kJ/mol[/tex]
27.77 kJ/mol is the activation energy for the reaction.
Hydrogen gas and nitrogen gas react to form ammonia gas. What volume of ammonia would be produced by this reaction if 6.9 m3 of hydrogen were consumed? Also, be sure your answer has a unit symbol, and is rounded to the correct number of significant digits.
Answer:
4600 Liters NH₃(g)
Explanation:
Applying Avogadro's law to the reaction of hydrogen and nitrogen gas forming ammonia, we find that 6.9 m3 of hydrogen will form about 4.6 m3 of ammonia, assuming that temperature and pressure remain constant.
Explanation:The subject in question relates to the application of Avogadro's law to chemical reactions involving gases. More specifically, we're investigating the reaction between hydrogen gas and nitrogen gas to produce ammonia gas. As per Avogadro's law, gases react in definite and simple proportions by volume, if all gas volumes are measured at the same temperature and pressure.
Focusing on the hydrogen to ammonia conversion, the reaction equation N₂(g) + 3H₂(g) turns into 2NH3(g). We can surmise that three volumes of hydrogen gas (H2) react to form two volumes of ammonia gas (NH3). Considering that 6.9 m3 of hydrogen gas is consumed, by the rule of three, we can infer that the reaction would result in approximately 4.6 m3 of ammonia gas.
Bear in mind, these calculations are assuming that the temperature and pressure remain constant during the reaction and that it goes to completion.
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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling point of ethanol from this data.
Answer : The normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]
Explanation :
The Clausius- Clapeyron equation is :
[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = vapor pressure of ethanol at [tex]30^oC[/tex] = 98.5 mmHg
[tex]P_2[/tex] = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
[tex]T_1[/tex] = temperature of ethanol = [tex]30^oC=273+30=303K[/tex]
[tex]T_2[/tex] = normal boiling point of ethanol = ?
[tex]\Delta H_{vap}[/tex] = heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
[tex]\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})[/tex]
[tex]T_2=348.67K=348.67-273=75.67^oC[/tex]
Hence, the normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]
The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 75% of the lead to decay? (Round your answer to two decimal places.)
Answer:
It will take 6.6 hours for 75% of the lead to decay.
Explanation:
The radioactive decay follows first order rate law
The half life and rate constant are related as
[tex]k=rate constant=\frac{0.693}{halflife}=\frac{0.693}{3.3}=0.21h^{-1}[/tex]
The rate law for first order reaction is
[tex]time=\frac{1}{k}(ln[\frac{A_{0}}{A_{t}}][/tex]
Where
A0 = initial concentration = 1 g
At= final concentration = 0.25 g (as 75% undergoes decay so 25% left]
[tex]time=\frac{1}{0.21}(ln(\frac{1}{0.25})=6.6hours[/tex]
When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 × 10−7 m and (b) 4.2 × 10−7 m. What are the frequencies of the two lines? What color do we see when we heat a rubidium compound?
Answer:
The frequencies of the two lines are:
a) [tex]3.79\times 10^{14} s^{-1}[/tex]
b)[tex]7.14\times 10^{14} s^{-1}[/tex]
When we heat rubidium compound we will see red color.
Explanation:
[tex]\nu=\frac{c}{\lambda }[/tex]
c = speed of light
[tex]\lambda [/tex] = wavelength of light
a) Frequency of the light when wavelength is equal to [tex]7.9\times 10^{-7} m[/tex]
[tex]\nu=\frac{c}{\lambda }[/tex]
[tex]\nu=\frac{3\times 10^8m/s)}{7.9\times 10^{-7}}[/tex]
[tex]\nu=3.79\times 10^{14} s^{-1}[/tex]
This frequency corresponds to red light
b) Frequency of the light when wavelength is equal to [tex]4.2\times 10^{-7} m[/tex]
[tex]\nu=\frac{c}{\lambda }[/tex]
[tex]\nu=\frac{3\times 10^8m/s)}{4.2\times 10^{-7}}[/tex]
[tex]\nu=7.14\times 10^{14} s^{-1}[/tex]
This frequency corresponds to violet light
When we heat rubidium compound we will see red color.
When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 × 10⁻⁷ m and (b) 4.2 × 10⁻⁷ m. The frequencies of the two lines are 3.8 × 10¹⁴ Hz and 7.1 × 10¹⁴ Hz. The color observed when we heat a rubidium compound is red and blue.
Explanation:When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 × 10⁻⁷ m and (b) 4.2 × 10⁻⁷ m. To find the frequencies of these lines, we can use the equation v = c/λ, where v is the frequency, c is the speed of light, and λ is the wavelength. By plugging in the given wavelengths, we can calculate the frequencies.
The frequency of line (a) is 3.8 × 10¹⁴ Hz, and the frequency of line (b) is 7.1 × 10¹⁴ Hz.
When a rubidium compound is heated, we observe two lines in the line spectrum. The first line has a wavelength of 7.9 × 10⁻⁷ m, which corresponds to a red color. The second line has a wavelength of 4.2 × 10⁻⁷ m, which corresponds to a blue color.
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2 N H 3 ( g ) ⟷ N 2 ( g ) + 3 H 2 ( g ) K p = 0.83 Consider your answers above, if the initial pressures for all three species is 1 atm what is the equilibrium pressure of H2? (Hint: Your quadratic will have two solutions, which one is impossible?)
The equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.
Equilibrium pressure of H2The equilibrium pressure of H2 is calculated by creating ICE table as follows;
2 N H3 ( g ) ⟷ N2( g ) + 3H2
I: 1 1 1
C: -2x x 3x
E: 1 - 2x 1 + x 1 + 3x
[tex]KP = \frac{(N_2)(H_2)^3}{(NH_3)^2} \\\\0.83 = \frac{(1 + x)(1 + 3x)^3}{(1 - 2x)^2}[/tex]
0.83(1 - 2x)² = (1 + x)(1 + 3x)³
0.83(1 - 4x + 4x²) = (1 + x)((1 + 3x)³)
0.83 - 3.32x + 3.32x² = (1 + x)((1 + 3x)³)
0.83 - 3.32x + 3.32x² = 1 + 10x + 36x² + 54x³ + 27x⁴
27x⁴ + 54x³ + 32.68x² + 13.32x + 0.17 = 0
x = -1.379 or - 0.013
Partial pressure of H2 = 1 + 3xH2 = 1 + 3(-1.379)
H2 = -3.13 atm
H2 = 1 + 3(-0.013)
H2 = 0.96 atm
Thus, the equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.
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A solution is prepared by dissolving 42.0 g of glycerin, C3H8O3, in 186 g of water with a final volume of 200.0 mL. a. Calculate the molarity and molality of the solution. b. What would be the molarity if 300.0 mL of water was added to the solution?
Answer:
For a: The molality and molarity of the given solution is 2.45m and 2.28 M respectively.
For b: The molarity of the solution when more water is added is 0.912 M
Explanation:
For a:To calculate the molality of solution, we use the equation:
[tex]Molarity=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]
Where,
[tex]m_{solute}[/tex] = Given mass of solute [tex](C_3H_8O_3)[/tex] = 42.0 g
[tex]M_{solute}[/tex] = Molar mass of solute [tex](C_3H_8O_3)[/tex] = 92.093 g/mol
[tex]W_{solvent}[/tex] = Mass of solvent (water) = 186 g
Putting values in above equation, we get:
[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 186}\\\\\text{Molality of }C_3H_8O_3=2.45m[/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] .....(1)
We are given:
Molarity of solution = ?
Molar mass of [tex](C_3H_8O_3)[/tex] = 92.093 g/mol
Volume of solution = 200 mL
Mass of [tex](C_3H_8O_3)[/tex] = 42 g
Putting values in above equation, we get:
[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 200}\\\\\text{Molality of }C_3H_8O_3=2.28M[/tex]
Hence, the molality and molarity of the given solution is 2.45m and 2.28 M respectively.
For b:Now, the 300 mL water is added to the solution. So, the total volume of the solution becomes (200 + 300) = 500 mL
Using equation 1 to calculate the molarity of solution, we get:
Molar mass of [tex](C_3H_8O_3)[/tex] = 92.093 g/mol
Volume of solution = 500 mL
Mass of [tex](C_3H_8O_3)[/tex] = 42 g
Putting values in equation 1, we get:
[tex]\text{Molality of }C_3H_8O_3=\frac{42\times 1000}{92.093\times 500}\\\\\text{Molality of }C_3H_8O_3=0.912M[/tex]
Hence, the molarity of the solution when more water is added is 0.912 M
Using the Br?nsted-Lowry concept of acids and bases, identify the Br?nsted-Lowry acid and base in each of the following reactions:
H2PO3?(aq)+H2O(l)?H3PO3(aq)+OH?(aq)
(CH3)2NH(g)+BF3(g)?(CH3)2NHBF3(s)
Drag the appropriate items to their respective bins.
H2PO3- H2O BF3 (CH3)2NH
Bronsted Lowry Acid Bronsted Lowry Base Neither
Answer:
[tex]H_2PO_3^-[/tex] is Bronsted Lowry base.
[tex]H_2O[/tex] is Bronsted Lowry acid.
Explanation:
According to the Bronsted Lowry conjugate acid-base theory:
An acid is defined as a substance which donates protons and form conjugate baseA base is defined as a substance which accepts protons and forms conjugate acid.[tex]H_2PO_3^-(aq)+H2O(l)\rightarrow H_3PO_3(aq)+OH^-(aq)[/tex]
[tex]H_2PO_3^-[/tex] is Bronsted Lowry base.It accepts protons and forms conjugate acid [tex]H_3PO_3[/tex]
[tex]H_2O[/tex] is Bronsted Lowry acid.It donates protons and forms conjugate base [tex]OH^-[/tex]
[tex](CH_3)_2NH(g)+BF_3(g)\rightarrow (CH_3)_2NHBF_3(s)[/tex]
There in no exchange of proton in an above reaction.Neither of the reactants and products are Bronsted Lowry acid or Bronsted Lowry base
For the gas phase decomposition of phosphine at 120 °C, the rate of the reaction is determined by measuring the appearance of H2. 4 PH3(g)P4(g) + 6 H2(g) At the beginning of the reaction, the concentration of H2 is 0 M. After 93.0 s the concentration has increased to 0.101 M. What is the rate of the reaction? (mol H2/L) /s
Answer:
Rate = 1.09*10^-3 (mol H2/L)/s
Explanation:
Given:
Initial concentration of H2, C1 = 0 M
Final concentration of H2, C2 = 0.101 M
Time taken, t = 93.0 s
To determine:
The rate of the given reaction
Calculation:
The decomposition of PH3 is represented by the following chemical reaction
[tex]4 PH3(g)\rightarrow P4(g) + 6 H2(g)[/tex]
Reaction rate in terms of the appearance of H2 is given as:
[tex]Rate = +\frac{1}{6}*\frac{\Delta [H2]]}{\Delta t}[/tex]
[tex]Rate = +\frac{1}{6}*\frac{C2[H2]-C1[H2]}{\Delta t}[/tex]
Here C1(H2) = 0 M and C2(H2) = 0.101 M
Δt = 93.0 s
[tex]Rate = \frac{(0.101-0.0)M}{93.0 s} =1.09*10^{-3} M/s[/tex]
Since molarity M = mole/L
rate = 1.09*10^-3 (mol H2/L)/s
The heat flux, or the rate of heat transfer per unit area, is measured in watts (Watts) only A. TRUE B. FALSE
Answer: The given statement is false.
Explanation:
Heat flux is defined as the flow of heat or energy per unit time in per unit area. S.I. unit of heat flux is watts per square meter.
Heat flux is represented by the symbol [tex]\phi _{q}[/tex].
So, it means heat flux is not measured in watts only as it includes per unit area also.
Therefore, we can conclude that the given statement heat flux, or the rate of heat transfer per unit area, is measured in watts (Watts) only, is false.
Consider the balanced chemical equation that follows. You are asked to determine how many moles of water you can form from 4 moles of hydrogen and excess oxygen. 2H2(g)+O2(g)→2H2O(l) Which of the following shows calculations for a correct way to solve this problem? View Available Hint(s) Consider the balanced chemical equation that follows. You are asked to determine how many moles of water you can form from 4 moles of hydrogen and excess oxygen. Which of the following shows calculations for a correct way to solve this problem? 4 mol H2×2 mol H2O2 mol H2=4 mol H2O 4 mol H2×2 mol H2O1 mol O2=8 mol H2O 2 mol H2×2 mol H22 mol H2O=2 mol H2O
Answer: The correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]
Explanation:
We are given:
Moles of hydrogen gas = 4 moles
As, oxygen is given in excess. Thus, is considered as an excess reagent and hydrogen is considered as a limiting reagent because it limits the formation of products.
For the given chemical equation:
[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex]
By Stoichiometry of the reaction:
2 moles of hydrogen produces 2 moles of water molecule.
So, 4 moles of hydrogen will produce = [tex]\frac{2molH_2O}{2molH_2}\times 4molH_2=4mol[/tex] of water.
Hence, the correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]
If the absolute temperature of a gas is quadrupled, what happens to the root‑mean‑square speed of the molecules? Nothing happens to the rms speed. The new rms speed is 16 times the original rms speed. The new rms speed is 4 times the original rms speed. The new rms speed is 2 times the original rms speed. The new rms speed is 1/4 the original rms speed.
Answer:
The new rms speed is 2 times the original rms speed.
Explanation:
The root‑mean‑square speed, rms, is related to temperature, , by the formula
[tex]_{rms}[/tex]= √3 / ℳ
For a given gas,
[tex]_{rms}[/tex] ∝ √
or
[tex]_{rms,2}[/tex] / [tex]_{rms,1}[/tex] = √[tex]_{2}[/tex] / [tex]_{1}[/tex]
In this case, is quadrupled.
√4 = 2
The new rms speed is 2 times the original rms speed.
The root-mean-square (rms) speed of gas molecules is proportional to the square root of the temperature. If the absolute temperature is quadrupled, the new rms speed is 2 times the original rms speed.
Effect of Temperature on Root Mean Square Speed of Gas Molecules
The root-mean-square (rms) speed, [tex]V_{rms[/tex] , of the molecules in a gas is related to the absolute temperature, T, by the equation:
[tex]V_{rms[/tex] = [tex]\sqrt((3kBT) / m)[/tex]
where kB is the Boltzmann constant and m is the mass of a molecule. This shows that Vrms is proportional to the square root of the temperature. If the temperature T is quadrupled, the new temperature T' is 4T.
Therefore, the new rms speed [tex]V_{rms[/tex]' becomes:
[tex]V_{rms[/tex]' = [tex]\sqrt((3kB * 4T) / m)[/tex]= [tex]\sqrt(4) * \sqrt((3kBT) / m)[/tex] = 2 * [tex]V_{rms[/tex]
This means that the new rms speed is 2 times the original rms speed.