Answer:
Explanation:
1 . The strongest force observed at the surface of glass is:__DIPOLE______
2. . Water is__POLAR_________ and interacts, generating adhesive interactions with the only weak dispersion strong hydrogen bonding polar glass.
3 . Hexane is_NON-POLAR _______ and interacts, generating__ONLY WEAK __________ adhesive interactions with the glass.
All the following are oxidation–reduction reactions except:________
a. H2(g) + F2(g) → 2HF(g).
b. Ca(s) + H2(g) → CaH2(s).
c. 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g).
d. 6Li(s) + N2(g) → 2Li3N(s).
e. Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).
Answer:
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g)
Explanation:
All the following are oxidation–reduction reactions except:________
a. H₂(g) + F₂(g) → 2HF(g). Redox. H is oxidized and F is reduced.
b. Ca(s) + H₂(g) → CaH₂(s). Redox. Ca is oxidized and H is reduced.
c. 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g). Redox. K is oxidized and H is reduced.
d. 6Li(s) + N₂(g) → 2Li₃N(s). Redox. Li is oxidized and N is reduced.
e. Mg₃N₂(s) + 6H₂O(l) → 3Mg(OH)₂(s) + 2NH₃(g). Not redox. All the elements have the same oxidation number
The reaction Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g) is not a redox reaction.
Redox reactions are those reactions in which there is a change in the oxidation number of species from left to right in the reaction. A specie is oxidized leading to increase in oxidation number while another specie is reduced leading to decrease in oxidation number.
The reaction in which there is no change in oxidation number of species from left to right is the reaction; Mg3N2(s) + 6H2O(l) → 3Mg(OH)2(s) + 2NH3(g).
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1. In the simple cubic unit cell, the centers of ____________ identical particles define the ____________ of a cube. The particles do touch along the cube's ____________ but do not touch along the cube's ____________ or through the center. There is/are ____________ particle per unit cell and the coordination number is ____________ .
2. In the body-centered cubic unit cell, the centers of ____________ identical particles define the ____________ of the cube plus ____________ particle at the ____________ of ____________ . The particles do not touch along the cube's ____________ or faces but do touch along the cube's ____________ . There is/are ____________ particles per unit cell and the coordination number is ____________ .
3. In the face-centered cubic cell, the centers of ____________ identical particles define the ____________ of the cube plus ____________ particle in the ____________ of ____________ . The particles on the ____________ do not touch each other but do touch those on the ____________ . There is/are ____________ particles per unit cell and the coordination number is ____________ .
Answer:please see below for answers in the spaces given.
Explanation:
There are three types of cubic-unit cells of a cubic system which include Simple cubic unit cell, body-centered cubic unit cell and face-centered cubic-unit cell and Thier characteristics are completed below.
1) In the simple cubic unit cell, the centers of _______eight _____ identical particles define the _________corners___ of a cube. The particles do touch along the cube's _______edges_____ but do not touch along the cube's ____diagonal_______ or through the center. There is/are _______one_____ particle per unit cell and the coordination number is
__six______ .
2. In the body-centered cubic unit cell, the centers of _______eight _____ identical particles define the _______corners_____ of the cube plus ______one______ particle at the _______center_____ of ______the cube______ . The particles do not touch along the cube's _______edges_____ or faces but do touch along the cube's ____diagonal________ . There is/are _____two_______ particles per unit cell and the coordination number is _____eight_______ .
3. In the face-centered cubic cell, the centers of ______eight______ identical particles define the _______corner____ of the cube plus ________one____ particle in the _____center_______ of ______each face______ . The particles on the _____corners_______ do not touch each other but do touch those on the ______faces____ . There is/are ________four___ particles per unit cell and the coordination number is _____twelve_______ .
Explain the term isomers?
Answer:
Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.
Which equation best represents the net ionic equation for the reaction that occurs when aqueous solutions of potassium phosphate and iron(II) nitrate are mixed? Question 8 options: 3Fe2+(aq) + 2PO43–(aq) → Fe3(PO4)2(s) 2K+(aq) + Fe(NO3)2(aq) → 2KNO3(aq) + Fe2+(aq) 3Fe2+(aq) + 2PO43–(aq) → Fe3(PO4)2(aq) 2K3PO4(aq) + 3Fe2+(aq) → Fe3(PO4)2(s) + (K+)6(aq) 2K3PO4(aq) + 3Fe(NO3)2(aq) → Fe3(PO4)2(s) + 6KNO3(aq)
Answer:
2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)
Explanation:
In a net ionic equation you list only the ions that are participating in the reaction.
When potassium phosphate, K₃PO₄, reacts with iron (II) nitrate, Fe(NO₃)₂ producing iron (II) phosphate, Fe₃(PO₄)₂ that is an insoluble salt. The reaction is:
2K₃PO₄ + 3 Fe(NO₃)₂ → Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺
The ionic equation is:
6K⁺ + 2PO₄³⁻ + 3Fe²⁺ + 6NO₃⁻→ Fe₃(PO₄)₂(s) + 6NO₃⁻ + 6K⁺
Subtracting the K⁺ and NO₃⁻ ions that are not participating in the reaction, the net ionic equation is:
2PO₄³⁻ + 3Fe²⁺ → Fe₃(PO₄)₂(s)of all the hydrogen nuclei in the ocean, 0.0156 how much deuterium could be obtained from 1.0 gal of ordinary tap water
Answer:
Poop Butt.
Explanation: Poop Butt.
Q1) How much heat is released when 6.38 grams of Ag(s) (m.m = 107.9 g/mol) reacts by the equation shown below at
standard state conditions?
4A9 (s) + 2H,Sq) + O2(g)
2Ag $(s) + 2H200)
Substance
AHof (kJ/mol)
-20.6
H259)
Ag2S (5)
H200
-32.6
-285.8
a)
8.80 KI
b) 69.9 kJ
C) 22.1 kJ
d) 90.8 kJ
e) 40.5 kJ
Answer:
The correct answer is -8.80 kJ.
Explanation:
The ΔH° can be determined by using the formula,
ΔH°rxn = ΔH°f (products) - ΔH°f(reactants)
Based on the given information, the ΔH°f of H2S(g) is -20.6, for Ag2S (s) is -32.6 and for H2O (l) is -285.8 kJ/mole.
Now putting the values we get,
= [2 molΔH°f (Ag2S) + 2 molΔH°f (H2O)] - [4 molΔH°f(Ag) + 2 molΔH°f(H2S) + 1 molΔH°f(O2)]
=[2 mol (-32.6 kJ/mol) + 2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol (-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)
= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]
= -595.6 kJ
Thus, the enthalpy change of -595.6 kJ takes place when 4 mol of Ag reacts by the equation mentioned.
The mass of Ag given is 6.38 grams, the molecular mass of Ag is 107.9 g/mol. The formula for calculating moles is,
Moles = mass/molar mass
= 6.38 g / 107.9 g/mol
= 0.0591 mol
Now the change in enthalpy when 0.0591 mol of Ag reacts by the given reaction is (-595.6 kJ/4 mol) × 0.0591 mol = -8.80 kJ
The negative sign indicates that the heat is released in the process. Therefore, the -8.80 kJ of heat is released by 6.38 grams of Ag in the given case.
Question 6 of 8 >
Calculate the standard enthalpy change for the reaction at 25 °C. Standard enthalpy of formation values can be found in this
list of thermodynamic properties.
H2O(g) + C(graphite)(s)
H2(g) + CO(g)
KJ
ΔΗΓκη
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8/2/2020
Answer:
131.29 kJ
Explanation:
Let's consider the following balanced equation.
H₂O(g) + C(graphite)(s) ⇄ H₂(g) + CO(g)
Given the standard enthalpies of formation (ΔH°f), we can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(H₂(g)) + 1 mol × ΔH°f(CO(g)) - 1 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C(graphite)(s))
ΔH°r = 1 mol × (0 kJ/mol) + 1 mol × (-110.53 kJ/mol) - 1 mol × (-241.82 kJ/mol) - 1 mol × (0 kJ/mol)
ΔH°r = 131.29 kJ
Not all bonds are "created equal". From the following molecules, which one contains the most easily broken carbon to carbon bond? Group of answer choices H3C—CH3 F2C=CF2 H2C=CH2 HCCH
Answer:
H3C—CH3
Explanation:
The strength of a bond is indicated by the value of its bond dissociation energy. Simply put, the bond dissociation energy is the energy required to break the bond.
Carbon forms single, double and triple bonds with itself. As a matter of fact, carbon atoms can link to each other indefinitely. This is known as catenation and has been attributed to the low bond energy of the carbon-carbon single bond.
The bond energy of the carbon-carbon single bond is about 90KJmol-1 while that of carbon-carbon double bond is about 174KJmok-1. The carbon-carbon triple bond has the highest bond dissociation energy of about 230KJmol-1.
Hence, it is easier to break carbon-carbon single bonds than double and triple bonds respectively, hence the answer.
According to the forces of attraction, the molecule which can be easily broken is CH₃-CH₃ as it has a single bond with low dissociation energy as compared to double or triple bonds.
Forces of attraction is a force by which atoms in a molecule combine. it is basically an attractive force in nature. It can act between an ion and an atom as well.It varies for different states of matter that is solids, liquids and gases.
The forces of attraction are maximum in solids as the molecules present in solid are tightly held while it is minimum in gases as the molecules are far apart . The forces of attraction in liquids is intermediate of solids and gases.
The physical properties such as melting point, boiling point, density are all dependent on forces of attraction which exists in the substances.Single bonds have least dissociation energy while triple bonds have the maximum dissociation energy.
Thus,the molecule which can be easily broken is CH₃-CH₃.
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The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that
The given question is incomplete. The complete question is given here :
The equilibrium constant for the reaction is [tex]1.1\times 10^6[/tex] M.
[tex]HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)[/tex]
This value indicates that
A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
B. HCN is a stronger acid than HONO
C. The conjugate base of HONO is [tex]ONO^-[/tex]
D. The conjugate acid of CN- is HCN
Answer: A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
Explanation:
Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.
When [tex]K_{p}>1[/tex]; the reaction is product favoured.
When [tex]K_{p};<1[/tex] ; the reaction is reactant favored.
[tex]When K_{p}=1[/tex]; the reaction is in equilibrium.
As, [tex]K_p>>1[/tex], the reaction will be product favoured and as it is a acid base reaction where [tex]HONO[/tex] acts as acid by donating [tex]H^+[/tex] ions and [tex]CN^-[/tex] acts as base by accepting [tex]H^+[/tex]
Thus [tex]HONO[/tex] is a strong acid thus [tex]ONO^-[/tex] will be a weak conjugate base and [tex]CN^-[/tex] is a strong base which has weak [tex]HCN[/tex] conjugate acid.
Thus the high value of K indicates that [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
In reaction NH3 →3H2 + N2, how many moles of N2 formed if 2.81 g NH3 dissociate? Show work!
Answer:
0.0826 mol (corrected to 3 sig. fig.)
Explanation:
First, balance the equation:
2NH3 →3H2 + N2
Take the atomic no. of N=14.0, and H=1.0,
no. of moles = mass / molar mass
So, no. of moles of NH3 dissociated = 2.81 / (14.0+1.0x3)
= 0.165294117mol
From the equation, the mole ratio of NH3:N2 = 2:1, meaning for every 2 moles of NH3 dissociated, one mole of N2 is formed.
So, using this ratio, the no. of moles of N2 formed will be 0.165294117 / 2
=0.0826 mol (corrected to 3 sig. fig.)
The gas in a 250. mL piston experiences a change in pressure from 1.00 atm to 2.55 atm. What is the new volume (in mL) assuming the moles of gas and temperature are held constant?
Answer:
[tex]\large \boxed{\text{0.980 L}}[/tex]
Explanation:
The temperature and amount of gas are constant, so we can use Boyle’s Law.
[tex]p_{1}V_{1} = p_{2}V_{2}[/tex]
Data:
[tex]\begin{array}{rcrrcl}p_{1}& =& \text{1.00 atm}\qquad & V_{1} &= & \text{250. mL} \\p_{2}& =& \text{2.55 atm}\qquad & V_{2} &= & ?\\\end{array}[/tex]
Calculations:
[tex]\begin{array}{rcl}\text{1.00 atm} \times \text{250. mL} & =& \text{2.55 atm} \times V_{2}\\\text{250. mL} & = & 2.55V_{2}\\V_{2} & = &\dfrac{\text{250. mL}}{2.55}\\\\& = &\textbf{98.0 mL}\\\end{array}\\\text{The balloon's new volume is $ \large \boxed{\textbf{0.980 L}}$}[/tex]
Calculate the moles of acetic acid used in each trial and record in Data Table 3.Volume acetic acid in Liters = (Mass/Density)/1000Moles = Volume ∗ Concentration
Calculate the moles of magnesium hydroxide (Mg(OH)2) used in each trial and record in Data Table 3.Moles Mg(OH2) = Moles acetic acid
Calculate the neutralization capacity of each trial and record in Data Table. Neutralization Capacity = Moles Mg(OH)2 / Mass of Milk of Magnesium
Data Table Trial 1 Trial 2
Mass of milk of magnesia 2.5g 2.5g
Density of milk of magnesia 1.14 g/mL 1.14 g/mL
Volume of acetic acid, initial 10mL 10mL
Volume of acetic acid, final 2.2mL 1.8mL
Volume of acetic acid, total 7.8mL 8.2mL
Concentration of acetic acid 0.88 M 0.88 M
Moles of acetic acid
Moles of Mg(OH)2
Moles Mg(OH)2 / g milk of magnesia
Answer:
Trial 1: Moles acetic acid = 0.00686 moles;
Moles of Mg(OH)₂ = 0.00343 moles
Neutralization capacity = 0.00137 mol/g
Trial 2: Moles acetic acid = 0.00722 moles
Moles of Mg(OH)₂ = 0.00361 moles
Neutralization capacity = 0.00144 mol/g
Explanation:
Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O
Trial 1:
Moles of acetic acid = concentration * volume in litres
concentration of acetic acid = 0.88 M
volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L
Moles acetic acid = 0.88 M * 0.0078 L
Moles acetic acid = 0.00686 moles
Moles of Mg(OH)₂:
From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂
Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂
Moles of Mg(OH)₂ = 0.00343 moles
Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia
Neutralization capacity = 0.00343 mole /2.5 g
Neutralization capacity = 0.00137 mol/g
Trial 2.
Moles of acetic acid = concentration * volume in litres
concentration of acetic acid = 0.88 M
volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L
Moles acetic acid = 0.88 * 0.0082
Moles acetic acid = 0.00722 moles
Moles of Mg(OH)₂:
From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂
Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂
Moles of Mg(OH)₂ = 0.00361 moles
Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia
Neutralization capacity = 0.00361 mole /2.5 g
Neutralization capacity = 0.00144 mol/g
Choose the substance with the lowest boiling point.
A. NBr3.
B. CI2H2.
C. H2O2.
D. H2S.
E. O2.
Answer:
E. O2
Explanation:
All substances has a simple molecular structure, where between their molecules are held by van der Waals' forces. But C must be incorrect because between the H2O2 molecules, they are mainly held by hydrogen bonds on top of van der Waals' forces. Hydrogen bonds are stronger than van der Waals' forces, so more energy is required to separate the H2O2 molecules.
In structures A and D, the molecules are polar. Their van der Waals' forces are stronger than Cl2H2 and O2, which are non-polar.
Between the Cl2H2 and O2, O2 has a smaller molecular size. The van der Waals' forces between the O2 molecules are hence the weakest. Least amount of energy is required to break the intermolecular forces between the O2 molecules therefore it has the lowest boiling point.
At what temperature is the following reaction feasible: Al2O3(s) + 3C(s) -> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol–1 Entropy (S) = +614 J K–1 mol–1 A. 2096.1 K B. 1273.8 K C. 477.1 K D. 1901.0 K
Answer:
Option A. 2096.1 K
Explanation:
The following data were obtained from the question:
Al2O3(s) + 3C(s) —> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol¯¹
Entropy (S) = +614 JK¯¹ mol¯¹
Temperature (T) =...?
Entropy, enthalphy and temperature are related by the following equation:
Change in Entropy (ΔS) = Change in Enthalphy (ΔH) /Temperature (T)
ΔS = ΔH/T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
Enthalpy (H) = +1287 kJ mol¯¹ = 1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =...?
ΔS = ΔH/T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K.
Describe the reactions during the electrolysis of water in an electrolytic cell. Describe the reactions during the electrolysis of water in an electrolytic cell. Oxygen and hydrogen are both reduced. Oxygen is oxidized and hydrogen is reduced. Neither oxygen or hydrogen are oxidized or reduced. Oxygen and hydrogen are both oxidized. Oxygen is reduced and hydrogen is oxidized.
Answer:
Oxygen is oxidized and hydrogen is reduced
Explanation:
In the electrolysis of water, a pair of Platinum electrodes are immersed in water. Th water has a small quantity of either an acid, salt or base, in most cases H2S04, added to it, to aid ionization. This is because water on its own does not posses enough ions to undergo electrolysis. At the platinum anode, water is oxidized to oxygen gas and hydrogen ions. At the platinum cathode, water is reduced to hydrogen gas and hydroxide ions. The proportion of oxygen and hydrogen produced should be theoretically 1 : 2 respectively, but is not usually so, due to competing side reactions. The hydrogen by product is usually used as a fuel source, and it usually combine with the hydroxide ion to form water back again.
In this experiment, you will analyze your sample by TLC by spotting pure benzophenone and your product and eluting with 5:1 hexanes/ethyl acetate. Choose the statement that BEST describes what you should observe for a successful experiment.
A) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value.
B) The benzophenone in the standard and reaction samples should travel the least and will have the same Rf value. No unreacted benzhydrol should be observed at a higher Rf value.
C) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. Unreacted benzhydrol should be observed at a lower Rf value.
D) The benzophenone in the standard and reaction samples should travel the farthest and will have the different Rf values. No unreacted benzhydrol should be observed at a lower Rf value.
Answer:
The correct answer is statement A.
Explanation:
In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.
For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.
The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.
What is TLC ?In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.
The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.
TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.
Thus, option A is correct.
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The migration of atoms or molecules through a material is called Choose one: biomineralization. precipitation from a gas. solidification of a melt. diffusion.
Answer:
diffusion
Explanation:
Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration in response to a concentration gradient. A concentration gradient simply means a difference in concentration.
Diffusion occurs in solids,liquids and gases. Diffusion is fastest in gases and slowest in solids. Diffusion of solid particles may take very many years while diffusion of gases takes a few milliseconds depending on the mass of the gas.
In materials, atoms and molecules also move from one part of the material to another. This is also refereed to as diffusion.
Modern atomic theory states that atoms are neutral. How is this neutrality achieved in atoms? (2 points)
Rank the compounds in each set in order of increasing acid strength.
(a) CH3CH2COOH CH3CHBrCOOH CH3CBr2COOH
(b) CH3CH2CH2CHBrCOOH CH3CH2CHBrCH2COOH CH3CHBrCH2CH2COOH
Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion [tex]H^+[/tex]).
With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.
See figure 1
I hope it helps!
Give the formulas for all of the elements that exist as diatomic molecules under normal conditions. See if you can do this without looking anything up.
Answer:
They are:
H2, N2, O2, F2, Cl2, Br2, and I2.
Note: whether the element At molecule is monoatomic or diatomic is incredibly arguable. While some say it exists as diatomic because it is a halogen like bromine, iodine etc, At is in fact extremely unstable and no one has ever really studied the molecules on it, so, when others say it is monoatomic, this is also based on calculations. But the other 7 elements listen above is for sure diatomic.
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2).
Hydrogen (H2) , Nitrogen (N2) , Oxygen (O2) , Fluorine (F2) , Chlorine (Cl2) , Iodine (I2) , carbonmonoxide (CO) and Bromine (Br2) are the formulas of the elements that is present as diatomic molecules under normal environmental conditions. Diatomic molecules refers to those molecules that is composed of only two atoms of the same or different elements. There are large number of diatomic molecules which is made up of two similar elements or different elements.
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During which part of the scientific method would error bars be used?
A. Conclusion
B. Analysis
C. Hypothesis
D. Research
please helppppppppppp
Answer:
The correct answer is B. Analysis
Explanation:
Error bars are part of the statistical analysis in the scientific method. Once the scientist have collected the data, he or she proceed to the data analysis. A very common way of comparing the data variability is to use error bars in ghaphical representations. From these bars, it can be estimated the error of a determination and experimental groups are compared.
The error bars would be used during B. Analysis.
What is an Error Bar?A blunders bar is a line through a factor on a graph, parallel to one of the axes, which represents the uncertainty or variant of the corresponding coordinate of the point. In IB Biology, the error bars most often represent the same old deviation of an information set.
When would error bars be used?Blunders bars can be used to examine visual quantities if various other situations preserve. This can decide whether or not differences are statistically sizable. Mistake bars can also propose the goodness of match of a given characteristic, i.e., how well the function describes the facts.
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Assuming an efficiency of 34.90%, calculate the actual yield of magnesium nitrate formed from 139.6 g of magnesium and excess copper(II) nitrate.Mg+Cu(NO3)2⟶Mg(NO3)2+Cu
Answer:
300.44 g
Explanation:
The balanced equation for the reaction is given below:
Mg + Cu(NO3)2 —> Mg(NO3)2 + Cu
Next, we shall determine the mass of Mg that reacted and the mass of Mg(NO3)2 produced from the balanced equation.
This is illustrated below:
Molar mass of Mg = 24 g/mol
Mass of Mg from the balanced equation = 1 x 24 = 24 g
Molar mass of Mg(NO3)2 = 24 + 2[14 + (16x3)]
= 24 + 2[ 14 + 48]
= 24 + 124 = 148 g/mol
Mass of Mg(NO3)2 from the balanced equation =
1 x 148 = 148 g
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Next, we shall determine the theoretical yield of Mg(NO3)2.
This can be obtained as follow:
From the balanced equation above,
24 g of Mg reacted to produce 148 g of Mg(NO3)2.
Therefore, 139.6 g of Mg will react to = (139.6 x 148)/24 = 860.87 g of Mg(NO3)2
Therefore, the theoretical yield of Mg(NO3)2 is 860.87 g
Finally, we shall determine the actual yield of Mg(NO3)2 as follow:
Theoretical of Mg(NO3)2 = 860.87 g
Percentage yield = 34.90%
Actual yield of Mg(NO3)2 =?
Percentage yield = Actual yield /Theoretical yield x 100
34.90% = Actual yield /860.87
Cross multiply
Actual yield = 34.90% x 860.87
Actual yield = 34.9/100 x 860.87
Actual yield = 300.44 g
Therefore, the actual yield of Mg(NO3)2 is 300.44 g
what is the differences between amorphous solid and crystalline solid
Answer:
Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points.
In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.
Draw structures for (a) a chain isomer, (b) a positional isomer, and (c) a functional isomer of hexan-1-ol
(i.e., 1-hexanol)
a. Chain isomer
b. Positional isomer
c. Functional isomer
Answer:
See attached picture.
Explanation:
Hello,
In this case, we should define each type of structural formula as shown below:
- Chain isomers: molecules with the same molecular formula, but different arrangements.
- Positional isomers are constitutional isomers that have the same carbon skeleton and the same functional groups but differ from each other in the location of the functional groups.
- Functional isomers are structural isomers that have the same molecular formula (that is, the same number of atoms of the same elements), but the atoms are connected in different ways so that the groupings are dissimilar.
Regards.
Carbon monoxide gas reacts with hydrogen gas to form methanol: CO (g_ + 2H2 (g) → CH3OH (g) A 1.50L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 397 mmHg. Identify the limiting reactant and determine the theoretical yield of methanol in grams.
:Answer : The limiting reactant is and the theoretical yield of methanol is, 0.96 grams.
Explanation :
First we have to calculate the moles of and .
where,
= pressure of CO gas = 232 mmHg = 0.305 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of CO gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
and,
where,
= pressure of gas = 374 mmHg = 0.492 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of
So, 0.0601 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 2 mole of react to give 1 mole of
So, 0.0601 moles of react with moles of
Now we have to calculate the mass of
Therefore, the theoretical yield of methanol is, 0.96 grams.
The theoretical yield of methanol is 0.496 g of methanol.
The reaction equation is CO (g) + 2H2 (g) → CH3OH (g).
From the partial pressures of each reactant, we can obtain the number of moles of reactants.
For CO;
P = 232 mmHg or 0.305 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.305 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.018 moles
For hydrogen;
P = 397 mmHg or 0.522 atm
V = 1.5 L
T = 305 K
n = ?
R = 0.082 atmL-1mol-1K-1
PV = nRT
n = PV/RT
n = 0.522 atm × 1.5 L/0.082 atmL-1mol-1K-1 × 305 K
n = 0.031 moles
From the reaction equation;
1 mole of CO reacted with 2 moles of H2
0.018 moles of CO will react with 0.018 moles × 2 moles/1 mole
= 0.036 moles of H2
We can see that there is not enough H2 to react with CO hence H2 is the limiting reactant.
2 moles of H2 yields 1 mole of methanol
0.031 moles of H2 yields 0.031 moles × 1 moles/2 mole
= 0.0155 moles of methanol
Mass of methanol produced = 0.0155 moles of methanol × 32 g/mol
= 0.496 g of methanol
Learn more: https://brainly.com/question/2286339
Consider the following reaction: Br2(g) + 3 F2(g) LaTeX: \rightarrow→ 2 BrF3(g) LaTeX: \Delta H_{rxn}Δ H r x n= ‒836 kJ/mol Bond Bond Energy (kJ/mol) Br–Br 193 F–F 155 Using the above bond dissociation energies, calculate the energy, in kJ/mol, of a Br–F bond.
Answer: The energy of a Br–F bond is 110 kJ/mol
Explanation:
The balanced chemical reaction is,
[tex]Br_2(g)+3F_2(g)\rightarrow 2BrF_3(g)[/tex]
The expression for enthalpy change is,
[tex]\Delta H=\sum [n\times B.E(reactant)]-\sum [n\times B.E(product)][/tex]
[tex]\Delta H=[(n_{Br_2}\times B.E_{Br_2})+(n_{F_2}\times B.E_{F_2}) ]-[(n_{BrF_3}\times B.E_{BrF_3})][/tex]
[tex]\Delta H=[(n_{Br_2}\times B.E_{Br-Br})+(n_{F_2}\times B.E_{F_F}) ]-[(n_{BrF_3}\times 3\times B.E_{Br-F})][/tex]
where,
n = number of moles
Now put all the given values in this expression, we get
[tex]\Delta H=[(1\times 193)+(3\times 155)]-[(2\times 3\times B.E_{Br-F})][/tex]
[tex]B.E_{Br-F}=110kJ/mol[/tex]
Thus the energy, in kJ/mol, of a Br–F bond is 110
Natural atom of the same element may have the same _________?
A)proton
B)neutron
C)electron
D)All
Answer:B
Explanation:
Answer: i think it is c
Explanation: i checked my textbook.
Limiting reagent problem. How many grams of H2O is produced from 40.0 g N2O4 and 25.0 g N2H4. N2O4 (l) + 2 N2H4 (l) → 3 N2 (g) + 4 H2O(g)
Answer:
28.13 g of H2O.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)
Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.
This is illustrated below:
Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol
Mass of N2O4 from the balanced equation = 1 x 92 = 92g
Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol
Mass of N2H4 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 4 x 18 = 72 g
Summary:
From the balanced equation above,
92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
92 g of N2O4 reacted with 64 g of N2H4.
Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.
From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.
Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.
Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.
In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.
The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:
From the balanced equation above,
64 g of N2H4 reacted to produce 72 g of H2O.
Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.
Therefore, 28.13 g of H2O were obtained from the reaction.
A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL) with excess benzaldehyde and NaOH to produce 79.4 g of (1E,4E)-1,5-diphenylpenta-1,4-dien-3-one (234.29 g/mol). What is the percent yield of this student's experiment
Answer:
% yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
Explanation:
given
volume of acetone= 43.8 mL
molar weight of acetone = 58.08 g/mol
density of acetone = 0.791 g/mL
A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)
43.8 mL = 43.8mL × 0.791g/mL
= 34.6458g ≈34.65g
1 mole of acetone = 58.08g
∴34.65g = 34.65g/58.08g
= 0.60mol
molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol
mole = mass/ molar weight
mole = 79.4g/ 234.29g/mol
mole(n) = 0.3389mol ≈ 0.34mol
1 mole of acetone will produce 1 mole of the product
∴0.60mol of acetone will produce 0.60mol of the product
but we get 0.34mol of the product
∴ % yield of the student's experiment is
[tex]\frac{0.34}{0.60}[/tex] ˣ 100 = 56.67%
At what pressure would 11.1 moles of a gas occupy 44.8 L at 300 K?
Answer:
[tex]P=6.10atm[/tex]
Explanation:
Hello,
In this case, we can study the ideal gas equation that relates temperature, volume, pressure and moles as shown below:
[tex]PV=nRT[/tex]
Thus, since we are asked to compute the pressure y simply solve for it as follows:
[tex]P=\frac{nRT}{V}=\frac{11.1mol*0.082\frac{atm*L}{mol*K}*300K}{44.8L}\\ \\P=6.10atm[/tex]
Best regards.