Answer:
Ground state
the state with the smallest amount of energy.
Calculate the molarity of the following solution.
45.7 g C10H12 in 5230 mL of solution
unit:
Answer:
Molarity = Moles of solute/Volume of solution(in Litres)
Solute = C10H12
moles of solute = Mass/Molar Mass
Molar Mass of C10H12 = 10(12) + 12(1)
= 132gmol-¹
Mole = 45.7/132 = 0.346mol
Molarity = 0.346/5.23
=0.066M
Calculate the amount of water (in grams) that must be added to (a) 6.80 g of urea [(NH2)2CO] in the preparation of a 9.95 percent by mass solution: g (b) 29.3 g of MgBr2 in the preparation of a 1.70 percent mass solution: g
Explanation:
Amount of water required in each case:
(a)The mass% of the solution is:9.95
Mass of solute that is urea is 6.80 g
To determine the mass of solvent water use the formula:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\9.95=(6.80g/mass of solution )x100\\mass of solution =(6.80 /9.95)x100\\=68.3 g[/tex]
Hence the mass of solvent = mass of solution - the mass of solute
=68.3 g - 6.80g
=61.5 g
Hence, the answer is mass of solvent water required is 61.5 g.
(b) Given mass%=1.70
mass of solute MgBr2 = 29.3 g
The mass of solvent water required can be calculated as shown below:
[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\1.70=(29.3g/mass of solution )x100\\mass of solution =(29.3 g /1.70)x100\\=1720 g[/tex]
The mass of the solution is 1720 g.
Mass of solvent water = mass of solution - mass of solute
=1720 g - 29.3 g
=1690.7 g
Answer: The mass of water required is 1690.7 g.
A 10.53 mol sample of krypton gas is maintained in a 0.8006 L container at 299.8 K. What is the pressure in atm calculated using the
van der Waals' equation for Kr gas under these conditions? For Kr, a = 2.318 L'atm/mol and b = 3.978x10²L/mol.
Answer:
-401.06 atm
Explanation:
Applying,
P = (nRT/V-nb)-(an²/V²)............... Equation 1
Where P = Pressure, R = Universal gas constant, V = molar Volume, T = Temperature in Kelvin, a = gas constant a , b = gas constant b, n = numbers of mole
From the question,
Given: T = 299.8 K, V = 0.8006 L, a = 2.318 L.atm/mol, b = 3.978×10²L/mol
Constant: R = 0.0082 atm.dm³/K.mol
Substitute these values into equation 1
P = [(0.0082×299.8×10.53)/(0.8006-(10.53×397.8)]-[(10.53²×2.318/0.8006²)]
P = (25.89/-4188.0334)-(400.995)
P = -0.0618-400.995
P = -401.06 atm
How many moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+?
Answer:
0.1 mol
Explanation:
6/(15*3+15)
0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+
What is mole?
The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.
The compound trimethylamine, (CH3 )3N, exists as a weak base when dissolved in water.
A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.
To find the amount of the substance (CH3)3NH+ to calculate its molar mass:
M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol
n((CH3)3NH+) = m/M
m((CH3)3NH+) = 6g
Thus,
n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol
Hence,
n((CH3)3NH+) = 0.1 mol
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1) Recall the two written definitions of an oxidation-reduction reaction provided in our lessons. Which of these definitions is
most inclusive of redox reactions? Explain your answer:
A redox reaction is where the oxidation and reduction reaction takes place at the same time, the oxidation half
reaction involves losing electrons and in the reduction half reaction involves gaining electrons. So in a redox
reaction an electron is lost by the reducing agent.
Explain how the reaction below meets these definitions. Which substance is being oxidized and which is
being reduced?
4Ag(s) + 2H2S(g) + O2(g)
2Ag2S(s) + 2H20(9)
Answer:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.
Explanation:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.
The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.
The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.
In the given reaction as shown below:
4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)
The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.
The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.
On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.
The oxidation numbers of hydrogen ion and sulfide ion do not change.
You are inside a room with a temperature of 11°C. You step outside and the temperature is 100°C. What is the AT?
A) 0.76°C
B) 157°C
C) 89°C
D) 6052°C
E) 1.31°C
F) 21°C
Answer:
89°c
Explanation:
i think this is the answer cause the temperature changed from 11 to 100 and so the atmospheric temperature would be the change in temperature
100-11=89°c
I hope this helps and sorry if it's wrong
A solution of acetic acid and water contains 205.0 g L-1 of acetic acid and 820.0 g L-1 of water. Compute the density of the solution ( report your answer in g per mL)
Answer:
[tex]\rho_t=1025000 gmL^{-1}[/tex]
Explanation:
From the question we are told that:
Density of acetic acid [tex]\rho_a=205 gL^{-1}[/tex]
Density of Water [tex]\rho_w=820 gL^{-1}[/tex]
Generally the equation for Solution Density is mathematically given by
[tex]\rho_t= \rho_w+\rho_a[/tex]
[tex]\rho_t=205+820[/tex]
[tex]\rho_t=1025 gL^{-1}[/tex]
[tex]\rho_t=1025000 gmL^{-1}[/tex]
Compounds such as butane and isobutane that have the same molecular formula but differ in the order in which the atoms are connected are called ____________
a. trans isomers
b. cis isomers
c. conventional isomers
d. constitutional isomers
Answer:
One compound, called n-butane, where the prefix n- represents normal, has its four carbon atoms bonded in a continuous chain. The other, called isobutane, has a branched chain. Different compounds that have the same molecular formula are called isomers.
Answer:
d. constitutional isomers
Explanation:
i hope it will help
For which of the following reactions is the enthalpy change equal to the second ionization energy of nitrogen?
Answer:
"[tex]N^+(g) \rightarrow N^{2+}(g) + e^-[/tex]" is the appropriate answer.
Explanation:
Whenever one electron or particle must be removed from some kind of gas atom or molecule, it requires that the very first amount of energy necessary.Two electrons must be removed from such a mono-positive exhaust gases structure or position of ion before they may become a dipositive gaseous ion.Thus the above is the correct answer.
According to the Vaporization Heat table, the heat needed for 1 mol of H2O to evaporate at 100°C is 40.7KJ and 44.0KJ/mol is needed to evaporate H2O at 25°C. Thus 44.0-40.7=3.7KJ is the energy needed to heat H2O to 100°C from 25°C.
However, according to the heat capacity of H2O, 3.7KJ will only warm the water by ~+43°C, which is not enough to reach 100°C starting from 25°C!
Am I missing something?!
Suppose you have a material in it's liquid phase. As you give energy to that liquid, the temperature of the liquid will increase gradually, and the relation between the increase of temperature and the given energy is the specific heat.
Now, there is a point, a critical point, where the temperature stops to increase, which means that we are near a change of phase. So from this point on, the energy is not used to increase the kinetic energy of the particles (which would increase the temperature), the energy is used to break the bonds and allow a change of phase, for example, from liquid to gas.
So, we know that if you have a mol of water at 100°C, then you need to add 40.7 kJ of energy to change the phase of the water from liquid to gas phase.
This means that if you have a mol of water and you give that exact energy, the temperature will not change, instead, you now will have a mol of water at the temperature of 100°C.
Similarly with the case at 25°C (which happens for a particular pressure only)
So the heat of vaporization can not really be related to increases in temperature as you thought.
For changes in temperature, you need to use the specific heat.
We know that for water it is:
c = 4.184 J/g*°C = 76.15 J/mol*°C
So, if you want to increase the temperature from 25° to 100°
This means an increase of 75°C of one mol of water.
We just need to multiply the above number by:
1mol*(75°C)
Energy needed = (76.15 J/mol*°C)*1mol*(75°C) = 5,711.25 J
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List the following substances in order of decreasing boiling point:
CO2, Ne, CH3OH, KF
The correct order of the substances in order of decreasing boiling point is,
KF , CH30OH , CO2 , Ne
What is boiling point?Boiling point, the temperature at which the force exerted by the surroundings upon a liquid exists equaled by the pressure exerted by the vapor of the liquid; under this situation, the addition of heat affects the transformation of the liquid into its vapor without increasing the temperature.
The main difference between the boiling point and the melting point stands that the melting point is determined as the temperature at which solid and liquid phases exist in equilibrium, whereas the boiling point stands as the temperature at which the vapor pressure of a liquid stands equal to the external pressure.
Hence, The correct order of the substances in order of decreasing boiling point is,
KF , CH30OH , CO2 , Ne
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The metal sample suspected of being aluminum is warmed and then submerged into water, which is near room temperature. The final temperature of the water and the metal is given below. The specific heat capacity of water is 4.18 J/g.oC. Calculate the specific heat capacity of the metal based on the data below. Remember heat lost = heat gained.
Type of metal used:
Trial 1 Trial 2 Trial 3
Mass of metal, g 2.746 g 2.750 g 2.900 g
Mass of water, g 15.200 g 15.206 g 15.201 g
Initial Temp. of Water, oC 24.7 oC 24.6 oC 24.5 oC
Initial Temp. of Metal, oC 72.1 oC 72.2 oC 71.9 oC
Final Temp of Water & Metal,oC 26.3 oC 26.2 oC 24.7 oC
ΔT for water, oC ______ ______ ______
ΔT for metal, oC ______ ______ ______
Specific heat capacity of metal, J/g.oC ______ ______ ______
Average specific heat capacity, J/g .oC ______ (use two significant figures due to ΔT of water)
Answer:
Average specific heat capacity of metal = 0.57 J/g°C
Explanation:
Heat lost = Heat gained
Heat energy gained or lost, H = mcΔT
where m = mass of substance, c = specific heat capacity, ΔT = temperature change
Trial 1:
Heat lost by metal = -[2.746 g × c × ΔT]
ΔT = (26.3 - 72.1) °C = -45.8 °C
Heat lost by metal = -[2.746 g × c × (-45.8 °C)] = c × (125.7688)g°C
Heat gained by water = 15.200 × 4.18 × ΔT
ΔT = (26.3 - 24.7) = 1.6 °C
Heat gained by water = 15.200 × 4.18 × 1.6 = 101.6576 J
From Heat lost = Heat gained
c × (125.7688)g°C = 101.6576 J
c = 101.6576 J / 125.7688 g°C
c = 0.8083 J/g°C
Trial 2:
Heat lost by metal = -[2.750 g × c × ΔT]
ΔT = (26.2 - 72.2)°C] = - 46 °C
Heat lost by metal = -[2.750 g × c × (-46 °C)
Heat lost by metal = c × (126.5) g°C
Heat gained by water = 15.206 × 4.18 × ΔT
ΔT = (26.2 - 24.6) = 1.6 °C
Heat gained by water = 15.206 × 4.18 × 1.6 = 101.697728 J
From Heat lost = Heat gained
c × (126.5)g°C = 101.6977 J
c = 101.697728 J / 126.5 g°C
c = 0.8039 J/g°C
Trial 3:
Heat lost by metal = -[2.900 g × c × ΔT]
ΔT = (24.7 - 71.9)°C] = - 47.2 °C
Heat lost by metal = -[2.900 g × c × (- 47.2 °C)
Heat lost by metal = -[2.900 g × c × (- 47.2)°C] = c × (136.88)g°C
Heat gained by water = 15.201 × 4.18 × ΔT
ΔT = (24.7 - 24.5) = 0.2 °C
Heat gained by water = 15.201 × 4.18 × 0.2 = 12.708036 J
From Heat lost = Heat gained
c × (136.88)g°C = 12.708036 J
c = 12.708036 J / 136.88 g°C
c = 0.0928 J/g°C
Average specific heat capacity of metal = (0.8083 + 0.8039 + 0.0928) J/g°C / 3
Average specific heat capacity of metal = 0.57 J/g°C
What are fires classified by?
Answer:
A fire class is a system of categorising fire with regard to the type of material and fuel for combustion. Class letters are often assigned to the different types of fire, but these differ between territories. There are separate standards for the United States, Europe, and Australia
tính ΔH° của phả ứng sau ở 200°C
CO+1÷2O=CO2
ΔH°
How many moles of Al2O3 can be formed from 10.0 g of Al?
Select an answer and submit. For keyboard navigation, use the up/down arrow keys to
select an answer.
Answer:
n Al= 10/27( mol)- >n Al2O 3 =5/27(mol)
Explanation:
2. Calculate the wavelength of the emitted photon from hydrogen for the transition from ni = 3 to nf = 2. What part of the visible spectrum is this wavelength? Visible wavelengths are: Red 700 - 620 nm, Yellow 620 - 560 nm, Green 560 - 500 nm, Blue 500 - 440 nm, and Violet 440 - 400 nm.
Answer:
The correct answer is "654.54 nm".
Explanation:
According to the question,
⇒ [tex]\frac{1}{\lambda}= Rh(\frac{1}{n1^2} -\frac{1}{n2^2} )[/tex]
By substituting the values, we get
[tex]=1.1\times 10^7(\frac{1}{4} -\frac{1}{9} )[/tex]
[tex]=1.1\times 10^7(\frac{9-4}{36} )[/tex]
[tex]=1.1\times 10^7(\frac{5}{36} )[/tex]
[tex]=654.54\ nm[/tex]
Thus the above is the right solution.
Write the balanced half-reactions and an overall cell reaction below. Be sure to include states of matter.
Answer: Hello your question is incomplete attached below is the missing image
answer:
Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq) + 2e⁻ ( occurs at anode )
Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s) ( occurs at the cathode )
Overall cell reaction ; Zn(s) + CO²⁺(aq) ⇒ Zn⁺² (aq) + CO (s)
Explanation:
stating the standard reduction potentials
E° zn²⁺/zn = -0.76 v
E°Co²⁺ / Co = -0.28 v
since ; -0.76 v < -0.28 v. Zn will be oxidized while Co²⁺ will be reduced .
Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq) + 2e⁻ ( occurs at anode )
Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s) ( occurs at the cathode )
hence the
Overall cell reaction ;
Zn(s) + CO²⁺(aq) ⇒ Zn⁺² (aq) + CO (s)
What alcohol is formed formed when the Alkene is treated with H2O in the presence of h2so4
Explanation:
Ethanol is made by the hydration of ethylene in the presence of a catalyst such as sulfuric acid (H 2SO 4).
The radioactivity due to carbon-14 measured in a piece of a wood from an ancient site was found to produce 20 counts per minute from a given sample, whereas the same amount of carbon from a piece of living wood produced 160 counts per minute. The half-life of carbon-14, a beta emitter, is 5730 y. The age of the artifact is closest to
Answer:
The answer is "17200 years".
Explanation:
Given:
[tex]A = 20 \ \frac{counts}{minute}\\\\A_{o} = 160\ \frac{counts}{minute}[/tex]
Let the half-life of carbon-14, is beta emitter, is [tex]T = 5730\ years[/tex]
Constant decay [tex]\ w = \frac{0.693}{ T}[/tex]
[tex]= 1.209 \times 10^{-4} \ \frac{1}{year}\\[/tex]
The artifact age [tex]t= ?[/tex]
[tex]A = A_{o} e^{-wt} \\\\e^{-wt} = \frac{A}{A_{o}}\\\\-wt = \ln \frac{A}{A_{o}}\\\\= -2.079\\\\t = 1.7199 \times 10^{4} \ years\\\\\sim \ 17200\ years\\[/tex]
Please help with all 3 parts!
Answer:
1:Part A.
[tex]\bold{42.2 g C_{12}H_{22}O_{11} \:in \:528 g H₂O}[/tex]
Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]
=[tex]\frac{42.2}{528}*100\%=\bold{\underline{7.99\: or \:8\%}}[/tex]
Part B.
[tex]\bold{198\:m g\: C_{6}H_{12}O_{6} \:in\:4.71 g\: H₂O}[/tex]
mass of solute: 198mg
mass of solvent :4.71g=4710g
Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]
=[tex]\frac{198}{4710}*100\%=\bold{\underline{4.20\%}} [/tex]
Part C.
[tex]\bold{8.85 g NaCl \:in \:190 g\: H₂O}[/tex]
Mass Percent=[tex]\bold{\frac{Mass\: of \:Solute}{Mass\: of \:Solution}×100\%}[/tex]
=[tex]\frac{8.85}{190}*100\%=\bold{\underline{4.66\%}}[/tex]
Answer:
It will help you !!!!!!!!!!
A Single Orbital With Two Lobes At 90°In A Single Plane And A Node In The Center Would Likely Be Found Where?
a.4s
b.4p
c.4d
d. it would not be found in any of these
e.4f
Predict the missing component in the nuclear equation.
238 92U → 234 90Th + X
A. 4 2He
B. 0 -1e
C. 0 0v
Answer:
A
Explanation:
helium (alpha particle)
If the concentration of [CO32−] were doubled (say, by adding a highly soluble carbonate salt such as Cs2CO3), what would be the new concentration of [Li+] in the saturated solution?
Answer:
ufvkoo5dvjutfbnjiiihhb
Which of the following is used in EBRT?
O Silver tube
O Gold tube
O Copper tube
O Iron tube
Copper tube is used in EBRT.
What is meant by EBRT?External Beam Radiation. Therapy (EBRT) is a type of radiation therapy that directs a beam of radiation from outside the body, toward cancerous tissues inside the body.External beam radiation therapy (EBRT) is the most common type of radiation therapy. It directs high-energy radiation beams at the cancer.Copper tube is used in EBRT.
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Which of the following statements is true about what happens in all chemical reactions? A. The ways in which atoms are joined together is not changed. B. Bonds between atoms are broken and new bonds are formed. C. The final substances are called reactants.
Answer:
B.bonds are broken and new bonds are formed
explain hydrogen dioxide
Answer:
Two molecules of hydrogen combine with two molecules of oxygen to form hydrogen peroxide. Hence, its chemical formula is H2O2. It is the simplest peroxide (since it is a compound with an oxygen-oxygen single bond). Hydrogen peroxide has basic uses as an oxidizer, bleaching agent and antiseptic
Which of the following reasons correctly explains the color changes that take place when ethylenediamine (C2N2H8) is added to the solution of cobalt(II) chloride?
a. Addition of the liquid ethylenediamine dilutes the concentration of cobalt(II) chloride in the solution resulting in a color change.
b. The ethylenediamine is oxidized and the resulting solution is deeply colored.
c. The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.
Answer:
The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.
Explanation:
The five d-orbitals are usually degenerate. Upon approach of a ligand, the d-orbitals split into two sets of orbitals depending in the nature of the crystal field.
The magnitude of crystal field splitting is affected by the nature of the ligand. Ligands having filled p-π orbitals such as ethylenediamine lead to greater crystal field splitting.
The change in the colour that takes place when ethylenediamine is added to the solution of cobalt(II) chloride occurs due to a different crystal field splitting pattern. Thus, the energy associated with electron transitions between the d-orbitals now differ for the two compounds showing a color change.
Tapeworm and Roundworn
Answer:
Tapeworms and roundworms both belong to the same phylum however, their families are different from one another.
Tapeworms are flat, segmented intestinal parasites of the cat and dog and humans sometimes. They are present in the intestines of pets and depend on them, therefore, are parasites. These parasites look like tape which gives it its name.
Roundworms can also infect humans and the most common cases are among children. When not treated immediately, they can cause severe damage to a human host and can even cause blindness. Tapeworms are white in color with a long, segmented body.
It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of by-product formed. What is the by-product
Answer:
Biphenyl
Explanation:
The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.
The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.
Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.
what is the mass of 4 moles of fluorine atoms?
Answer:
Vien, sometimes we make problems like this harder than they need to be. Suppose I asked you, “How many dozen wheels are on four dozen automobiles?” You would have no trouble answering 4 x 4 = 16 dozen.
A mole is just a quantity like a dozen. And in this case, instead of 4 wheels, each C2F6 molecule bears 6 fluorine atoms, right?
So 4 x 6 = 24 moles of fluorine atoms. You’ve got this, Vien!
