Answer:
The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)
Explanation:
The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)
what happens when you combine Mg2 and NO3
Answer: they blow up
Explanation: add them together and they will blow up
Answer:
Magnesium nitrate Reactions
Magnesium nitrate has a high affinity towards water. Therefore, heating it results to decompose into magnesium oxide, nitrogen oxides, and oxygen. 2 Mg(NO3)2 → 2 MgO + 4 NO2 + O2.
CHEMISTRY HELP!
using only the periodic table, determine the charge on the ion that is formed by arsenic.
The ion charge is:
a. -3
b. -2
c. -1
d. 0
e. +1
f. +2
g. +3
also what is it for elements lithium and strontium?
Answer:
A
Explanation:
Arsenic is in the same group as Nitrogen - group 5. They all have 5 valence electrons in their outermost shell. To achieve its most stable state - 8 valence electrons (octet rule - elements are most stable when the entire shell is filled) - arsenic needs to gain 3 electrons. Since electrons have a negative charge, the charge of an As ion would be -3.
Try observing the periodic table and how many valence electrons that each element has. From there, you can determine the charges of the elements lithium and strontium. You can guess, I'll help you with those once you attempt to find the charge of those ions.
A gaseous hydride of Nitrogen
contains its own volume of Nitrogen
and twice its volume of Hydrogen
and has vapour density 16. The
formula of the hydride is.
Select one:
a. NH2
b. NH3
c. N3H
• d. N2H4
Answer:
N2H4
Explanation:
A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.
Now consider the hydride N2H4.
N2H4(g) -----> N2(g) + 2H2(g)
The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.
The molecular mass of the compound is;
N2H4= 2(14) + 4(1)= 28+4= 32
Since
molecular mass= 2 vapour density
Vapour density= molecular mass/2
Vapour density= 32/2
Vapour density = 16
Therefore the hydride of nitrogen referred to in the question is N2H4
Classify the following unbalanced chemical reaction Fe(OH)2(s) + HCl(aq) = FeCl2(aq) + H2O(l)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction
Answer:
1. Acid-Base Reaction
Explanation:
Fe(OH)2(s) + HCl(aq) = FeCl2(aq) + H2O(l)
base acid
This a reaction between base and acid.
Ferrous hydroxide is an inorganic alkaline compound whereas hydrochloric acid is an acid. The reaction between Fe(OH)₂and HCl is an acid-base reaction. Thus, option 1 is correct.
What is an acid-base reaction?An acid-base reaction is a chemical change that occurs and takes place when the reactant constitutes an acid and a base. They are characterized by the exchange of protons that results in the formation of conjugate bases and acids or salt.
The acid-base chemical reaction is shown as,
Fe(OH)₂(s) + HCl(aq) ⇒ FeCl₂(aq) + H₂O(l)
Here, ferrous hydroxide is a base with hydroxide ions and hydrochloric acid is an acid with hydrogen ions. HCl donates its proton to form water molecules with hydroxide ions of ferrous hydroxide.
Therefore, in option 1. the reaction is an acid-base reaction.
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what type of bond is most likely form between 2 gold atoms
Answer:
Metallic
Explanation:
"Metallic bonding is a type of chemical bonding that rises from the electrostatic attractive force between conduction electrons and positively charged metal ions. It may be described as the sharing of free electrons among a structure of positively charged ions." -Wikipedia
In which of these statements are protons, electrons, and neutrons correctly compared?
Quarks are present in protons and neutrons but not in electrons.
Quarks are present in protons, neutrons, and electrons.
Quarks are present in neutrons and electrons but not in protons.
Quarks are present in protons and electrons but not in neutrons
the second statement is the correct one quarks are needed to balance charges in all subatomic particles such as neutrons, protons and electrons
2. In a paper chromatography analysis, three pigments, A, B, and C, were dissolved in a polar solvent. A is slightly polar, B is highly polar, and C is moderately polar. List in order how these will appear on the surface of the chromatography
If 196L of air at 1.0 atm is compressed to 26000ml, what is the new pressure
Answer:
7.5 atm
Explanation:
Initial pressure P1 = 1.0 ATM
Initial volume V1= 196 L
Final pressure P2= the unknown
Final volume V2= 26000ml or 26 L
From Boyle's law we have;
P1V1= P2V2
P2= P1V1/V2
P2= 1.0 × 196/26
P2 = 7.5 atm
Therefore, as the air is compressed, the pressure increases to 7.5 atm.
Identify the particle represented by each symbol as an alpha particle, a beta particle, a gamma ray, a positron, a neutron, or a proton.
a. 11P
b. 42He
c. +10e
Answer:
[tex]_1^{1} {P}[/tex] is symbol for proton emission in the nucleus.
[tex]_2^{4} {He}[/tex] symbolises alpha emission, equivalent to helium atom emission of a radioactive particle
[tex]+_1^{0} {e}[/tex] is the radiation symbol for positiron particle. which occurs when beta + radioactive decay occurs
A sample of 6.022 x 1023 particles of gas has a volume of 22.4 L at 0°C and a pressure of 1.000 atm. Although it may seem silly to contemplate, what volume would 1 particle of gas occupy?
pv=nRT
Answer:
1 particle of the gas would occupy a volume of 3.718*10⁻²³L
Explanation:
Hello,
1. The sample has a particle of 6.022×10²²particles
2. Volume of the sample = 22.4L
3. Temperature of the sample = 0°C = (0 +273.15)K = 273.15K
4. Pressure of the sample = 1.0atm
What volume would 1 particle of the gas occupy?
But we remember that 1 mole of any substance = 6.022×10²² molecules or particles or atoms
What would be the number of moles for 1 particule?
1 mole = 6.022×10²² particles
X moles = 1 particle
X = (1 × 1) / 6.022×10²² particles
X = 1.66×10⁻²⁴ moles
Therefore, 1 particle contains 1.66×10⁻²⁴ moles
Since we know our number of moles, we can proceed to use ideal gas equation,
Ideal gas equation holds for all ideal gas and is defined as
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles of the gas
R = ideal gas constant = 0.082 L.atm / mol.K
T = temperature of the gas
PV = nRT
Solving for V,
V = nRT/ P
We can now plug in our values into the above
equation.
V = (1.66*10⁻²⁴ × 0.082 × 273.15) / 1
V = 3.718*10⁻²³L
Therefore, 1 particule of the gas would occupy a volume of 3.718*10⁻²³L.
A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C
Answer:
FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL
Explanation:
From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:
k = Ae^ -Ea/RT
At initial temperature T1, the initial rate constant is (k1)
At final temperature T2, the final rate constant is k2
For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.
That is, k2 / k1 = 2 (rate is doubled)
Equating this into the Arrhenius equation, we have:
k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)
2 = e^ (-Ea / R) (1 / T2 = 1 / T1)
Taking the natural logarithm of both sides:
ln 2 = - (Ea / R) (1 / T2 - 1 / T1)
Making Ea the subject of the formula, we obtain:
Ea = - (ln 2 R / (1 / T2- 1 / T1))
Let T1 = 25 C = 25 + 273 K = 298 K
T2 = 35 C = 35 + 273 K = 308 K
R = 8.314
So,
Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))
Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)
Ea = - 5.7616 / -0.00011
Ea = 52 378,18 J / mol
So therefore, the activation energy Ea is 52.4 kJ/mol.
which element causes burning when me mix it with oxygen
Answer:
Hydrogen peroxide is ans
The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 500 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg
Answer:
Explanation:
a )
m = m₀ [tex]e^{-\lambda t[/tex]
m is mass after time t . original mass is m₀ , λ is disintegration constant
λ = .693 / half life
= .693 / 1590
= .0004358
m = m₀ [tex]e^{- 0.0004358 t}[/tex]
b )
m = 50 x [tex]e^{-.0004358\times 500}[/tex]
= 40.21 mg .
c )
40 = 50 [tex]e^{-.0004358t[/tex]
.8 = [tex]e^{-.0004358t[/tex]
[tex]e^{.0004358t[/tex] = 1.25
.0004358 t = .22314
t = 512 years .
When 106 g of water at a temperature of 21.4 °C is mixed with 64.3 g of water at an unknown temperature, the final temperature of the resulting mixture is 46.8 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.)
Answer:
THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K
Explanation:
Mass of first sample of water = 106 g
Initial temp of first sample = 21.4 °C = 21.4 + 273 K = 294.4 K
Mass of second sample = 64.3 g
Final temp of theresulting mixture = 46.8 °C = 46.8 + 273 K = 319.8 K
Specific heat capacity of water = 4.184 J/g K
It is worthy to note that;
Heat gained by the first sample = Heat lost by the second sample
Since heat = mass * specific heat capacity * change in temperature, we have
Mass * specific heat * change in temp of the first sample = Mass * specific heat * change in temp. of the second sample
MC (T2 - T1) = MC (T2-T1)
106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)
106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)
11 265.0016 = 269.0312 (319.8 - T1)
Since the change in temperature = 319.8 -T1
Change in temperature =11265.0016 / 269.0312
Change in temperature = 41.87
Change in temperature = 319.8 -T1
41.87 = 319.8 - T1
T1 = 319.8 - 41.87
T1 = 277.93 K
T1 = 4.93 °C
So therefore, the initial temperature of the sacond sample is 4.73 °C or 277.93 K
Using the following balanced chemical equation 8 H2 + S8à 8 H2S. Determine the mass of the product (molar mass = 34.08g/mol) if you start with 1.35 g of hydrogen and 6.86 g of S8 (Molar mass = 256.5 g/mole).
Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]
[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]
[tex]8H_2+S_8\rightarrow 8H_2S[/tex]
According to stoichiometry :
1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2[/tex]
Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2S[/tex]
Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]
Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
The type of nuclear decay an unstable nucleus will undergo depends on its ratio of neutrons to protons. The radioisotope cobalt-65 has a ratio of neutrons to protons of 1.41, which is too high for a nucleus of this size. What nuclear changes could reduce this ratio
Answer:
Explanation:
In cobalt - 65 ,
no of protons is 27 ( p )
no of neutron = 65 - 27 ( n )
= 38
n / p ratio
= 38 / 27
= 1.41
If case of emission of alpha particle
no of proton p = 27 - 2 = 25
no of neutrons = 38 - 2 = 36
n / p ratio = 36 / 25
= 1.44
So it increases
In case of emission of beta particle
No of neutron n = 38 - 1 = 37
No of proton = 27 + 1 = 28
n / p ratio = 37 / 28
= 1.32
Hence ratio decreases.
Hence beta ray decay will result in decrease in n / p ratio.
The solubility of N2 in water at a particular temperature and at a N2 pressure of 1 atm is 6.8 × 10–4 mol L–1. Calculate the concentration of dissolved N2 in water under normal atmospheric conditions where the partial pressure of N2 is 0.78 atm.
Answer:
The correct answer is 5.30 * 10^-4 mol per L.
Explanation:
Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.
Therefore, it can be written as,
C2/C1 = P2/P1
Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,
C2/(6.8*10^-4) = 0.78/1
C2 = 0.78 * 6.8*10^-4
C2 = 5.30 * 10^-4 mol per L.
Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.
Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a tank with of ammonia gas, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 13. mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The concentration equilibrium constant is [tex]K_c = 14.39[/tex]
Explanation:
The chemical equation for this decomposition of ammonia is
[tex]2 NH_3[/tex] ↔ [tex]N_2 + 3 H_2[/tex]
The initial concentration of ammonia is mathematically represented a
[tex][NH_3] = \frac{n_1}{V_1} = \frac{29}{75}[/tex]
[tex][NH_3] = 0.387 \ M[/tex]
The initial concentration of nitrogen gas is mathematically represented a
[tex][N_2] = \frac{n_2}{V_2}[/tex]
[tex][N_2] = 0.173 \ M[/tex]
So looking at the equation
Initially (Before reaction)
[tex]NH_3 = 0.387 \ M[/tex]
[tex]N_2 = 0 \ M[/tex]
[tex]H_2 = 0 \ M[/tex]
During reaction(this is gotten from the reaction equation )
[tex]NH_3 = -2 x[/tex](this implies that it losses two moles of concentration )
[tex]N_2 = + x[/tex] (this implies that it gains 1 moles)
[tex]H_2 = +3 x[/tex](this implies that it gains 3 moles)
Note : x denotes concentration
At equilibrium
[tex]NH_3 = 0.387 -2x[/tex]
[tex]N_2 = x[/tex]
[tex]H_2 = 3 x[/tex]
Now since
[tex][NH_3] = 0.387 \ M[/tex]
[tex]x= 0.387 \ M[/tex]
[tex]H_2 = 3 * 0.173[/tex]
[tex]H_2 = 0.519 \ M[/tex]
[tex]NH_3 = 0.387 -2(0.173)[/tex]
[tex]NH_3 = 0.041 \ M[/tex]
Now the equilibrium constant is
[tex]K_c = \frac{[N_2][H_2]^3}{[NH_3]^2}[/tex]
substituting values
[tex]K_c = \frac{(0.173) (0.519)^3}{(0.041)^2}[/tex]
[tex]K_c = 14.39[/tex]
Clues about the history of the earth have been obtained from the study of
fossil fuels
rain forest materials
soll samples
O synthetic plastics
Answer: Fossil fuels
Explanation:
Fossil fuels such as petroleum, oil, and natural gas, are non-renewable energy resources which are formed from the remains of prehistoric ancient plants and animals beneath layers of rock of the earth surface.
By analyzing and studying fossil fuels using Radiocarbon analyses by archaeologists, earth scientists etc, Information about the history of the earth can be obtained from the decomposition of dead organisms present in fossil fuels.
Which compound would you expect to be least soluble in water? Explain.
a. CCl4
b. CH3Cl
c. NH3
d. KF
Answer: a.CCl4 aka carbon tetrachloride
Explanation:
ionic compounds and polar molecules can be dissolved in water which is a polar solvent.
choice d (KF) is a salt (an ionic compound) and can be dissolved in water /(K+ and F- ions would be formed in water).
choice c (NH3 or ammonia) is a very polar molecule and thus can be dissolved in water(Hydrogen bonding).
choice b (CH3Cl) is slightly polar because the atoms surrounding the central carbon atom are different(3 H atoms and 1 chlorine atom) and can be dissolved in water(Dipole-dipole interaction).
choice a is nonpolar and cannot be dissolved in water.
WILL GIVE BRAINLIEST!!!!! will give brainliest!!!! ******According to the reaction in question I above, how many grams of solid copper will theoretically be
produced when 14.4 g of aluminum are reacted with 14.4 g of copper (II) sulfate? Which reactant is the
limiting reactant? Show your work. Be sure to include units!
Answer:
5.73 g Cu
Explanation:
M(CuSO4) = 159.6 g/mol
M(Al) = 27.0 g/mol
M(Cu) = 63.5 g/mol
14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al
14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 2 mol 3 mol
given (0.5333 mol )x 0.0902 mol
needed 0.0601 mol
x= 2*0.0902/3 = 0.0601 mol Al
Al is excess, CuSO4 is a limiting reactant.
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 3 mol 3 mol
given 0.0902 mol x mol
x = 0.0902 mol Cu
0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu
Based on VSEPR theory and your observations from the Molecular Geometry lab consider the following questions What is the predicted hybridization at an atom which is surrounded by a double bond and two single bonds?
a) Sp
b) sp^2
c) sp^3
Answer:
b) sp^2
Explanation:
Hybridization refers to the concept that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding properties. In chemistry, orbital hybridisation (or hybridization) is the implies the mixing of atomic orbitals to form hybrid orbitals (with different energies, shapes, etc., different from that of the component atomic orbitals) suitable for the pairing of electrons to form chemical bonds according to the principles of the valence bond theory.
In 1931 Linus Pauling proposed the idea of “mixing” the orbitals or “hybridizing” them to account for certain observed bonding patterns. Pauling proposed a sort of a combination of the orbitals giving you an orbital that has partial characters.
Hybridization is merely a mathematical construct. It is never an actual “process” that occurs within orbitals . Hybridization is a mathematical model that describes how the atomic orbitals would’ve looked like based on the observable molecular orbitals.
sp2 hybridization leads to the formation of a double bond. sigma bonds may also be formed depending on the valency of the central atom. In alkenes, an sp2 hybridized carbon atom forms a double bond in addition to two sigma bonds to other atoms.
The predicted hybridization is:
b) [tex]sp^2[/tex]
What does Hybridization tell us?It is the integration of atomic orbitals to shape new orbitals with exclusive energies and shapes than the unique orbitals.
Given: An atom that's surrounded with the aid of using a double bond and unmarried bonds.
[tex]sp^2[/tex] hybridization ends in the formation of a double bond. sigma bonds can also be shaped relying at the valency alkenes, an [tex]sp^2[/tex]sigma bonds to different atoms.
Thus, correct option is b.
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An unknown compound, B, has the molecular formula C7H12. On catalytic hydrogenation 1 mol of B absorbs 2 mol of hydrogen and yields 2-methylhexane. B has significant IR absorption band at about 3300 and 2200 cm-1. Which compound best represents B?
a. 5-methyl-1,3-hexadiene
b. 5-methyl-1-hexyne
c. 3-methyl-1-hexyne
d. 5-methyl-2-hexyne
e. 2-methyl-1,5-hexadiene
Answer:
B and D
Explanation:
If we use the info given we have a band a 3300 cm-1 and 2200 cm-1 this indicates that we have an alkyne functional group. Additionally, the hydrogenation of the unknown molecule will consume two moles of hydrogens this fits with the 2 pi bonds in the alkyne functional group. So, we can discard "a" and "e". The product of this hydrogenation is 2-methylhexane therefore we can discard c because the methyl group is placed on carbon 3. Structures b and d can work.
See figure 1
I hope it helps!
n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a phenolphthalein end point. Calculate the molarity of the acetic acid solution, and the percent (by weight) of acetic acid in the solution (assuming its density to be 1.00 g/mL).
Answer:
Molarity: 0.111M
% (w/w): 0.666
Explanation:
The reaction of NaOH with acetic acid (CH₃COOH) is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.
As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:
0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:
4.169x10⁻³ moles of CH₃COOH.
Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:
4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M
As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:
4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid
Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.
Thus, percent by weight is:
0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)
The molarity of acetic acid is 0.11M and the percent by weight is 0.666%.
How we calculate molarity?Molarity of any solution is used to define their concentration and it will be calculated as:
M = n/V, where
n = moles
V = volume
Molarity of acetic acid will be calculated as:
M₁V₁ = M₂V₂, where
M₁ = molarity of acetic acid = ?
V₁ = volume of acetic acid = 37.54mL = 0.037L
M₂ = molarity of NaOH = 0.1062M
V₂ = volume of NaOH = 39.26mL = 0.039L
On putting all these values on the above equation we can calculate the molarity as:
M₁ = (0.1062)(39.26) / (37.54) = 0.11M
Now we calculate the moles of acetic acid by using the molarity formula as:
n = 0.11M × 0.037L = 0.00407 moles
Molar mass of acetic acid = 60g/mole
Mass of 0.00407 moles of acetic acid = 4.1x10⁻³ moles×(60g / mol) = 0.2501 g
Density of solution = 1.00 g/mL
So, 37.54mL in 1g/mL = 37.54g/mL
Percent by weight will be calculated as:
%w/w = 0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)
Hence, molarity and %(w/w) of acetic acid is 0.11M and 0.666% respectively.
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Fractionation of Crude Oil Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first: butane, heptadecane, dodecane, ethane, decane Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first:
butane, heptadecane, dodecane, ethane, decane
A. ethane > butane > decane > dodecane > heptadecane
B. heptadecane > > dodecane > decane butane > ethane
C. ethane > butane > decane> heptadecane >
D. dodecane butane > ethane > decane > dodecane > heptadecane
Answer:
A. ethane > butane > decane > dodecane > heptadecane
Explanation:
In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.
The product with the least weight rises to top height while the product with highest weight will move down.
For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is
ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).
Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane
For the following reaction, draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be incorrect.
Select the correct IUPAC name for the organic reactant:
a) 2-methylbutene
b) 2-methyl-1-butene
c) 3-methyl-3-butened) 3-methylbutene
Answer:
The correct IUPAC name for the organic reactant is :
d) 3-methylbutene
Explanation:
Firstly the missing diagram is attached in the diagram below.
The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.
The correct IUPAC name for the organic reactant is : 3-methylbutene
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2 and I2 are 0.453 M, what will the concentration of HI be at equilibrium
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
please help!!!! Chem question
Answer : The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The given balanced ionic equation will be,
[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]
In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
differentiate between sol,aerosol and solid soluti
Answer:
Sol is a colloidal suspension with solid particles in a liquid. Foam is formed when many gas particles are trapped in a liquid or solid. Aerosol contains small particles of liquid or solid dispersed in a gas. While solid solution contain solid as solute in either solid, liquid or gas.
An attempt at synthesizing a certain optically active compound resulted in a mixture of its enantiomers. The mixture had an observed specific rotation of 14.1°. If it is known that the specific rotation of the R enantiomer is –28.4°, determine the percentage of each isomer in the mixture. g
Answer:
The percentage of the R-enantiomer is 26.18% while the percentage of the S-enantiomer is 73.82%
Explanation:
If the specific rotation of R enantiomer = -28.4, then the specific rotation of S = +28.4
Now, let us have x = % R, thus
% S = 100-x =y
Hence;
{- 28.4x + 28.4( 100 -x)}/100= 14.1
Thus;
-28.4x + 2840 -28.4x = 1410
-56.8x + 2840 = 1410
-56.8x = 1410-2840
-56.8x = -1430
x = 1430/56.8
x = 26.18%
y = 100-26.18% = 73.82%
