Answer:2C4H10+2C12+12O2 4CO2+CC14+H20
Arrange these compounds by their expected boiling point. Highest boiling point Lowest boiling point CH3OH, CH3CI CH4.
Answer: The given compounds are arranged according to decreasing boiling point as [tex]CH_{3}OH > CH_{3}Cl > CH_{4}[/tex].
Explanation:
The temperature at which vapor pressure of a substance becomes equal to the atmospheric pressure is called boiling point.
Stronger is the intermolecular forces present the atoms of a molecule more heat will be required by it to break the bond between its atoms. Hence, more will the boiling point of the molecule.
In [tex]CH_{3}OH[/tex] (methanol), there is hydrogen bonding present which is a stronger force. So, it will have highest boiling point as compared to [tex]CH_{3}Cl[/tex] and [tex]CH_{4}[/tex].
In [tex]CH_{3}Cl[/tex] (chloroform), there is more electronegative atom attached (Cl) is attached to less electronegative atom (C and H). So, electrons are more pulled towards the chlorine atom. So, boiling point of [tex]CH_{3}Cl[/tex] is more than methane [tex](CH_{4})[/tex].
Thus, we can conclude that given compounds are arranged according to decreasing boiling point as [tex]CH_{3}OH > CH_{3}Cl > CH_{4}[/tex].
Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?
a. Al (s)
b. H2O (l)
c. HCN (g)
d. CH3COOH (l)
e. C2H6 (g)
Answer:
Rank the following chemical species from lowest absolute entropy (So) (1) to highest absolute entropy (5) at 298 K?
a. Al (s)
b. H2O (l)
c. HCN (g)
d. CH3COOH (l)
e. C2H6 (g)
Explanation:
Entropy is the measure of the degree of disorderness.
In solids, the entropy is very less compared to liquids and gases.
The entropy order is:
solids<liquids<gases
Among the given substances, water in liquid form has a strong intermolecular H-bond.
So, it has also less entropy.
Next acetic acid.
Between the gases, HCN, and ethane, ethane has more entropy due to very weak intermolecular interactions.
HCN has slight H-bonding in IT.
Hence, the entropy order is:
Al(s) < CH3COOH (l) <H2O(l) < HCN(g) < C2H6(g)
he equation for the dissociation of pyridine is
C5H5N(aq) + H2O(l) ⇌ C 5H5NH+(aq) + OH-(aq) Kb = 1.9 × 10-9
Calculate the pH of a pyridine solution that has a concentration of 9.2 M. Round your answer to two decimal places.
Answer:
10.10
Explanation:
Step 1: Write the basic dissociation reaction for pyridine
C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.9 × 10⁻⁹
Step 2: Calculate [OH⁻]
For a weak base, we will use the following expression.
[OH⁻] = √(Cb × Kb) = √(9.2 × 1.9 × 10⁻⁹) = 1.3 × 10⁻⁴ M
Step 3: Calculate pOH
We will use the definition of pOH.
pOH = -log [OH⁻] = -log 1.3 × 10⁻⁴ = 3.9
Step 4: Calculate pH
We will use the following expression.
pH = 14 - pOH = 14 - 3.9 = 10.10
according to the kinetic molecular theory what happens to a liquid when it is transferred from one container to another
Answer:
See explanation
Explanation:
Liquids are known to have a definite volumes but not a definite shape. This means that a liquid takes on the volume of the container in which it is found.
Hence, when a liquid is transferred from one container to another, the volume of the liquid remains the same but the shape of the liquid changes.
This happens when the two containers do not possess the same shape.
How is stoichiometry used to calculate energy released when a mass of liquid freezes?
A. Grams liquid x mol/g x delta Hreaction
B. Grams liquid x mol/g x delta Hfusion
C. Grams liquid x mol/g x delta Hvap
D. Grams liquid x mol/g x delta Hf
The formula that we can use to calculate energy released when a mass of liquid freezes is Grams liquid x mol/g x delta Hfusion.
What is stoichiometry?Stoichiometry deals with the relationship between mass and moles or volume which can be used to make calculations involving chemical reactions. Most chemical calculations involving stoichiometry use the mole concept.
Hence, the formula used to obtain the energy released when a mass of liquid freezes is Grams liquid x mol/g x delta Hfusion.
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True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
Answer:
True
Explanation:
The valence orbitals of boron are 2s2 2p1. We have to recall that all the valence orbitals whether full or empty are involved in the formation of molecular orbitals.
The number of molecular orbitals formed is equal to the number of atomic orbitals that are combined.
Since there are two valence orbitals and there is only one p orbital among the valence orbitals, it is true that only one p orbital is needed to form molecular orbitals in boron.
the force of attraction between non polar molecules are what
Answer:
dispersion force
Explanation:
Mava=mbvb ma x5.0ml = 5.2ml x 0.10m
Answer:
[tex]M_{a}[/tex] = 0.104 m
Explanation:
This expression is used to determine either the mass or volume of an acid or a base used during titration process.
So that;
[tex]M_{a}[/tex][tex]V_{a}[/tex] = [tex]M_{b}[/tex][tex]V_{b}[/tex]
[tex]M_{a}[/tex] is the mass of the acid
[tex]V_{a}[/tex] is the volume of the acid
[tex]M_{b}[/tex] is the mass of the base
[tex]V_{b}[/tex] is the volume of the base
Given that:
[tex]M_{a}[/tex] x 5.0 m l = 5.2 ml x 0.10 m
[tex]M_{a}[/tex] x 5.0 = 0.52 m
[tex]M_{a}[/tex] = [tex]\frac{0.52}{5.0}[/tex]
= 0.104
[tex]M_{a}[/tex] = 0.104 m
The mass of the acid used is 0.104 m.
Someone can you please please help me
Answer:
FalseExplanation:
Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.
For each of the following circumstances, indicate whether the calculated molarity of NaOH would be lower, higher or unaffected. Explain your answer in each case. a. The inside of the pipet used to transfer the standard HCl solution was wet with water.b. you added 40 mL of water to the titration flask rather than 25ml. c. The buret, wet with water, was not rinsed with NaOH solution before filling the buret with NaOH solution. d. Five (5) drops of phenolphthalein were added to the solution to be titrated rather than three (3) drops.
Answer:
a)calculated molarity of NaOH would be lower
b) calculated molarity of NaOH would be lower
c) calculated molarity of NaOH would be lower
d) calculated molarity of NaOH would be unaffected
Explanation:
Let us recall that the reaction of NaOH and HCl is as follows;
NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)
Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.
When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.
If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.
If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.
For which of the following transitions would a hydrogen atom absorb a photon with the longest wavelength?
a. n = 1 to n = 2
b. n = 3 to n = 2
c. n = 5 to n = 6
d. n = 7 to n = 6
Answer:
Hence among the options a and c, option d is that the correct answer because it has rock bottom energy ( as n value increases, energy decreases as energy levels come closer).
Explanation:
The relation between energy and wavelength is:
[tex]\Lambda = hC/E[/tex]
From this equation, it's clear that wavelength and energy are inversely proportional to every other. The Lower the energy of a specific transition, the longest will the wavelength be of that specific transition.
Among the given options, options b and d are often ruled out, since those transitions produce to release of a photon because it is coming down from an excited state.
important of organic chemistry in our daily life
Answer:
Human life has become very simple by using organic chemicals. The importance of organic chemicals in the daily life and industrial area can be explained as follows. (i) Food Vitamins, proteins, sugar, flour, fats etc. ... (ii) Agriculture is an important place for organic chemical for the growth of agricultural production.
What quantity of heat is transferred when a 150.0g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of the heat flow?
The direction of heat flow is increased which means blocks temperature is higher and hotter than it was before
Q2.Which is true about potassium?
Extremely unreactive
Not very reactive
Slightly reactive
Very reactive
2- A 0.60 sample an unknown organic acid found in muscle cells is burned in air and found to contain 0.24 grams of carbon, 0.040 grams of hydrogen, with the rest being oxygen. If the molecular weight of the substance is 90 grams/n, what is the molecular formula
Answer:
C₃H₆O₃
Explanation:
To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).
With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:
Moles C -Molar mass:12.0g/mol-
0.24g * (1mol/12.0g) = 0.020 moles C
Moles H = Mass H because molar mass = 1g/mol:
0.040 moles H
Moles O -Molar mass: 16g/mol-
Mass O: 0.60g - 0.24g - 0.040g = 0.32g O
0.32g O * (1mol/16g) = 0.020 moles O
Ratio of atoms (Dividing in moles of C: Lower number of moles):
C = 0.020 moles C / 0.020 moles C = 1
H = 0.040 moles H / 0.020 moles C = 2
O = 0.020 moles O / 0.020 moles C = 1
Empirical formula:
CH₂O.
Molar mass CH2O:
12g/mol + 2*1g/mol + 16g/mol = 30g/mol
As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:
C₃H₆O₃The molecular formula of the organic acid would be C3H6O3
Molecular formulaMolecular formula = [empirical formula]n
Where n = molar mass/mass of empirical formula
Empirical formula
C = 0.24/12 = 0.02
H = 0.040/1 = 0.04
O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02
Divide by the smallest
C = 1
H = 2
O = 1
Empirical formula = CH2O
Empirical formula mass = 12 + 2 + 16 = 30
n = 90/30 = 3
Molecular formula = [CH2O]3
= C3H6O3
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Calculate the number of ATOMS in 1.0 mole of O2. blank x 1024
Answer:
6.023*10^23........ .......
Water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the
bottom.
ANSWER:
True
False
Answer:
False
Explanation:
An object with a density greater than 1.00g/mL (greater than the density of water) will sink. An object with a density less than the density of water, will float.
If the water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the bottom, this statement is false.
What is density?The density of an actual content is its mass per unit volume. The most common symbol for density is d, but the Latin letter D can also be used.
Three of an object's most fundamental properties are mass, volume, and density. Mass describes how heavy something is, volume describes its size, and density is defined as mass divided by volume.
The density of something is a measure of how heavy it is in relation to its size. When an artifact is more dense than water, it plunges; when an object is less dense than water, it floats.
Density is a property of a substance that is independent of the amount of substance.
As in the given scenario, water is having density 1 g/mL and object in having density less then it so it will float on water.
Thus, the given statement is false as the material will not sink, rather it will float on water.
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define saturated and unsaturated fats
Explanation:
Saturated fats are defined as the fat where fatty acid chains contain only single bonds.
For example, stearic acid, palmitic acid etc.
Unsaturated fats are defined as the fat where fatty acids contain one or more number of double bonds on the carbon atoms.
For example, oleic acid, palmitoleic acid etc.
In a solution, the solvent is
ANSWER:
A. always water
B. dissolved in the solute
C. present in larger amount than the solute is
D. always nonpolar
Answer:
dissolved in the solute
Explanation:
A solvent is the component that dissolves the solute and is present in larger amount. The type of solution is determined by the state of the solute and solvent. If you have NaCl, a solid, dissolved in water, a liquid, the type of solution is a solid/liquid solution.
Which piece of equipment would be BEST to measure 5 mL of a liquid? A 50 mL beaker filled half way between 0 and 10 mL markings A 50 mL Erlenmeyer flask filled half way between 0 and 10 ml markings A 10 mL graduated cylinder A disposable dropper A 100 mL graduated cylinder
Answer:
Which piece of equipment would be BEST to measure 5 mL of a liquid?
A 50 mL beaker filled halfway between 0 and 10 mL markings
A 50 mL Erlenmeyer flask filled halfway between 0 and 10 ml marking
A 10 mL graduated cylinder
A disposable dropper
A 100 mL graduated cylinder
Explanation:
To measure 5 mL of a liquid, the best equipment is
a 10 mL graduated cylinder.
The remaining apparatus do not give an accurate measurement because readings are not marked on them.
1) Write a balanced equation to show the reaction of gaseous ethane with gaseous oxygen to form carbon monoxide gas and water vapor.
Answer:
C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
Explanation:
Let's consider the unbalanced equation in which gaseous ethane reacts with gaseous oxygen to form carbon monoxide gas and water vapor. This is an incomplete combustion reaction.
C₂H₆(g) + O₂(g) ⇒ CO(g) + H₂O(g)
We will balance it using the trial and error method.
First, we will balance C atoms by multiplying CO by 2 and H atoms by multiplying H₂O by 3.
C₂H₆(g) + O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
Then, we get the balanced equation by multiplying O₂ by 2.5.
C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
A 250 mL sample of gas at 1.00 atm and has the temperature increased to and the volume increased to 500 mL. What is the new pressure
Answer:
0.53atm = P2
Explanation:
Gas at 1.00atm and 20°C. Temperature increased to 40°C...
We can solve this question using combined gas law:
P1*V1 / T1 = P2*V2 / T2
Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state.
Compunting the values of the problem:
P1 = 1.00atm
V1 = 250mL
T1 = 20°C + 273.15 = 293.15K
P2 = ?
V2 = 500mL
T2 = 40°C + 273.15 = 313.15K
1.00atm*250mL / 293.15K = P2*500mL / 313.15K
0.53atm = P2
(4.184 J——> 1 calorie ? Take your answer from number 1 and convert to the answer into calories?
Answer:
There are 1.195 calories in 5 J.
Explanation:
The relation between calorie and Joules is as follow :
4.184 J = 1 calorie
or
1 J = (1/4.184) calorie
Let the taken number is 5 J
So,
5 J = (5/4.184) calorie
5 J = 1.195 calorie
So, there are 1.195 calories in 5 J.
Which diagram correctly depicts the trend in electronegativity?
a.
b.
c.
d.
The electronegativity increases across the period and decreases down the group. Thus, option B is correct.
Electronegativity can be defined as the tendency of an atom to gain or attract an electron. The electronegativity has been dependent on the size of the atom, as well as the atomic number and valence electrons.
The atom with the requirement of a less number of atoms to complete its octet can easily gain the electron and thereby have high electronegativity. The atomic size also plays a role in the electronegativity of the atom.
The atom with a bigger size has the lesser force of attraction from the nucleus and thus has difficulty attracting the electron, however, the smaller size atom can easily attract the electron with the attraction force from the nucleus.
Thus, the elements with smaller sizes and a high number of valence electrons are more electronegative. In the periodic table, on moving from left to right the valence electrons increase, thus the electronegativity increases.
On moving down the group, the element size increase, thus the electronegativity decreases down the group.
The electronegativity increases across the period and decreases down the group. Thus, option B is correct.
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Typhoon signals rise due to what? wind speed or wind strength or both?
When the pH of a solution is 12.83, what is [H +]?
A 7.4 × 10 -12 M
B 9.7 10 -11 M
C 1.5 X 10 -13 M
D 12.8 x 10 -2 M
Answer:
B.9.710-11M
Explanation:
plss tell me if im wrong3. Does entropy increase or decrease in the following processes?
A. Complex carbohydrates are metabolized by the body, converted into simple sugars.
Answer: Increase
es-lesund
B. Steam condenses on a glass surface.
Answer:
decreare
-->
MgCl2(s)
C. Mg(s) + Cl2(g)
correct
Answer:
Answer:
hope it helps much as you can
If 12.3 g of Cu is deposited at the cathode of an electrolytic cell after 5.50 h, what was the current used?
Answer:
1.88 A
Explanation:
Let's consider the reduction of copper in an electrolytic cell.
Cu²⁺ + 2 e⁻ ⇒ Cu
We can calculate the charge used to deposit 12.3 g of Cu using the following relations.
The molar mass of Cu is 63.55 g/mol.1 mole of Cu is deposited when 2 moles of electrons circulate.1 mole of electrons has a charge of 96486 C (Faraday's constant).The charge used is:
[tex]12.3 g \times \frac{1 molCu}{63.55gCu} \times \frac{2molElectron}{1molCu} \times \frac{96486C}{1molElectron} = 3.73 \times 10^{4} C[/tex]
We can convert 5.50 h to seconds using the conversion factor 1 h = 3600 s.
5.50 h × 3600 s/1 h = 1.98 × 10⁴ s
The current used is:
I = q/t = 3.73 × 10⁴ C/1.98 × 10⁴ s = 1.88 A
Which pairs of aqueous solutions will not produce a precipitate when mixed AgNo3(aq) and NaCl(aq)?
Answer:
CHCI3
Explanation:
there are no free CI ions hence it doesnt precipitate with an aqeous solution of AQUO33
Cis-4-tertButylcyclohexyl bromide (compound 1) and Trans-4 tert Butylcyclohexylbromide (compound 2) are reacted with Potassium Tertiary butoxide in Tertiary butanol to produce 4-tertbutylcyclohexene. The following statement is completely true?A. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
B. In compound 1 the Tert butyl group occupies the axial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
C. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the equatorial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compound 1 reacts faster than compound 2.
D. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.
E. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl occupies the equatorial and the bromine occupies axial position. Compound 1 reacts faster than compound 2.
F. In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 both the Tert butyl and the bromine occupy equatorial positions. Compond 2 reacts faster than compound 1.
Answer:
In compound 1 the Tert butyl group occupies the equatorial position and the Bromine occupies the axial position and in compound 2 the Tert butyl occupies the axial and the bromine occupies equatorial positions. Compound 1 reacts faster than compound 2.
Explanation:
In cyclic organic compounds, substituents may occupy the axial or equatorial positions. The axial positions are aligned parallel to the symmetry axis of the ring while the equatorial positions are around the plane of the ring.
Bulky substituents have more room in the equatorial than in the axial position. This means that compound 1 is more stable than compound 2.
This is clear on the basis of stability of the molecules because compound 1 will react faster than compound 2 since the bulky tertiary butyl group in compound 1 occupy equatorial and not axial positions.