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.
What mass of carbon dioxide is produced from the complete combustion of 7.30×10−3 g of methane?
Benzene can be converted to 1,3,5-tribromobenzene in five reaction steps and four intermediate compounds. Select the appropriate reagent from the followings.
Br2, R2O2
CH3Cl, AlCl3
CH3COCl, AlCl3
NaNO2, HCl
HNO3, H2SO4
H3PO2
H3PO4
KMnO4
Answer:
The appropriate reagent is: H3PO2.
Explanation:
H3PO2 is in charge of eliminating the amino group by diazotization, remember that the amino group had previously achieved bromination at positions m; that is to say that it achieved in the beginning that the three bromine atoms of 1,2,4 tribromobenzene were introduced in the meta positions among themselves, which finally corresponds as part of the last reaction to the 1,3,5-tribromobenzene position.
Aspirin is usually packaged with
A. acetic anhydride
B. salicylic acid
C. buffering agents
Answer:
Aspirin is usually packaged with C. buffering agents.
Explanation:
How many moles of H2 are needed to produce 34.8 moles of NH3?
2 i hope this helps
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An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, molality, and mass percent of the solution. (Assume the density of 1.05 g/mL for the solution.)
Answer:
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of the solution es 3.45%
Explanation:
Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:
[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Being:
K: 39 g/moleN: 14 g/moleO: 16 g/moleThe molar mass of KNO₃ is:
KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole
You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?
[tex]moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}[/tex]
moles of KNO₃= 0.718
So you have:
moles of KNO₃= 0.718volume= 2 LApplying this quantity in the definition of molarity:
[tex]molarity=\frac{0.718 moles}{2 L}[/tex]
Molarity= 0.359[tex]\frac{moles}{L}[/tex]
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.
Then the molality is calculated by:
[tex]Molality=\frac{moles of solute}{mass of solvent in kilograms}[/tex]
Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?
[tex]mass=\frac{2000 mL*1.05 grams}{1 mL}[/tex]
mass= 2100 grams
Since mass solution = mass water + mass KNO₃
then mass water = mass solution - mass KNO₃
Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams
mass water= 2,027.5 grams
Then, being:
moles of KNO₃= 0.718mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)Replacing in the definition of molality:
[tex]molality=\frac{0.718 moles}{2.0275 kg}[/tex]
molality= 0.354 [tex]\frac{moles}{kg}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.
[tex]mass percent=\frac{mass of solute}{mass of solution} *100[/tex]
So, in this case:
[tex]mass percent=\frac{72.5 grams}{2100 grams} *100[/tex]
mass percent= 3.45 % KNO₃ by mass
The mass percent of the solution es 3.45%
The molarity of the solution is 0.36 mol/L. The molality of the solution is 0.34 m. The mass percent of the solution is 3.33%.
Number of moles of KNO3 = mass/molar mass = 72.5 g/101 g/mol = 0.72 moles
Molarity = Number of moles / volume = 0.72 moles/ 2.00 L = 0.36 mol/L
The molality = Number of moles of solute/Mass of solution in kilograms
mass of solution = 1.05 g/mL × 2000 mL = 21000 g or 2.1Kg
Molality of solution = 0.72 moles/2.1 Kg = 0.34 m
Mass percent of solution = mass of solute/mass of solution × 100/1
Mass percent of solution = 72.5 g/ (72.5 g + 2100 g) × 100/1
= 3.33%
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Describe why some acids are strong while other acids are weak
Answer:
I hope this help you. Mark me as brainliest and rate pleaseExplanation:
the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.
It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.
As you will see below, the strength of an acid is related to the proportion of it which has reacted with water to produce ions. The concentration tells you about how much of the original acid is dissolved in the solution.
It is perfectly possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid.
Question 5 of 20:
Select the best answer for the question.
5. Which of the following is a homonuclear diatomic molecule?
O A. NH3
O B. 2002
O C. Hz
O D. CO
Answer:
Homo nuclear molecule mean having atoms of only one element,
I cant see clearly the option B and C can you correct them , 2002? Hz?
Explanation:
Answer:
H2
Explanation:
What is the equilibrium constant for the following reaction:HCO2H(aq) + CN–(aq) HCO2–(aq) + HCN(aq)Does the reaction favor the formation of reactants or products? The acid dissociation constant, Ka, for HCO2H is 1.8 x 10–4and the acid dissociation constant for HCN is 4.0 x 10–10.(A) K = 1.00. The reaction favors neither the formation of reactants nor products.(B) K = 2.2 x 10–6. The reaction favors the formation of products.(C) K = 2.2 x 10–6. The reaction favors the formation of reactants.(D) K = 4.5 x 105. The reaction favors the formation of products.(E) K = 4.5 x 105. The reaction favors the formation of reactants.
Answer:
(D) K = 4.5 x 10⁵. The reaction favors the formation of products
Explanation:
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
K = [HCOO⁻] [ HCN ] / [ HCOOH] [ CN⁻]
HCOOH ⇄ H ⁺ + COO⁻
K₁ = [ H⁺] [ COO⁻ ] / [HCOOH ]
HCN ⇆ H⁺ + CN⁻
K₂ = [ H⁺] [ CN⁻] / [ HCN ]
K₁ / K₂
= [ H⁺] [ COO⁻ ] / [HCOOH ] X [ HCN ] / [ H⁺] [ CN⁻]
= [ COO⁻ ][ HCN ] / [HCOOH ] [ CN⁻]
= K
K = K₁ / K₂
= 1.8 x 10⁻⁴ / 4 x 10⁻¹⁰
= 4.5 x 10⁵
So equilibrium constant of the reaction
HCOOH + CN⁻ ⇆ HCOO⁻ + HCN
is very high . Hence reaction favours the formation of product.
option (D) is correct.
A gas has a volume of 6.6 L at a temperature of 40 C. What is the volume of
the gas if the temperature changes to 15 C?
Answer:
6.07 L
Explanation:
It appears that the reading has been made at constant pressure .
At constant pressure , the gas law formula is
V/T = constant V is volume and T is temperature of the gas.
V₁ / T₁ = V₂ / T₂
V₁ = 6.6 L ,
T₁ = 40°C
= 273 + 40
= 313 K
T₂ = 15+ 273
= 288K
V₂ = ?
Putting the values in the formula above
6.6 / 313 = V₂ / 288
V₂ = 6.07 L.
Nitrogen forms more oxides than any other element. The percents by mass of in three different nitrogen oxides are (1) (II) and (III) 25.94 For each compound, determine (a) the simplest whole-number ratio of to and (b) the number of grams of oxygen per 1.00 of nitrogen.
Complete question;
Nitrogen forms more oxides than any other element. The percents by mass of N in three different nitrogen oxides are (|) 46.69%;(II) 36.85 %; (III) 25.94%. For each compound, determine (a) the simplest whole-number ratio of N to O, and (b) the number of grams of oxygen per 1.00 g of nitrogen.
Answer:
a. (i) The ratio is 1:1 , the formula = NO (ii)The ratio is 1 : 1.5 which is 2 : 3, the formula = N₂O₃ (iii) The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅
b. (i)number of grams of oxygen = 53.31/46.69 = 1.14 g
(ii)number of grams of oxygen = 63.15/36.8 = 1.71 g
(iii)number of grams of oxygen = 74.06/25.94 = 2.855 g
Explanation:
a.
(i) The percentage by mass of the nitrogen in Nitrogen oxide (i) is 46.69% which is taken as 46.69 grams . Since the other element is oxygen the mass of oxygen will be 100 - 46.69 = 53.31 grams.
The relative atomic mass of Nitrogen and oxygen is 14 amu and 16 amu respectively.
Therefore, to know the whole number ratio of N and O we find the number of moles.
number of moles of N = 46.69/14 = 3.335
number of moles of O = 53.31/16 = 3.332
The ratio is 1:1 , the formula = NO
(ii)
number of moles of N = 36.85/14 = 2.632
number of moles of O = 63.15/16 = 3.947
The ratio is 1 : 1.5 which is 2 : 3, the formula = N₂O₃
(iii)
number of moles of N = 25.94/14 = 1.85
number of moles of O = 74.06/16 = 4.63
The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅
b.
(i) 46.69 g of nitrogen = 53.31 g of oxygen
1 g of nitrogen = ? of Oxygen
number of grams of oxygen = 53.31/46.69 = 1.14 g
(ii)
Using similar method in b(i)
number of grams of oxygen = 63.15/36.8 = 1.71 g
(iii)
Using similar method in b(i)
number of grams of oxygen = 74.06/25.94 = 2.855 g
The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW
Answer:
Explanation:
From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer
0.25 L elution buffer = 250 mL elution butter
The breaking buffer that we use this week contains
10mM Tris = 0.01 M
150mM NaCl = 0.15 M
300mM imidazole. = 0.3 M
The stock concentration of Tris in 1M
Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;
[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 2.5 \ mL[/tex]
In NaCl, The amount of stock concentration is 5 M
so; using the same formula; we have:
[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]
[tex]V_1 = 7.5 \ mL[/tex]
From Imidazole ; the amount of stock concentration is
[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 75 \ mL[/tex]
Thus; we can have a table as shown as :
Stock concentration volume to be added Final concentration
1 M of Tris 2.5 mL 10 mM
5 M of NaCl 7.5 mL 150 mM
1 M of Imidazole 75 mL 300 mM
In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.
Which of the following best describe an atomic number? An element identity is defined by its atomic number this means it represents the number of
a)Atomic number is the number of protons in the nucleus of an atom.
b)atomic number represents the number of protons
An element's identity is defined by its atomic number; this means it represents the number of protons in its nucleus.
The question is incomplete, the complete question is;
Which of the following best describes an atomic number? An element's identity is defined by its atomic number; this means it represents the number of
A) protons plus neutrons in its nucleus.
B) electrons in the element.
C) protons in its nucleus.
D) neutrons in its nucleus.
The subatomic particles in the atom are;
ElectronsProtonsNeutronsNeutrons and protons are contained in the nucleus hence they are collectively called nucleons.
Electrons are found in the orbits. The number of protons must be equal to the number of electrons for the atom to be electrically neutral.
The number of protons in an atom or atomic number serve as a means of identifying an atom.
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What is Hess‘s law please help
The correct answer is D. Hess's law states than the enthalpy of a reaction does not depend on the reaction path
Explanation:
In a chemical reaction, the enthalpy refers to the internal energy in a system and how this increases or decreases during the reaction. According to Hess's law proposed by German Hess in 1940, the enthalpy does not depend on the reaction path or the number of steps in a reaction. This means one reaction of only one step will have the same enthalpy that if the reaction occurs in several steps because the energy that requires all the process is the same. Thus, the "Hess's law states than the enthalpy of a reaction doe s not depend on the reaction path".
Which of the following errors could cause your percent yield to be falsely high, or even over 100%?
Select ALL that apply.
A.) Heating the sample too vigorously.
B.) Handling the crucible directly with your hands.
C.) Failing to completely decompose the sodium bicarbonate sample.
D.) Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements.
E.) Taking the mass of all samples with the lid included.
Answer:
B.Handling the crucible directly with your hands.
D.Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements.
E.Taking the mass of all samples with the lid included.
Explanation:
When observed critically , the measures associated with the errors which could cause your percent yield to be falsely high, or even over 100% are those which increase the weight of the substance with the individual neglecting.
Handling the crucible directly with your hands,Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements and taking the mass of all samples with the lid included will all increase the weight of the substance. Instead the substance should be placed alone without any form of support or contamination.
Unscramble the following words to form a complete
sentence about the cycles of nature:
limited is through environment Matter recycled the on Earth is and
Answer:
recycled is limited through enviroment and matter on earth
Explanation:
In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102. In the living E. coli cells, [ATP] = 7.9 mM; [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
[tex]q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}[/tex]
[tex]q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}[/tex]
so,
[tex]q<<K_e_q[/tex] ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
Propane (C3H8) is widely used in liquid form as a fuel for barbecue grills and camp stoves. For 67.7 g of propane, determine the following.(a) Calculate the moles of compound.mol(b) Calculate the grams of carbon.g
Answer:
A. 1.54 mole.
B. 55.39g of carbon
Explanation:
A. Determination of the number of mole in 67.7g of C3H8.
Mass of C3H8 = 67.7g
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Number of mole of C3H8 =..?
Number of mole = Mass/Molar Mass
Number of mole of C3H8 = 67.7/44
Number of mole of C3H8 = 1.54 mole
B. Determination of the mass of carbon in the compound.
This is illustrated below:
The mass of C in compound can be obtained as follow:
=> 3C/C3H8 x 67. 7
=> 3x12 / 44 x 67.7
=> 36/44 x 67.7
=> 55.39g
Therefore, 55.39g of carbon is present in the compound.
For the iodine trichloride molecule: a. Determine the number of valence electrons for each atom in the molecule b. Draw the Lewis Dot structure c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?) d. Show the polarity of each bond and for the molecule by drawing in the dipole +à
Answer:
Explanation:
a. Determine the number of valence electrons for each atom in the molecule
In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.
b. Draw the Lewis Dot structure
In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1
c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)
In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).
d. Show the polarity of each bond and for the molecule by drawing in the dipole +d
The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".
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Chemical formula for copper gluconate I have 1.4g of Copper gluconate. There is .2g of copper within the copper gluconate. Determine the chemical formula for Copper gluconate with the given information: Copper Gluconate: Cu(C6H11O?)? Cu = 63.55 g/mol H = 12.01 g/mol O = 1.008 g/mol Cu = 63.55 g/mol
Answer:
The simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂
Explanation:
Given mass of sample = 1.4 g
mass of copper in the sample = 0.2 g
mass of the gluconate =1.4 - 0.2 = 1.2 g
The mole ratio is determined first using the formula;
mole ratio = reacting mass / atomic mass
atomic mass of copper = 63.55
mass of gluconate, C₆H₁₁O₇ = 12*6 + 1*11 + 16*7 = 195 g/mol
mole ratio ( copper : gluconate) = 0.2/63.55 : 1.4/195
mole ratio ( copper : gluconate) = 0.003 : 0.007
convert to whole number ratios by dividing with the smallest ratio
mole ratio ( copper : gluconate) = 0.003/0.003 : 0.007/0.003
mole ratio ( copper : gluconate) = 1 : 2
Therefore, the simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂
What is the molar mass of CH2O2 ? ( C= 12.01 g/mol, H=1.008 g/mol, O=16.00)
Answer:
Molar Mass of CH2O2 is 46.026
Explanation:
What is the molar mass of CH2O2 ? ( C= 12.01 g/mol, H=1.008 g/mol, O=16.00)
C = 12.01g/mol
H = 1.008g/mol
O = 16g/mol
CH2O2 = 12.01+1.008x2+16x2 = 46.026g/mole
which statement describes the reactions in an electrochemical cell
Answer & explanation:
Summary on electrochemical cells and redox reactions:
Electrochemical cells (or batteries) can be defined as devices capable of transforming chemical energy into electrical energy through spontaneous reactions of redox, in which electron transfer occurs.
Redox it is a chemical reaction in which there is the occurrence of oxidation and reduction of atoms of substances (chemical species) present in the process.
Oxidation is the loss of electrons by an atom of a chemical species, while reduction is the gain of electrons by an atom of a chemical species.
Thus, during an oxirreduction reaction, electrons move from the species that loses them towards the species that will receive them. This "movement" results in the formation of an electric current (or electrical energy) as occurs with batteries, for example.
A sample of thallium(III) peroxide, Tl2(O2)3, contains 2.45 mol of thallium(III) ions. The number of moles of peroxide ions in the sample is
Answer:
The correct answer is 3.675 moles.
Explanation:
Based on the question, the reaction taking place is,
Tl₂(O₂)₃ ⇒ 2Tl⁺³ + 3O₂⁻²
Thus, 1 mole of thallium peroxide comprise 2 moles of thallium and 3 moles of peroxide ions.
However, based on the given question, a sample of thallium peroxide comprise 2.45 moles of thallium ions. The moles of peroxide ions present in the sample will be,
= 2.45 × 3 / 2
= 3.675 moles.
Hence, the moles of peroxide ions present in the given sample is 3.675.
The calculated yield for the production of carbon dioxide was 73.4g. When the
experiment was performed in the lab, a yield of 72.3g was produced. What is the
percent yield of carbon dioxide?
Answer:10 grams of CO2
Explanation:
Yeild= exp. yeild÷ thoretical yeild × 100
Yeild= 73.3÷73.4 × 100
Yeild= 0.1 ×100
Yeild= 10
During lab, you evaluated the bond order and bond length of a series of carbon-carbon bonds. Use the same concepts to predict the bond order and bond length of a series of nitrogen-nitrogen bonds.(a) Which of the structures below have a nitrogen-nitrogen bond order of 3?(b) Which of the structures below have the shortest nitrogen-nitrogen bond?
Answer:
N≡N
Explanation:
The image attached shows the nitrogen compounds that are being referred to in the question.
There are certain things we ought to know in order to answer the question accurately.
The bond order of a compound is equal to the number of bonds between two atoms. The greater the bond order, the shorter the bond length between the two atoms.
N≡N has a bond order of three, this is the highest bond order among all the species listed in the question. Hence it has the shortest bond length among the trio. Hence the answer.
Which of the following is an example of a mechanical wave?
O A. A light ray
B. A seismic wave
C. A radio wave
D. An X-ray
Answer:
A seismic wave
Explanation:
It requires a medium for its propagation.
Ba(OH)2:_______.
A. 1 barium atom, 1 oxygen atom and 1 hydrogen atom.
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
C. 1 barium atom, 2 oxygen atoms and 2 hydrogen atoms.
D. 1 barium atom, 2 oxygen atoms and 1 hydrogen atom.
Answer: D
Explanation: Expand this (OH)2 you will get 2O, 2H
Hence 1Ba, 2O, 2H
Answer:
B. 1 barium atom, 1 oxygen atom and 2 hydrogen atoms.
A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.
Answer:
C3H6.
Explanation:
Data obtained from the question:
Mass of the compound = 8g
Mass of CO2 = 24.01g
Mass of H2O = 13.10g
Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01
=> 12/44 x 24.01 = 6.5g
Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1
=> 2x1/18 x 13.1 = 1.5g
Mass of O in the compound = Mass of compound – (mass of C + Mass of H)
=> 8 – (6.5 + 1.5) = 0
Next, we shall determine the empirical formula of the compound. This is illustrated below:
C = 6.5g
H = 1.
Divide by their molar mass
C = 6.5/12 = 0.54
H = 1.4/1 = 1.
Divide by the smallest
C = 0.54/0.54 = 1
H = 1/0.54 = 2
Therefore, the empirical formula is CH2
Finally, we shall determine the molecular formula as follow:
The molecular formula of a compound is a multiple of the empirical formula.
Molecular formula = [CH2]n
[CH2]n = 44
[12 + (2x1)]n = 44
14n = 44
Divide both side by 14
n = 44/14
n = 3
Molecular formula = [CH2]n = [CH2]3 = C3H6
Therefore, the molecular formula of the compound is C3H6
The fluoride ion is the conjugate base of the weak acid hydrofluoric acid. The value of Kb for F-, is 1.39×10-11. Write the equation for the reaction that goes with this equilibrium constant.
Answer:
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)
Explanation:
According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺ ions. In this sense, hydrofluoric acid is an acid according to the following equation.
HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq)
According to Brönsted-Lowry acid-base theory, a base is a substance that accepts H⁺ ions. In this sense, the fluoride ion is a base according to the following equation.
F⁻(aq) + H₂O(l) ⇄ HF(aq) + OH⁻(aq)
The equilibrium constant for this reaction is Kb = 1.39 × 10⁻¹¹.
The emission line used for zinc determinations in atomic emission spectroscopy is 214 nm. If there are 6.00×1010 atoms of zinc emitting light in the instrument flame at any given instant, what energy (in joules) must the flame continuously supply to achieve this level of emission?
A blood sample is left on a phlebotomy tray for 4 hours before it is delivered to the laboratory. Which group of tests could be performed:
