Answer:
C. The mass stays the same because the total number of atoms does not change
Explanation:
According to the law of conservation of matter/mass, matter cannot be created nor destroyed, hence, the amount of matter in the reactants must be the same amount in the products.
Using the photosynthetic reaction as a case study, carbon dioxide (CO2) and water (H2O) are the compounds that go into the reaction (reactants) while glucose and oxygen (O2) are the products of the reaction.
Using the law of conservation of matter to explain, the total mass of both the reactants and products stays the same because the total number of atoms does not change i.e. if 6 atoms of Carbon starts the reaction, 6 atoms of carbon will end it.
Chemistry Grade 11: Hi, I don't know what this is, please help?
CALCULATING % YIELD STEPS:
1.
2.
3.
4.
Answer:
1. Write the balanced chemical equation for the reaction
2. Identify all important information provided in the word problems or data table.
3. Solve for the theoretical yield of the reaction, following all the steps of a stoichiometry calculation organizer. Use two calculations if both reactants are provided.
4. Use the percent yield equation to calculate the percent yield of the reaction.
Explanation:
its comes right from the 5.06 lesson
Organic compounds undergo a variety of different reactions, including substitution, addition, elimination, and rearrangement. An atom or a group of atoms in a molecule is replaced by another atom or a group of atoms in a substitution reaction. In an addition reaction, two molecules combine to yield a single molecule. Addition reactions occur at double or triple bonds. An elimination reaction can be thought of as the reverse of an addition reaction. It involves the removal of two atoms or groups from a molecule. A rearrangement reaction occurs when bonds in the molecule are broken and new bonds are formed, converting it to its isomer. Classify the following characteristics of the organic reactions according to the type of organic reaction.
a. Reactions involving the replacement of one atom or group of atoms.
b. Reactions involving removal of two atoms or groups from a molecule.
c. Products show increased bond order between two adjacent atoms.
d. Reactant requires presence of a π bond.
e. Product is the structural isomer of the reactant.
1. Substitution reaction
2. Addition reaction
3. Elimination reaction
4. Rearrangement reaction
Answer:
Reactions involving the replacement of one atom or group of atoms. - Substitution reaction
Reactions involving removal of two atoms or groups from a molecule - Elimination reaction
Products show increased bond order between two adjacent atoms - Elimination reaction
Reactant requires presence of a π bond - Addition reaction
Product is the structural isomer of the reactant - Rearrangement reaction
Explanation:
When an atom or a group of atoms is replaced by another in a reaction, then such is a substitution reaction. A typical example is the halogenation of alkanes.
A reaction involving the removal of two atoms or groups from a molecule resulting in increased bond order of products is called an elimination reaction. A typical example of such is dehydrohalogenation of alkyl halides.
Any reaction that involves a pi bond is an addition reaction because a molecule is added across the pi bond. A typical example is hydrogenation of alkenes.
Rearrangement reactions yield isomers of a molecule. Rearrangement may involve alkyl or hydride shifts in molecules.
Reactions involving the replacement of one atom or group of atoms is substitution reaction, reactions involving removal of two atoms or groups from a molecule and products show increased bond order between two adjacent atoms is elimination reaction, reactant requires presence of a π bond in addition reaction and product is the structural isomer of the reactant is rearrangement reaction.
What is chemical reaction?Chemical reactions are those reactions in which reactants undergoes through a variety of changes for the formation of new product.
Substitution reaction: In this reaction any atom or molecule of reactant is replaced by any outside atom or molecule.Addition reaction: In this reaction addition of any reagent takes place across the double or triple bond of any reactant for the formation of product.Elimination reaction: In this reaction any molecule or two atoms will eliminate from the reactant as a result of which we get a bond order increased product.Rearrangement reaction: In this reaction atoms or bonds of a reactant get rearranged for the formation of new product.Hence, classification of above points are done according to their characteristics.
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n a combination redox reaction, two or more ____________ , at least one of which is a(n) ____________ , form a(n) ____________ . General Reaction: ____________ In a decomposition redox reaction, a(n) ____________ forms two or more ____________ , at least one of which is a(n) ____________ . General Reaction: ____________ In double-displacement (metathesis) reactions, such as precipitation and acid-base reactions, ____________ of two ____________ exchange places; these reactions ____________ redox processes.General Reaction: ____________ In solution, single-displacement reactions occur when a(n) ____________ of one ____________ displaces the ____________ of another. Since one of the ____________ is a(n) ____________ , a
Answer:
In a combination redox reaction, two or more reactants, at least one of which is a(n) element, form a(n) compound. General Reaction: X + Y > Z
In a decomposition redox reaction, a(n) compound forms two or more products, at least one of which is a(n) element. General Reaction: Z>X+Y
In double-displacement (metathesis) reactions, such as precipitation and acid-base reactions, atoms (or ions) of two compounds exchange places; these reactions are not redox processes. General Reaction: AB+CD>AD+CB
In solution, single-displacement reactions occur when a(n) atom of one element displaces the atom of another. Since one of the reactants is a(n) element, all single-displacement reactions are redox processes. General Reaction: X+YZ>XY+Z
Explanation:
In a combination redox reaction, two or more reactants, at least one of which is a(n) element, form a(n) compound.
General Reaction: X + Y > Z
In the reaction scheme above, X combines with Y to give Z as a product.
In a decomposition redox reaction, a(n) compound forms two or more products, at least one of which is a(n) element.
General Reaction: Z>X+Y
In the reaction scheme above, Z decomposes to X and Y
In double-displacement (metathesis) reactions, such as precipitation and acid-base reactions, atoms (or ions) of two compounds exchange places; these reactions are not redox processes since there are no changes occurring in the oxidation number of the atoms (or ions) involved.
General Reaction: AB+CD>AD+CB
In the reaction scheme above, B and D exchange places in their respective compounds
In solution, single-displacement reactions occur when a(n) atom of one element displaces the atom of another. This type of reaction is due to the difference in the reactivities of the elements. The more reactive atom of one element displaces the least reactive atom of another element from its solution.
Since one of the reactants is a(n) element, all single-displacement reactions are redox processes.
General Reaction: X+YZ>XY+Z
In the reaction scheme above, X displaces Z from the compound YZ.
A weak acid is titrated with 0.1236 M NaOH. From the titration curve you determine that the equivalence point occurs after exactly 12.42 mL of NaOH have been added. What is the volume of NaOH at the half-equivalence point (a.k.a. the midpoint)
Answer: The volume of NaOH required at the half-equivalence point is 6.21 mL
Explanation:
The chemical equation for the reaction of a weak acid with NaOH follows:
[tex]HA+ NaOH\rightarrow NaA+H_2O[/tex]
From the equation, we can say that NaOH and weak acid is present in a 1 : 1 ratio.
We are given:
Volume of NaOH required at equivalence point = 12.42 mL
The volume of NaOH required at half-equivalence point will be = [tex]\frac{12.42mL}{2}=6.21mL[/tex]
Hence, the volume of NaOH required at the half-equivalence point is 6.21 mL
The volume of NaOH at the half-equivalence point is 6.21 mL
What is equivalence point?The equivalence point is the point at which equal amount of the acid and base have reacted.
How to determine the half-equivalence pointVolume at equivalence point = 12.42 mLVolume at half-equivalence point =?Half equivalence point = Equivalence point / 2
Half equivalence point = 12.42 / 2
Half equivalence point = 6.21 mL
Therefore, we can conclude that the volume of NaOH at the half-equivalence point is 6.21 mL.
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Rank the following alkenes in order of increasing stability of the double bond towards addition of HBr:
2,3-dimethyl-2-butene, cis-3-hexene, 3-methyl-3-hexene, 1-hexene
Answer:
2,3-dimethyl-2-butene > 3-methyl-3-hexene > cis-3-hexene > 1-hexene
Explanation:
According to Saytzeff rule, the more highly substituted an alkene is, the more stable it is. Since this is so, 2,3-dimethyl-2-butene will be the most stable of all the alkenes listed because it is the most substituted alkene.
Let us also note that terminal alkenes are the least stable because the pi bonds of the alkenes are least stabilized by alkyl groups. This implies that 1-hexene is the least stable alkene among the listed alkenes.
If a sample is said to contain a 98.0% enantiomeric excess of the compound (+)-camphor, what would be the observed specific rotation if the pure (+)-camphor was found to have a rotation of 44.1?
Answer:
the observed specific rotation is 43.218
Explanation:
Given the data in the question;
percentage of enantiomeric excess = 98.0%
observed specific rotation = ? { represented by x }
specific rotation of pure compound = 44.1
Now, we know that;
% of enantiomeric excess = ( observed specific rotation / specific rotation of pure compound ) × 100%
so we substitute
98.0 % = ( x / 44.1 ) × 100%
0.98 = x / 44.1
x = 0.98 × 44.1
x = 43.218
Therefore, the observed specific rotation is 43.218
How can you identify ethane from ethene
Course Home P Acceptable units x + courseld=16709491&OpenVellumHMAC=f5c9929f4e4da0b5529475e262c91d79=10001 1 Review art A alculate the heat change in calories for condensation of 11.0 g of steam at 100°C. xpress your answer as a positive value using three significant figures and inc 2 MIKIN M HA Value CS
Answer:
The heat change in calories for condensation of 11.0 g of steam at 100°C is 5930 calories
Explanation:
Latent heat of condensation is the heat released when one mole of steam or water vapor condenses to form liquid droplets. The heat of condensation of water at 100° C is about 2,260 kJ/kg, which is equal to 40.68 kJ/mol. Since condensation of steam and vaporization of water occur at the same temperature and require the same amount of energy to occur, the heat of condensation is exactly equal to the heat vaporization, but has the opposite sign. In the vaporization, heat energy is absorbed by the substance, whereas in condensation heat energy is released by the substance.
The specific latent heat of vaporization of steam at 100° C = 40.68 kJ/mol
Number of moles of moles of water in 11.0 g of steam = mass/ molar mass
Molar mass of water = 18.0 g/mol
Number of moles of steam = 11.0 g / 18.0 g/mol = 0.61 moles
Heat released = 40.68 K/mol × 0.61 moles = 24.815 kJ
Converting to kcal by dividing 24.815 kJ by 4.184 = 5.93 kcal or 5930 calories
Therefore, the heat change in calories for condensation of 11.0 g of steam at 100°C is 5930 calories
A certain first-order reaction is 27.5 percent complete in 8.90 min at 25°C. What is its rate constant?
Answer:
[tex]k= 0.145min^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out necessary for us remember that the first-order kinetics is given by:
[tex]ln(A/A_0)=-kt[/tex]
Whereas the 27.5% complete means A/Ao=0.275, and thus, we solve for the rate constant as follows:
[tex]k=\frac{ln(A/A_0)}{-t}[/tex]
Then, we plug in the variables to obtain:
[tex]k=\frac{ln(0.275)}{-8.90min}\\\\k= 0.145min^{-1}[/tex]
Regards!
Which statement
about Niels Bohr's atomic model is true?
Higher orbits have lower energies.
Each orbit has a specific energy level.
&
Electrons can exist in any energy level.
Orbits close to the nucleus have no energy.
Answer:
b. each orbit has a specific energy level
Explanation:
edge
Briefly explain in your own words why the bond angle increases as the number of electron groups decreases
Answer:
i) The bond angle decreases due to the presence of lone pairs, which causes more repulsion on the bond pairs and as a result, the bond pairs tend to come closer. ii) The repulsion between electron pairs increases with an increase in electronegativity of the central atom and hence the bond angle increases.
Explanation:
The bond angle increases as the number of electron groups increases due to less repulsion between the bonded groups.
We know that in a molecule, repulsion between electron pairs affects the bond angle in the molecule. The magnitude of repulsion depends on the number of electron groups in the molecule.
The more the number of bonded electron groups in the molecule, the lesser the repulsion between electron pairs and the higher the observed bond angle.
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d=ut+5 make u the subject
Explanation:
d=ut+5
d-5=ut
d-5/t=u
!!!!!!!
For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 100.0%, what is the actual mass of hydrazine (N2H4) produced when 59.20 g of nitrogen reacts with 6.750 g of hydrogen?
a. Molar mass of N2 = 28.01 g/mol
b. Molar mass of H2 = 2.016 g/mol
c. Molar mass of N2H4 = 32.05 g/mol
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
Calcular el pH de una disolución cuando 10 g de NaOH se diluyen con agua hasta un volumen final de 250 ml
Answer:
My nettttttworkkkk is slowww
A solution of the primary standard potassium hydrogen phthalate (KHP), KHC8H4O4 , was prepared by dissolving 0.4877 g of KHP in about 50 mL of water. Titration of the KHP solution with a KOH solution of unknown concentration required 28.49 mL to reach a phenolphthalein end point. What is the concentration of the KOH solution?
Answer:
0.08382 M
Explanation:
Step 1: Write the balanced neutralization equation
KHC₈H₄O₄ + KOH ⇒ K₂C₈H₄O₄+ H₂O
Step 2: Calculate the moles corresponding to 0.4877 g of KHC₈H₄O₄
The molar mass of KHC₈H₄O₄ is 204.22 g/mol.
0.4877 g × 1 mol/204.22 g = 2.388 × 10⁻³ mol
Step 3: Calculate the moles of KOH that react with 2.388 × 10⁻³ moles of KHC₈H₄O₄
The molar ratio of KHC₈H₄O₄ to KOH is 1:1. The moles of KOH that react are 1/1 × 2.388 × 10⁻³ mol = 2.388 × 10⁻³ mol.
Step 4: Calculate the molar concentration of KOH
2.388 × 10⁻³ moles of KOH are in 28.49 mL of solution.
2.388 × 10⁻³ mol / 0.02849 L = 0.08382 M
A 0.15 M solution of BaCl2 contains: Group of answer choices 0.30 M Ba2 ions and 0.30 M Cl- ions. 0.15 M Ba2 ions and 0.15 M Cl- ions. 0.30 M Ba2 ions and 0.15 M Cl- ions. 0.15 M Ba2 ions and 0.30 M Cl- ions. none of the above
Answer:
0.15 M Ba⁺² ions and 0.30 M Cl⁻ ions
Explanation:
The dissociaton of barium chloride is as follows:
BaCl₂ → Ba²⁺ + 2Cl⁻
By observing the stoichiometric coefficients, we can tell that the number of moles of Ba²⁺ is the same as the number of moles of BaCl₂, while the number of moles of Cl⁻ is the double of that.
5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:
2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)
The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 oC. The heat capacity of water = 4.18 J/g/oC and the heat capacity of the calorimeter = 10.5 kJ/oC. (1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process.
Answer:
A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction:
2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)
The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 oC. The heat capacity of water = 4.18 J/g/oC and the heat capacity of the calorimeter = 10.5 kJ/oC. (1) what is the ΔH of the reaction?
Explanation:
The heat energy released by the reaction = heat absorbed by calorimeter + heat absorbed by water
Heat absorbed by water = mass of water x specific heat capacity of water x change in temperature
Heat absorbed by water = 500 g x 4.18 J/g. oC x (53.13-25.00)oC
= 58791.7 J
Heat absorbed by calorimeter = heat capacity of calorimeter x change in temperature
Heat absorbed by calorimeter = 10.5 x 10^3 J /oC x (53.13-25.00)oC
=295365 J
Total heat energy absorbed = 58791.7 J + 295365 J = 354156.7 J
Number of moles of benzene given is:
number of moles = goven mass of benzene /its molar mass
=7.05 g / 78.0 g/mol
=0.0903mol
Hence, the heat released by the reaction is:
= 354156.7 J / 0.0903 mol
= 3922.00 kJ/mol
Answer:
The heat released during the combustion of 7.05g of benzene is 3922.00kJ/mol.
Help please! this is timed
Answer:
C, P, P, C, P
Explanation:
is it still the same thing but the physical property change or did the thing change too? that's what it's asking
Electron sharing can be depicted by a Lewis dot structure, in which element symbols are surrounded by dots that represent the valence electrons (electrons in the ___________ shell). A ______________ bond is the sharing of a pair of valence electrons by _____________ atoms. Hydrogen has _____________ valence electron(s) in the first shell, but the capacity of the shell is ______________ electron(s). When a hydrogen atom comes close enough to a carbon atom for their orbitals to overlap, they can share their electrons. The hydrogen atom is now associated with _______________ electron(s) and a ______________ bond is formed. As a result, one of the structures does not make sense because hydrogen has only ____________ valence electron(s) to share, so it cannot form bonds with two atoms.
Answer:
Outermost
Covalent
Two
One
Two
Two
Covalent
One
Explanation:
A covalent bond is formed when an atom shares two electrons with another atom. These shared electrons could be contributed by each of the bonding atoms or by only one of the bonding atoms.
Hydrogen has the electronic configuration of 1s1. This implies that it has only one electron in its valence shell although the 1s shell can accommodate two electrons. When the atomic orbitals of carbon and hydrogen overlap, they share two electrons and hydrogen is now associated with two electrons in a covalent bond.
Since hydrogen possesses only one valence electron, it can not be bonded to two atoms.
Which of the following is an oxidation reduction reaction
Oxidation is lost of electrons. Reduced is gain of electrons. you can also remember them as OIL RIG.
Suppose you need to prepare 21.0 mL of formate buffer with a ratio of 4 of [sodium formate]/[formic acid] by mixing 0.10 M formic acid and 0.10 M sodium formate. How many milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid)
Answer: A volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
Explanation:
Given: Total volume of the buffer = 21.0 mL
[tex]\frac{[HCOONa]}{[HCOOH]} = 4[/tex] ... (1)
It is assumed that the volume of HCOONa is x. Hence, volume of HCOOH is (21.0 - x) mL.
Hence,
[HCOONa] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] x
= 0.1x mmol
Similarly, [HCOOH] = Molarity [tex]\times[/tex] Volume
= 0.10 [tex]\times[/tex] (21.0 - x) mmol
Using equation (1),
[tex]\frac{[HCOONa]}{[HCOOH]} = 4\\\frac{0.1x}{(21.0 - x)} = 4\\0.1x = 84.0 - 4x\\4.1x = 84.0\\x = 20.49 mL[/tex]
As x is the volume of sodium formate. Hence, 20.49 mL of sodium formate is required to make the buffer.
Thus, we can conclude that a volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).
f. . A metal cylinder has a mass of 100.00 g is heated to 95.50 celcius and then put in 245.5 g of water whose initial temperature is 22.50 Celsius. The final temperature of the mixture is 24.17 Celsius what is the specific heat of the metal.
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1. Explain why food lasts longer when placed in the refrigerator as opposed to on the counter in the kitchen.
2. List and describe four factors that affect the rate of a reaction.
3. Sketch a potential energy diagram for the following reaction: N2(g)+3H2(g)-->2NH3(g)
Use Table I in the Chemistry Reference Table to identify the following on your graph: potential energy of reactants, potential energy of products, activation energy, and heat of reaction. State the value of the enthalpy and whether the reaction is endothermic or exothermic. Label the y-axis and potential energy and x-axis as reaction coordinate.
4. Describe and explain the effect of (a) the increase in temperature, (b) the increase in concentration of C2H4, and (c) a decrease in pressure to the following system at equilibrium: 2C + 2H2 + heat ⇌ C2H4
It last longer because it slows down the spread of bacteria.
Temperature, concentration, particle size, use of a catalyst.
Write any two drawbacks of the octet theory.
Answer:
Octet rule fails to explain the following:
(1) The stability of incomplete octet molecules, i.e., the molecules with the central atom containing less than eight electrons. (2) The stability of expanded octet molecules, i.e., the molecules with the central atom containing more than eight electrons.
What is the concentration of HNO3 if 5.00×10−2 mol are present in 905 mL of the solution?
Answer:
0.05525 M or 55.25 mM
Explanation:
Concentration = moles/volume
*Note that volume is expressed in L so you will need to convert mL > L here
[tex]C= \frac{n}{V}\\C= \frac{0.05}{0.905}\\C=0.05525[/tex]
For a particular first-order reaction, it takes 48 minutes for the concentration of the reactant to decrease to 25% of its initial value. What is the value for rate constant (in s -1) for the reaction
Answer: The value for rate constant for a reaction is [tex]4.81\times 10^{-4} s^{-1}[/tex]
Explanation:
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(1)
Let the initial concentration of reactant be 100 g
Given values:
a = initial concentration of reactant = 100 g
a - x = concentration of reactant left after time 't' = 25 % of a = 25 g
t = time period = 48 min = 2880 s (Conversion factor: 1 min = 60 s)
Putting values in equation 1:
[tex]k=\frac{2.303}{2880s}\log (\frac{100}{25})\\\\k=4.81\times 10^{-4} s^{-1}[/tex]
Hence, the value for rate constant for a reaction is [tex]4.81\times 10^{-4} s^{-1}[/tex]
Which of the following is true about oxidation-reduction reactions?
=============================================================
One atom is oxidized and one is reduced
Both atoms are oxidized and reduced
The total number of electrons changes
One atom can be oxidized without one being reduced
Answer:
the last one probably
Explanation:
Selenium, an element used in the manufacture of solar energy devices, forms an oxide that contains only one atom of selenium (SeOx) and is 37.8% oxygen by mass. What is the molecular formula of the oxide? (Hint: find “x”)
Answer: The molecular formula of the compound will be [tex]SeO_3[/tex]
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Let the mass of the compound be 100 g
Given values:
% of O = 37.8%
% of Se = [100 - 37.8] = 62.2%
Mass of O = 37.8 g
Mass of Se = 62.2 g
We know:
Molar mass of Se = 79 g/mol
Molar mass of O = 16 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of Se}=\frac{62.2g}{79g/mol}=0.787 mol[/tex]
[tex]\text{Moles of O}=\frac{37.8g}{16g/mol}=2.36 mol[/tex]
Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.787 moles
[tex]\text{Mole fraction of Se}=\frac{0.787 }{0.787 }=1[/tex]
[tex]\text{Mole fraction of O}=\frac{2.36}{0.787 }=2.99\approx 3[/tex]
Taking the mole ratio as their subscripts.
The ratio of Se : O = 1 : 3
Hence, the molecular formula of the compound will be [tex]SeO_3[/tex]
If this experiment was performed again, but this time, 5.0 g of the mixture were used, then, assuming the same mass percentages (5 % cellulose, 47.5 % caffeine, and 47.5 % benzoic acid), what is the theoretical mass (in g) of cellulose in this mixture
Answer:
the theoretical mass of cellulose in this mixture is 0.25 grams
Explanation:
Given the data in the question;
mass of the mixture = 5.0 gram
mass percentage of cellulose = 5%
mass percentage of caffeine = 47.5%
mass percentage of benzoic acid = 47.5%
so mass of caffeine in the mixture will be;
⇒ ( mass percentage of cellulose ) × ( mass of the mixture )
= 5% × 5.0 gram
= ( 5 / 100 ) × 5.0 grams
= 0.05 × 5.0 grams
= 0.25 grams
Therefore, the theoretical mass of cellulose in this mixture is 0.25 grams
red litmus paper was used to test toothpaste and it turns blue.Explain what this tells about the toothpaste