Answer:
1. 17.5 g of CO₂
2. The limiting reactant is carbon (graphite), and its formula is C(graphite)
3. 3.7 g of O₂
Explanation:
First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:
Reactants: carbon(graphite) ⇒ C(graphite) ; oxygen gas ⇒ O₂(g)Products: carbon dioxide ⇒ CO₂(g)Thus, we write the chemical equation:
C(graphite) + O₂(g) → CO₂(g)
The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).
Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:
Mw(C) = 12 g/mol
moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol
Mw(O₂) = 16 g/mol x 2 = 32 g/mol
moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol
Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):
stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂
actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂
We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).
The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).
Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:
moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂
Now, we convert the moles of CO₂ to mass by using the Mw:
Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g
Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.
Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:
remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂
Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :
mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g
Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.
How to balance this equation KClO4 → KCl + ?O2(g) and what type of reaction occurs?
Explanation:
here is the answer to your question
Why does increasing the temperature of two reactants in solution make a
reaction proceed more quickly?
Answer:
-The particles of the two reactants will gain kinetic energy and collide with one another more frequently and forcefully, which makes the reaction take place more quickly
how does lead resemble chromium?
Write the cell notation for an electrochemical cell consisting of an anode where Mn (s) is oxidized to Mn2 (aq) and a cathode where Co2 (aq) is reduced to Co (s) . Assume all aqueous solutions have a concentration of 1 mol/L.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Write the balanced equation showing the decomposition of carbonic acid and sulfurous acid.
Explanation:
here's the answer to your question
Decomposition of Sulfurous Acid (H₂SO₃):
H₂SO₃ → H₂O + SO₂
In this reaction, sulfurous acid decomposes into water (H₂O) and sulfur dioxide (SO₂).
The decomposition of carbonic acid (H₂CO₃) and sulfurous acid (H₂SO₃) can be represented by the following balanced chemical equations:
Decomposition of Carbonic Acid (H₂CO₃):
H₂CO₃ → H₂O + CO₂
In this reaction, carbonic acid decomposes into water (H₂O) and carbon dioxide (CO₂).
Decomposition of Sulfurous Acid (H₂SO₃):
H₂SO₃ → H₂O + SO₂
In this reaction, sulfurous acid decomposes into water (H₂O) and sulfur dioxide (SO₂).
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Write the equation for the reaction described: The organic material 9,10-anthracenedione, C14H8O2, is oxidized, reacting with oxygen, forming carbon dioxide and water.
Answer:
[tex]C_{14}H_8O_2+15O_2\rightarrow 14CO_2+4H_2O[/tex]
Explanation:
Hello there!
In this case, according to the given description, it turns out possible for us to set up the required chemical reaction as follows:
[tex]C_{14}H_8O_2+O_2\rightarrow CO_2+H_2O[/tex]
However, it turns out mandatory to balance it according to the law of conservation of mass, which states that the atoms must the same on the reactants side as on the products side; thus, we proceed as follows:
[tex]C_{14}H_8O_2+15O_2\rightarrow 14CO_2+4H_2O[/tex]
In order to have 14 C atoms, 8 H atoms and 32 O atoms on both sides of the chemical equation.
Regards!
please help!
What is the definition of thermal chemistry?
a.The study of change that involves warm objects
b.The study of change that involves heat
c.The study of change that involves cool objects
d.The study of change that involves temperature
balance the following reaction using LCM method by showing each steps Pb (N3)2 + Cr(MnO4)2 Cr2O3 + MnO2 + Pb3O4+ NO
here's the answer to your question
How much BaSO4 can be formed from 196.0 g of H2SO4?
Answer:
a) You can form 466 g of BaSO₄.
Explanation:
a) Mass of BaSO4
196 g H₂SO4 × 1 mol H₂SO4
98.08 g H₂SO4
1 mol BaSO 1 mol H₂SO4 X X
466 g BaSO4
233.39 g BaSO4
1 mol BaSO4
Describe A Simple experiment that can be prepared in the laboratory to demonstrate the formation of Iron (III) Chloride from iron fillings
Answer:
Anhydrous iron(III) chloride may be prepared by treating iron with chlorine:[11]
{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl2_{(}g)->2FeCl3_{(}s)}}}
Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.
Dissolving iron ore in hydrochloric acid
{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}{\displaystyle {\ce {Fe3O4_{(}s){+~}8HCl_{(}aq)->FeCl2_{(}aq){+~}2FeCl3_{(}aq){+~}4H2O_{(}l)}}}
Oxidation of iron(II) chloride with chlorine
{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}{\displaystyle {\ce {2FeCl2_{(}aq){+~}Cl2_{(}g)->2FeCl3_{(}aq)}}}
Oxidation of iron(II) chloride with oxygen
{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}{\displaystyle {\ce {4FeCl2_{(}aq){+~}O2{+~}4HCl->4FeCl3_{(}aq){+~}2H2O_{(}l)}}}
Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride. Hydrated iron(III) chloride can be converted to the anhydrous form by treatment with thionyl chloride.[12] Similarly, dehydration can be effected with trimethylsilyl chloride:[13]
{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}{\displaystyle {\ce {FeCl3.6H2O + 12 Me3SiCl -> FeCl3 + 6 (Me3Si)2O + 12 HCl}}}
Anhydrous iron(III) chloride may be prepared by treating iron with chlorine.
What is an iron filling?
Iron filings are small shavings of ferromagnetic material.
[tex]{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}{\displaystyle {\ce {2{Fe_{(}s)}+3Cl_2_{(}g)- > 2FeCl_3_{(}s)}}}[/tex]
Solutions of iron(III) chloride are produced industrially both from iron and from ore, in a closed-loop process.
Dissolving iron ore in hydrochloric acid.
Oxidation of iron(II) chloride with chlorine.
[tex]{\displaystyle {\ce {2FeCl_2_{(}aq){+~}Cl_2_{(}g)- > 2FeCl_3_{(}aq)}}}\\[/tex]
Oxidation of iron(II) chloride with oxygen.
Heating hydrated iron(III) chloride does not yield anhydrous ferric chloride. Instead, the solid decomposes into hydrochloric acid and iron oxychloride.
Hydrated iron(III) chloride can be converted to an anhydrous form by treatment with thionyl chloride. Similarly, dehydration can be affected by trimethylsilyl chloride.
[tex]{\displaystyle {\ce {FeCl_3.6H2O + 12 Me_3SiCl - > FeCl3 + 6 (Me_3Si)2O + 12 HCl}}}[/tex]
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For a particular catalyzed reaction, the change in enthalpy is 26kJmole and the activation energy is 67kJmole. Which can be the change in enthalpy and the activation energy for the uncatalyzed reaction
Answer:
26kJmole,84kJmole
Explanation:
A catalyst lowers the activation energy between reactants and products. Once this energy barrier is lowered, reactants are converted into products faster.
However, the enthalpy change of a reaction is constant both in the catalysed and uncatalysed reaction.
The activation energy of the uncatalysed reaction must be higher than the activation energy of the catalysed reaction. Hence the answer above.
Which of the following are combustion reactions?
NaNO3 (aq) arrow Na+(aq) + NO3-(aq)
C2H6 (g) + O2 (g) arrow CO2 (g) + H2O (l)
Mg(s) + HCl (aq) arrow H2 (g) + MgCl2 (aq)
HCl (aq) + NaOH (aq) arrow HOH (l) + NaCl (s)
Answer:
C₂H₆(g) + O₂(g) ⇒ CO₂(g) + H₂O(l)
Explanation:
Which of the following are combustion reactions?
NaNO₃(aq) ⇒ Na⁺(aq) + NO₃⁻(aq)
NO. This is a dissociation reaction.
C₂H₆(g) + O₂(g) ⇒ CO₂(g) + H₂O(l)
YES. This is a combustion reaction because a compound reacts with oxygen to form carbon dioxide and water.
Mg(s) + HCl (aq) ⇒ H₂(g) + MgCl₂(aq)
NO. This is a single displacement reaction.
HCl(aq) + NaOH (aq) ⇒ HOH(l) + NaCl(s)
NO. This is a neutralization reaction.
5. How many grams of tin metal can be produced from smelting (heating) of a 4.5 kilograms of tin (IV) oxide? (Note: Elemental tin and oxygen gas are the only products of this reaction).
Answer:
About 3500 grams of tin.
Explanation:
We want to determine amount of tin metal (in grams) that can be produced from smelting 4.5 kilograms of tin(IV) oxide.
First, write the chemical compound. Since our cation is tin(IV), it forms a 3+ charge. Oxygen has a 2- charge, so we will have two oxygen atoms. Hence, tin(IV) oxide is given by SnO₂.
By smelting it, we acquire elemental tin and oxygen gas. Hence:
[tex]\text{SnO$_2$}\rightarrow \text{Sn} + \text{O$_2$}[/tex]
(Note: oxygen is a diatomic element.)
The equation is balanced as well.
To convert from SnO₂ to only Sn, we can first convert from grams of SnO₂ to moles, use mole ratios to convert to moles of Sn, and then from there convert to grams.
Since Sn has a molar mass of 118.71 g/mol and oxygen has a molar mass of 15.999 g/mol, the molar mass of SnO₂ is:
[tex](118.71)+2(15.999) = 150.708\text{ g/mol}[/tex]
Therefore, given 4.5 kilograms of SnO₂, we can first convert this into grams using 1000 g / kg and then using the ratio:
[tex]\displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}[/tex]
We can convert this into moles.
Next, from the chemical equation, we can see that one mole of SnO₂ produces exactly one mole of Sn (and also one mole of O₂). So, our mole ratio is:
[tex]\displaystyle \frac{1\text{ mol Sn}}{1\text{ mol SnO$_2$}}[/tex]
With SnO₂ in the denominator to simplify units.
Finally, we can convert from moles Sn to grams Sn using its molar mass:
[tex]\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
With the initial value and above ratios, we acquire:
[tex]\displaystyle 4.5\text{ kg SnO$_2$}\cdot \frac{1000 \text{ g SnO$_2$}}{1\text{ kg SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol SnO$_2$}}{150.708\text{ g SnO$_2$}}\cdot \displaystyle \frac{1\text{ mol Sn}}{1 \text{ mol SnO$_2$}} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1\text{ mol Sn}}[/tex]
Cancel like units:
[tex]=\displaystyle 4.5\cdot \frac{1000}{1}\cdot \displaystyle \frac{1}{150.708}\cdot \displaystyle \frac{1}{1} \cdot\displaystyle \frac{118.71\text{ g Sn}}{1}[/tex]
Multiply. Hence:
[tex]\displaystyle = 3544.5696...\text{ g Sn}[/tex]
Since we should have two significant figures:
[tex]=3500 \text{ g Sn}[/tex]
So, about 3500 grams of tin is produced from smelting 4.5 kg of tin(IV) oxide.
Answer:
3546g
Explanation:
start w/ tin (IV) oxide n elemental tin and oxygen gas are the only products of this reaction
SnO2 -> Sn + O2
Sn molecular wt: 119
O2 molecular wt: 32
SnO2 molecular wt: 119+32 = 151
so Sn / SnO2 wt ratio = 119 / 151
4.5 kilograms of tin (IV) oxide will produce:
= 4.5 * 119 / 151
= 3.546 kg
or 3546 grams of tin metal
no need to involve moles ;)
K always has the same value
at a given temperature
regardless of the amounts of
reactants or products that
are present initially.
Select one:
True
False
hope it will help you
How Many KJ
are in 1500 cal.
Answer: [tex]6.276\ kJ[/tex]
Explanation:
It is known that 1 cal is equivalent to 4.184 J
1500 cal will be equivalent to [tex]1500\times 4.184=6276\ J[/tex]
Also, 1 kJ is equivalent to 1000 J
So, 6276 J is equal to [tex]6.276\ kJ[/tex]
A process will definitely be spontaneous if: Select the correct answer below: the entropy of the system increases in the process the entropy of the system decreases in the process the entropy of the universe increases in the process the entropy of the universe decreases in the process
Answer:
the entropy of the universe increases
Explanation:
The second law of thermodynamics can be stated in terms of entropy, in this statement of the law; ''a spontaneous process increases the entropy of the universe''.
Given that; ∆Suniverse = ∆Ssystem + ∆Ssurroundings, the definition of a spontaneous process is one in which ∆Suniverse >0.
Hence, a process is spontaneous when the entropy of the universe increases.
How many molecules (or formula units) are in 138.56 g C4H10 Express your answer using four significant figures.
Answer:
dont buy cheap and off we went
Where is CO2 concentration greatest in the body?
Answer:
I believe it might be blood exiting the lungs, not positive though.
Explanation:
i think it is blood exiting the lungs
If the temperature of a volume of dieal gas ncreases for 100 to 200, what happens to the average kinetic energy of the molecules?
Answer:
It increases but less than double
Explanation:
As the temperature of a gas increase, the average kinetic energy of the gas increases. The kinetic energy of a gas is the thermal energy that the gas contains.
We know, the kinetic energy of an ideal gas is given by :
[tex]$V_{avg} = \sqrt{\frac{8R}{\pi M}}$[/tex]
where, R = gas constant
T = absolute temperature
M = molecular mass of the gas
From the above law, we get
[tex]$V_{avg} \propto \sqrt{T}$[/tex]
Thus, if we increase the temperature then the average kinetic energy of the ideal gas increases.
In the context, if the temperature of the ideal gas increases from 100°C to 200°C, then
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{T_2}{T_1}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{\frac{473.15}{373.15}}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =\sqrt{1.26}$[/tex]
[tex]$\frac{(V_{avg)_2}}{(V_{avg)_1}} =1.12$[/tex]
[tex]$(V_{avg})_2 = 1.12\ (V_{avg})_1$[/tex]
Therefore, [tex]$(V_{avg})_2 > (V_{avg})_1$[/tex]
Thus the average kinetic energy of the molecule increases but it increases 1.12 times which is less than the double.
Thus, the answer is " It increases but less that double".
Choose all the answers that apply. Silicon (Si) has 14 protons and an atomic mass of 28. Silicon has _____. three electron shells 14 electrons 14 neutrons two electron shells 28 electrons
Answer:
three electron shells
14 electrons
14 neutrons
Explanation:
Silicon has three electron shells arranged as follows; 2, 8, 4. This corresponds to the fact that silicon is a member of group 14 of the periodic table.
Note that, the number of protons in an atom is the same as the number of electrons in the neutral atom. Since Silicon has 14 protons, it also has 14 electrons likewise.
The mass number of silicon is 28 but number of neutrons= mass number - number of protons. Since mass number = 28, then there are 14 neutrons in silicon.
10g of a non-volatile and non-dissociating solute is dissolved in 200g of benzene.
The resulting solution boils At temperature of 81.20oC. Find the molar mass of solute.
Given that the BP of pure benzene is 80.10oC and Its elevation boiling point constant = 2.53 oC/m.
Answer: The molar mass of solute is 115 g/mol.
Explanation:
Elevation in the boiling point is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.
The expression for the calculation of elevation in boiling point is:
[tex]\text{Boiling point of solution}-\text{boiling point of pure solvent}=i\times K_b\times m[/tex]
OR
[tex]\text{Boiling point of solution}-\text{Boiling point of pure solvent}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)
where,
Boiling point of pure solvent (benzene) = [tex]80.10^oC[/tex]
Boiling point of solution = [tex]81.20^oC[/tex]
i = Vant Hoff factor = 1 (for non-electrolytes)
[tex]K_b[/tex] = Boiling point elevation constant = [tex]2.53^oC/m[/tex]
[tex]m_{solute}[/tex] = Given mass of solute = 10 g
[tex]M_{solute}[/tex] = Molar mass of solute = ? g/mol
[tex]w_{solvent}[/tex] = Mass of solvent = 200 g
Putting values in equation 1, we get:
[tex]81.20-80.10=1\times 2.53\times \frac{10\times 1000}{M_{solute}\times 200}\\\\M_{solute}=\frac{1\times 2.53\times 10\times 1000}{1.1\times 200}\\\\M_{solute}=115g/mol[/tex]
Hence, the molar mass of solute is 115 g/mol.
One and a half gallons of paint are used on refrigeration units which are then baked dry in a 275F oven for 2 hours. If the paint contains 5% solids and uses hexane as the solvent, what is the evaporation rate in pints per minute
Answer:
Hence the evaporation rate in pints per minute is 0.001583.
Explanation:
Now,
1.5 gallons of paint are used.
It has 5% solid, so total 95% of 1.5 gallon can evaporate(only the solvent part will evaporate).
95% of 1.5 gallon=(95/100)*1.5 = 1.425 gallons.
1gallon = 8 pints.
So 1.425 gallons=8*1.425 pints
=11.4 pints.
Evaporation requires 2 hours. That is 3600*2 seconds = 7200 seconds.
So evaporation rate= 11.4 pints/7200 seconds.
=0.001583 pints per second
Classify each of the reactions listed below as a single-displacement, double-displacement, synthesis,
decomposition, oxidation reduction or combustion reaction.
Reaction Type
: 2Na + Cl2 → 2NaCl
: C2H4 + 3O2 → 2CO2 + 2H2O
: 2Ag2O-> 4Ag + O2
: BaCl2 + Na2SO4->BaSO4 +2NaCl
: 2AI + Fe2O3-> 2Fe + Al2O3
The illustration on the left represents a mixture of I2 (purple) molecules and F2 (green) molecules. If these were to react to form IF3 molecules, what is the maximum number of IF3 molecules that could form? Hint: Atoms must be conserved in an ordinary chemical reaction.
Answer:
Depends on the reacting molecules.
Explanation:
More molecules of IF3 are formed when more I2 (purple) molecules and F2 (green) molecules combine or react with each other. If one molecule of I2 (purple) react with three F2 (green) molecules then two molecules of IF3 are formed. The equation for this reaction is given below:
I2 + 3F2 -----------> 2IF3. In this reaction, we can see that one molecule of I2 react with three molecules of F2 forming two molecules of IF3.
What is the electron domain geometry around N in N2CL4
Answer:
trigonal bipyramidal.
define surface are tension of liquid
The hydrological cycle refers to the circulation of water within the earth's hydrosphere in different from I. e. the liquid, solid and the gaseous forms.
1. Which of the following are covalent compounds?
Select all that apply:
Potassium Chloride: K CI
Octadecanol: C18H380
Dimethyl Sulfoxide: CHOS
Lithium Bromide: LiBr
Answer:
Octadecanol: C18H380
Dimethyl sulfoxide: CHOS
Explanation:
Covalent compounds are formed between non-metallic elements
7. There are 7. 0 ml of 0.175 M H2C2O4 , 1 ml of water , 4 ml of 3.5M KMnO4 what is the molar concentration ofH2C2O4 ?
8. Using the data from question 7 what is the molar concentration of KMnO4 ?
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Answer:
7. 0.1021 M
8. 1.167 M
10. Increase in volume of water would lower the rate of reaction
Explanation:
7. What is the molar concentration of H₂C₂O₄ ?
Since we have 7.0 ml of 0.175 M H₂C₂O₄, the number of moles of H₂C₂O₄ present n = molarity of H₂C₂O₄ × volume of H₂C₂O₄ = 0.175 mol/L × 7.0 ml = 0.175 mol/L × 7 × 10⁻³ L = 1.225 × 10⁻³ mol.
Also, the total volume present V = volume of H2C2O4 + volume of water + volume of KMnO4 = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of H₂C₂O₄, M = number of moles of H₂C₂O₄/volume = n/V
= 1.225 × 10⁻³ mol/12 × 10⁻³ L
= 0.1021 mol/L
= 0.1021 M
8. Using the data from question 7 what is the molar concentration of KMnO₄ ?
Since we have 4.0 ml of 3.5 M KMnO₄, the number of moles of KMnO4 present n' = molarity of KMnO₄ × volume of KMnO₄ = 3.5 mol/L × 4.0 ml = 3.5 mol/L × 4 × 10⁻³ L = 14 × 10⁻³ mol.
Also, the total volume present V = volume of KMnO₄ + volume of water + volume of KMnO₄ = 7.0 ml + 1 ml + 4 ml = 12 ml = 12 × 10⁻³ L
So, the molar concentration of KMnO₄, M' = number of moles of KMnO₄/volume = n'/V
= 14 × 10⁻³ mol/12 × 10⁻³ L
= 1.167 mol/L
= 1.167 M
10. From question number 7, what effect increasing the volume of water has on the reaction rate?
Increase in volume of water would lower the rate of reaction because, the particles of both substances would have to travel farther distances to collide with each other, since there are less particles present in the solution and thus, the concentration of the particles would decrease thereby decreasing the rate of reaction.
Which statement best summarizes the second law of thermodynamics?
A. The total disorder of a system and it’s surroundings tend to increase
B. Energy cannot be created or destroyed
C. Energy is lost when it changes form
D. A system and it’s surroundings tend toward a state of increased order
Answer:
B
Explanation:
A p e x
Name of this product
Answer:
Explanation:
ethyl 3-methylbenzoate
