Answer::
Lower rate of reaction
Explanation:
Lower concentration of reactant in an experiment is indicative of fewer ions or atoms present, which means a slower rate of reaction. In a titration reaction for instance, the end point would take longer to be substantiated, thereby increasing volume of titres. This is further backed up by collision theory which states that more particles in a system improves combinations of molecules.
This however may not be the case if the iodine in question is catalyst, in which case the change in concentration has no effect whatsoever on the reaction rte.
I hope this explanation fits the brief.
You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.18. You determine that the concentration of the unknown acid was 0.2230 M. You also determined that the acid was monoprotic (HA). What is the pKa of your unknown acid
Answer:
[tex]pKa=3.70[/tex]
Explanation:
Hello,
In this case, given the information, we can compute the concentration of hydronium given the pH:
[tex]pH=-log([H^+])\\[/tex]
[tex][H^+]=10^{-pH}=10^{-2.18}=6.61x10^{-3}M[/tex]
Next, given the concentration of the acid and due to the fact it is monoprotic, its dissociation should be:
[tex]HA\rightleftharpoons H^++A^-[/tex]
We can write the law of mass action for equilibrium:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
Thus, due to the stoichiometry, the concentration of hydronium and A⁻ are the same at equilibrium and the concentration of acid is:
[tex][HA]=0.2230M-6.61x10^{-3}M=0.2164M[/tex]
As the concentration of hydronium also equals the reaction extent ([tex]x[/tex]). Thereby, the acid dissociation constant turns out:
[tex]Ka=\frac{(6.61x10^{-3})^2}{0.2164}\\ \\Ka=2.02x10^{-4}[/tex]
And the pKa:
[tex]pKa=-log(Ka)=-log(2.02x10^{-4})\\\\pKa=3.70[/tex]
Regards.
Eugenol is a molecule that contains the phenolic functional group. Which option properly identifies the phenol in eugenol
Answer:
Explanation:
Hello,
Among the options given on the attached document, since phenolic functional group is characterized by a benzene ring bonded with a hydroxyl group (C₆H₅OH) we can see that the first option correctly points out such description. Thus, answer is on the second attached picture. Other options are related with other sections found in eugenol that are not phenolic.
Best regards.
The first option identified the phenol in eugenol.
Phenolic functional groupAccording to the attached image, since the phenolic functional group should be characterized by a benzene ring bonded along with a hydroxyl group (C₆H₅OH) so here we can see that the first option correctly points out such description. However, other options are related to other sections found in eugenol that are not phenolic.
learn more about molecule here: https://brainly.com/question/13127022
Mass of the condensed unknown liquid: 0.3175 g Temperature of the water bath: 99.00 oC Pressure of the gas: 748.2 mmHg Volume of the flask (volume of the gas): 145.0 mL Given : Kelvin = t oC + 273.15 1 L = 1000 mL 1 atm = 760 mmHg Gas constant: R = 0.08206 atm L / mole K; Ideal Gas Law: PV = nRT 1. What is the pressure of the gas in atm? (1 points) 2.
Answer:
1. 0.98 atm
Explanation:
The following data were obtained from the question:
Mass of unknown liquid (m) = 0.3175 g
Temperature (T) = 99 °C
Pressure (P) = 748.2 mmHg
Volume (V) = 145.0 mL
Gas constant (R) = 0.08206 atm.L/Kmol
1. Determination of the pressure in atm.
760 mmHg = 1 atm
Therefore,
748.2 mmHg = 748.2/760 = 0.98 atm
Therefore, the pressure in atm is 0.98 atm.
“Denitrifying” bacteria return molecular nitrogen gas (N2) back into the biosystem by a series of reductions. Identify the correct sequence. Select the correct answer below: A. NO−3→NO−2→N2O→N2 B. N2O→NO−3→NO−2→N2 C. N2O→NO−2→NO−3→N2 D. NO−3→N2O→NO−2→N2
Answer:
NO−3→NO−2→N2O→N2
Explanation:
Denitrification is the process by which nitrogen is returned to the atmosphere by denitrifying bacteria. The process of denitrification involves a sequence of reduction reactions in the sequence; NO3−→NO2−→N2O→N2.
Nitrogen is usually present in soil in the form of soil nitrates which are soluble in water and can be absorbed by plant roots. These denitrifying bacteria reduce soil nitrates to nitrites, then to nitrogen I oxide and finally to molecular nitrogen as shown in the sequence above.
Denitrification can release N2O, is an ozone-depleting substance and
greenhouse gas into the atmosphere with its attendant consequence on global warming.
Two moles of neon gas enclosed in a constant volume system receive 4250 J of heat. If the gas was initially at 293 K, what is the final temperature of the neon
Answer:
=355.5K
Explanation:
Specific heat, Q = mcΔT
where
Q= 4250JΔT= change in temp = final temp - initial tempc = specific heat capacity = 1.7m = mass of substance in grams[1 mole of Ne = 20g; 2 moles of Ne = 2 × 20 = 40g]
4250 = 40 × 1.7 × (final - 293K)
final - 293k = 4250 / ( 40 × 1.7)
Final temp = 62.5 + 293
=355.5K
I hope this steps are simple to follow and understand.
If a diatomic molecule has a vibrational force constant of k=240 kg s-2 and a reduced mass of 1.627x10-27 kg, its vibrational frequency should be (in cm-1):
A. 2040
B. 4079
C. 2885
D. 5770
E. 1443
Answer:
2040 cm-1
Explanation:
The vibrations frequency is obtained from;
v=1/2πc √k/μ
Where;
k= force constant = 240kgs-2
μ= reduced mass = 1.627×10^-27 kg
c= speed of light= 3×10^10cms-1
v= 1/2×3.142×3×10^10√240/1.627×10^-27
v= 5.3×10^-12 × 3.84×10^14
v= 20.4×10^2
v= 2040 cm-1
The s orbital can hold
Answer:
2 electrons
Explanation:
Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. do you think it is reasonable to exclude these types of substances as contaminants that would give a false positive when you tested your reaction residue to verify that it is KCl?
Answer:
-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.
-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.
With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.
What is a major product of the reaction in the box?
Answer:
Molecule C
Explanation:
In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.
I hope it helps!
Write the net ionic equation for any precipitation reaction that may be predicted when aqueous solutions of manganese(II) nitrate and sodium hydroxide are combined.
Answer:
Explanation:
Mn( NO₃ )₂ + 2Na OH = Mn( OH)₂ (s) ↓ + 2Na NO₃
Converting into ions
Mn⁺ + 2 NO₃⁻ + 2 Na⁺ + 2 OH⁻ = Mn( OH)₂ + 2 Na⁻ + 2 NO₃⁻
Cancelling out common terms
Mn⁺ + 2 OH⁻ = Mn( OH)₂
this is net ionic equation required.
Use bond energies provided to estimate 2Br2
For a reaction, what generally happens if the temperature is increased? a) A decrease in k occurs, which results in a faster rate. b) A decrease in k occurs, which results in a slower rate. c) An increase in k occurs, which results in a faster rate.
Answer:
an increase in K occurs,which results in a faster rate
if the temperature is increased for a reaction, An increase in k occurs, which results in a faster rate of reaction. Hence, Option (D) is correct.
What is Rate constant ?
A coefficient of proportionality relating the rate of a chemical reaction at a given temperature to the concentration of reactant (in a unimolecular reaction) or to the product of the concentrations of reactants.
It is represented as 'K'
The negative exponential relationship between k and the temperature indicates that as temperature increases, the value of k also increases.
Since the rate constant can be determined experimentally over a range of temperatures, the activation energy can be calculated using the Arrhenius equation.
Therefore, if the temperature is increased for a reaction, An increase in k occurs, which results in a faster rate of reaction. Hence, Option (D) is correct.
Learn more about Chemical kinetics here ;
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D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match the words in the left column to the appropriate blanks in the sentences on the right. Fill in the blanks.
a ketone
carbon 3
carbon 2
carbon 1
an aldehyde
carbon 4
In D-glucose, there is__________ functional group, and the carbonyl group is at___________ when looking at the Fischer projection.
In D-tructose. there is functional group, and the carbonyl group is at when looking at______ the Fischer projection.
Answer:
aldehyde
carbon-1
ketone
carbon-2
Explanation:
Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.
In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.
In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.
In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.
How many molecules of ATP are formed during the complete catabolism of a saturated fatty acid with the chemical formula CH3(CH2)20CO2H?
Answer:
[tex]148~ATP[/tex]
Explanation:
In case, we can start with the structure of [tex]CH_3(CH_2)_2_0CO_2H[/tex]. When we draw the molecule, we will obtain a total amount of 22 carbons. So, in order to calculate the total amount of ATP we have to make several questions:
How many Acetyl CoA is produced?
To calculate the total Acetyl Coa we have to use the equation:
[tex]Number~of~Acetyl~Coa=\frac{n}{2}[/tex], where n is the amount of carbons, so:
[tex]Number~of~Acetyl~Coa=\frac{22}{2}=11~carbons[/tex]
How many rounds take place?
To calculate the rounds we have to use the equation:
[tex]Number~of~rounds=\frac{n}{2}-1[/tex], where n is the amount of carbons, so:
[tex]Number~of~Acetyl~Coa=\frac{22}{2}-1=10~carbons[/tex]
How many [tex]FADH_2[/tex] and [tex]NADH[/tex] are produced for this fatty acid?
For each round, we will have 1 [tex]FADH_2[/tex] and 1 [tex]NADH[/tex], if we have 10 rounds. In total, we will have 10
How many ATP are formed?
The ATP would be formed in the electron transport chain and each coenzyme will have a different yield of ATP. So for the total calculation, we have to keep in mind the following relationships:
-) [tex]1~FADH_2~=~1.5~ATP [/tex]
-) [tex]1~NADH~=~2.5~ATP [/tex]
-) [tex]Acetyl~CoA~=~10~ATP[/tex]
So, know we can do the total calculation:
[tex](10*1.5)+(10*2.5)+(11*10)=150[/tex]
We have to subtract "2 ATP" molecules that correspond to the activation of the fatty acid, so:
[tex]150-2=148~ATP[/tex]
In total, we will have 148 ATP.
See figure 1
I hope it helps!
An astronomy research team has been studying the atmosphere of a moon orbiting a newly discovered exoplanet. The team has determined that the moon has an average temperature of 95K on the surface, with an average pressure of 1.6atm. Remote analysis of this moon's atmosphere has revealed it has a molar mass of 28.6 g/mol. Calculate the density (g/L) of 1 mole of the moon's atmosphere under the given conditions.
Answer:
5.81 g/L
Explanation:
Let's apply the Ideal Gases Law to determine this:
P . V = n . R . T
Pressure = 1.6 atm
Volume = ?
Mol = 1 mol
Temperature = 96 K
In order to find the density, we should know the volume of the atmosphere which is a mixture of gases so, we consider all the atmosphere as a unique ideal gas.
1. 6 atm . V = 1 mol . 0.082 L.atm/mol.K . 96K
V = (1 mol . 0.082 L.atm/mol.K . 96K) / 1.6 atm
V = 4.92 L → As this is the volume for the whole atmosphere and the mass of 1 mol is 28.6 g, density should be:
28.6 g / 4.92L = 5.81 g/L
Density → mass / volume
g Calculate the time (in min.) required to collect 0.0760 L of oxygen gas at 298 K and 1.00 atm if 2.60 A of current flows through water. (Hint: Ideal gas law)
Answer:
7.67 mins.
Explanation:
Data obtained from the question include the following:
Volume (V) = 0.0760 L
Temperature (T) = 298 K
Pressure (P) = 1 atm
Current (I) = 2.60 A
Time (t) =?
Next, we shall determine the number of mole (n) of O2 contained in 0.0760 L.
This can be obtained by using the ideal gas equation as follow:
Note:
Gas constant (R) = 0.0821 atm.L/Kmol
PV = nRT
1 x 0.0760 = n x 0.0821 x 298
Divide both side by 0.0821 x 298
n = 0.0760 / (0.0821 x 298)
n = 0.0031 mole
Next, we shall determine the quantity of electricity needed to liberate 0.0031 mole of O2.
This is illustrated below:
2O²¯ + 4e —> O2
Recall:
1 faraday = 1e = 96500 C
4e = 4 x 96500 C
4e = 386000 C
From the balanced equation above,
386000 C of electricity liberated 1 mole of O2.
Therefore, X C of electricity will liberate 0.0031 mole of O2 i.e
X C = 386000 X 0.0031
X C = 1196.6 C
Therefore, 1196.6 C of electricity is needed to liberate 0.0031 mole of O2
Next, we shall determine the time taken for the process. This can be obtained as follow:
Current (I) = 2.60 A
Quantity of electricity (Q) = 1196.6 C
Time (t) =?
Q = It
1196.6 = 2.6 x t
Divide both side by 2.6
t = 1196.6/2.6
t = 460.23 secs.
Finally, we shall convert 460.23 secs to minute. This can be achieved by doing the following:
60 secs = 1 min
Therefore,
460.23 secs = 460.23/60 = 7.67 mins
Therefore, the process took 7.67 mins.
Nitrogen has different oxidation states in the following compounds: nitrite ion, nitrous oxide, nitrate ion, ammonia, and nitrogen gas. Arrange these species in order of increasing nitrogen oxidation state. Select the correct answer below: A. ammonia, nitrogen gas, nitrite, nitrous oxide, nitrate B. nitrogen gas, ammonia, nitrous oxide, nitrite, nitrate C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate D. ammonia, nitrogen gas, nitrate, nitrite, nitrous oxide
Answer:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Explanation:
To establish the oxidation number of nitrogen in each compound, we know that the sum of the oxidation numbers of the elements is equal to the charge of the species.
Nitrite ion (NO₂⁻)
1 × N + 2 × O = -1
1 × N + 2 × (-2) = -1
N = +3
Nitrous oxide (NO)
1 × N + 1 × O = 0
1 × N + 1 × (-2) = 0
N = +2
Nitrate ion (NO₃⁻)
1 × N + 3 × O = -1
1 × N + 3 × (-2) = -1
N = +5
Ammonia (NH₃)
1 × N + 3 × H = 0
1 × N + 3 × (+1) = 0
N = -3
Nitrogen gas (N₂)
2 × N = 0
N = 0
The order of increasing nitrogen oxidation state is:
C. ammonia, nitrogen gas, nitrous oxide, nitrite, nitrate
Identify a homogeneous catalyst:
a. SO2 over vanadium (V) oxide
b. H2SO4 with concentrated HCl
c. Pd in H2 gas
d. N2 and H2 catalyzed by Fe
e. Pt with methane
Answer:
b, H2SO4 with HCl, as they are both liquid acids
What are the concentrations of [K+], [OH-], [CO32-] and [H+], in a 1.2 M solution of K2CO3 ? (Note: Question is asking for concentrations and not pH) g
Answer:
The concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.
Explanation:
The dissociation equation of K₂CO₃ in water is:
K₂CO₃(aq) ⇄ K⁺(aq) + CO₃²⁻(aq) (1)
Also, the CO₃²⁻ will react with water as follows:
CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (2)
The constant of the reaction (2) is:
[tex] Kb = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} = 2.08 \cdot 10^{-4} [/tex]
The solution of K₂CO₃ is 1.2 M, and since the mole ratio of K₂CO₃ with K⁺ and CO₃²⁻ is 1:1, then we have:
[tex] [K_{2}CO_{3}] = [K^{+}] = [CO_{3}^{-2}] = 1.2 M [/tex]
Now, from equation (2) we have:
CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (3)
1.2 - x x x
[tex] 2.08 \cdot 10^{-4} = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} [/tex]
[tex] 2.08 \cdot 10^{-4} = \frac{x^{2}}{1.2 - x} [/tex]
[tex] 2.08 \cdot 10^{-4}*(1.2 - x) - x^{2} = 0 [/tex] (4)
By solving equation (4) for x we have:
x = 0.016 M = [HCO₃⁻] = [OH⁻]
Hence, the CO₃²⁻ concentration is:
[CO₃²⁻] = 1.2 M - 0.016 M = 1.18 M
Finally, the concentration of [H⁺] is:
[tex] [H^{+}][OH^{-}] = 10^{-14} [/tex]
[tex][H^{+}] = \frac{10^{-14}}{[OH^{-}]} = \frac{10^{-14}}{0.016} = 6.25 \cdot 10^{-13} M[/tex]
Therefore, the concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.
I hope it helps you!
A crystal lattice formed by positive and negative ions is called a
Answer:
Ionic Crystal
Explanation:
The rate law for the reaction 2NO2 + O3 → N2O5 + O2 is rate = K[NO2][O3].
Which one of the following mechanisms Is consistent with this rate law?
A. NO2 + NO2 → N2O2 (fast)
N2O4 + O3 → N2O5 + O2 (slow)
B. NO2 + O3 → NO5 (fast)
NO5 + NO5 → N2O5 + (5/2)O2 (slow)
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
D. NO2 + NO2 → N2O2 + O2 (slow)
N2O2 + O3 → N2O5 (fast)
Answer:
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
Explanation:
A reaction mechanism represents an amount of elementary steps that explain how a reaction proceeds. The mechanism must explain the experimental rate law. Also, the slow step is the rate determining step.
This rate law is obtained from the multiplication of the reactants in the slow step, thus:
A. NO2 + NO2 → N2O2 (fast)
N2O4 + O3 → N2O5 + O2 (slow)
Rate law:
rate = k [N2O4] [O3]
This mechanism is not consistent with rate law.
B. NO2 + O3 → NO5 (fast)
NO5 + NO5 → N2O5 + (5/2)O2 (slow)
Rate law:
rate = k [NO5]²
This mechanism is not consistent with rate law.
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
Rate law:
rate = k [NO2] [O3]
This mechanism is consistent with rate law.D. NO2 + NO2 → N2O2 + O2 (slow)
N2O2 + O3 → N2O5 (fast)
Rate law:
rate = k [NO2]²
This mechanism is not consistent with rate law.
Thus, right solution is:
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
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AI(NO3)3 CH3NH3CN NaCIO
CH3NH3CI NaNO3
Acidic Basic pH-neutral
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Answer:
AI(NO₃)₃ → Acidic pH < 7
CH₃NH₃CN → Neutral pH = 7
NaCIO → Basic pH > 7
CH₃NH₃CI → Acidic pH < 7
NaNO₃ → Neutral pH = 7
Explanation:
First of all we dissociate the salts:
Al(NO₃)₃ → Al³⁺ + 3NO₃⁻Nitrate anion comes from the nitric acid which is strong, so the anion is the conjugate weak base. It does not react to water, but the Al is an special case. Aluminum as a cathion comes from the Al(OH)₃ which is a base but this compound can also react as an acid, it is called amphoterous.
Al³⁺ + H₂O ⇄ Al(OH)²⁺ + H⁺
Aluminium cathion reacts to water in order to produce a complex and to give protons to the medium, so the salt is acid.
CH₃NH₃CN → CH₃NH₃⁺ + CN⁻Both ions come from a weak base and a strong acid, so both ions are the conjugate strong base and acid, respectively. They can make hydrolysis to water so the salt is neutral.
CH₃NH₃⁺ + H₂O ⇄ CH₃NH₂ + H₃O⁺ Ka
CN⁻ + H₂O ⇄ HCN + OH⁻ Kb
NaCIO → Na⁺ + ClO⁻Sodium cathion, comes from the strong base NaOH so it is does not react to water. It is the conjugate weak acid. Hypochlorite comes from the weak acid, so it can hydrolyse to water.
ClO⁻ + H₂O ⇄ HClO + OH⁻ Kb
Hypochlorous acid is formed giving OH⁻ to medium, so the salt is basic.
CH₃NH₃CI → CH₃NH₃⁺ + Cl⁻Chloride comes from the strong acid HCl. It does not react to water.
Methylammonium comes from the weak base, methylamine so it can react to water in order to make hydrolysis. The salt will be acidic.
CH₃NH₃⁺ + H₂O ⇄ CH₃NH₂ + H₃O⁺ Ka
NaNO₃ → Na⁺ + NO₃⁻
Both ions come from a strong base and acid, so they are the conjugate base and acid, respectively. As they do not make hydrolisis in water, the salt will be neutral.
Anita heats a beaker containing water. As the temperature of the
water increases, which change to the water molecules occurs?
Select one:
The molecules separate into atoms of hydrogen and oxygen.
The molecules move at a faster rate.
The molecules become more massive.
The molecules expand and become wider.
Answer:
The molecules expand and become wider.
Explanation:
Answer:
The molecules move at a faster rate.
Explanation:
Temperature is directly proportional to the average kinetic energy of molecules, so when the temperature of water molecules increases, they have a higher average kinetic energy.
They move at a faster rate. The distance between the molecules only increases when they become gas where they break apart from the bondings with the other water molecules.
The combustion of propane (C 3H 8) in the presence of excess oxygen yields CO 2 and H 2O: C 3H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2O (g) When 2.5 mol of O 2 are consumed in their reaction, ________ mol of CO 2 are produced.
Answer:
1.5 mol of CO₂
Explanation:
Use the mole ratio to find how many moles of CO₂ are produced from the reaction.
For every 5 moles of O₂, three moles of CO₂ is produced.
2.5 mol O₂ × 3 mol CO₂ ÷ 5 mol O₂
= 2.5 mol O₂ × 0.6
= 1.5 mol CO₂
When 2.5 mol of O₂ is consumed in the reaction, 1.5 mol of CO₂ is produced.
Hope that helps.
For each reaction, write the chemical formulae of the oxidized reactants in the space provided. Write the chemical formulae of the reduced reactants.
reactants oxidized _____
reactants reduced _____
a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)
b. AgNO3(aq)+Cu(s)→2Ag(s)+CuNO)2(a)
c. 3AgNO(aq)+Al()→3Ags)+Al(NO3)3(aq)
Answer:
a. Oxidized: Fe(s)
Reduced: Pb(NO3)2
b.Oxidized: Cu(s)
Reduced: AgNO3
c. Oxidized: Al(s)
Reduced: AgNO3
Explanation:
In a redox reaction, one reactant is been oxidized whereas the other is reduced.. The reduced reactant is the one that is gaining electrons and the oxidized one is loosing electrons.
In the reactions:
a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)
The Fe is as reactant as Fe(s) (Oxidation state 0) and the product is +3 (Because NO3, nitrate ion, is always -1). That means Fe is oxidized. The Pb as reactant is +2 and as product 0 (Gaining 2 electrons). Pb(NO3)2 is reduced
b. 2AgNO3(aq)+Cu(s)→2Ag(s)+Cu(NO3)2(a)
AgNO3 is +1 and Ag(s) is 0. AgNO3 is reduced. Cu(s) is 0 as reactant and +2 as product. Cu(s) is been oxidized
c. 3AgNO3(aq)+Al(s)→3Ag(s)+Al(NO3)3(aq)
Here, in the same way, AgNO3 is +1 as reactant and 0 as product. AgNO3 is reduced. And Al(s) is 0 as reactant but + 3 as product. Al(s) is oxidized.
Match each property of a liquid to what it indicates about the relative strength of the intermolecular forces in that liquid.
Strong intermolecular forces
Weak intermolecular forces
Answer:
Strong intermolecular forces: an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point.
Weak intermolecular forces: a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point.
Explanation:
Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces inclde dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.
The strength of the intermolecular forces in a liquid usually affects the various properties of the liquid such as viscosity, surface tension, vapour pressure and boiling point.
Strong intermolecular forces in a liquid results in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.
Weak intermolecular forces in a liquid results in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point of that liquid.
Strong intermolecular force is defined as the increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point while weak intermolecular forces define as the decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in boiling point.
Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces include as follows:-
Dispersion forcesDipole-dipole interactionsHydrogen bondingion-dipole forces.
Strong intermolecular forces in a liquid result in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.
Weak intermolecular forces in a liquid result in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in the boiling point of that liquid.
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which statement describes the use of a flowchart?
Answer:
A flowchart is a type of diagram that represents a workflow or process
Explanation:
Answer: orders in which steps in a process happen
Explanation:
formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer
Answer:
pH = 3.95
Explanation:
It is possible to calculate the pH of a buffer using H-H equation.
pH = pka + log₁₀ [HCOONa] / [HCOOH]
If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:
3.77 = pka + log₁₀ [0.50] / [0.50]
3.77 = pka
Knowing pKa, the NaOH reacts with HCOOH, thus:
HCOOH + NaOH → HCOONa + H₂O
That means the NaOH you add reacts with HCOOH producing more HCOONa.
Initial moles of 100.0mL = 0.1000L:
[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH
[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa
After the reaction, moles of each species is:
0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH
0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa
With these moles of the buffer, you can calculate pH:
pH = 3.77 + log₁₀ [0.0600] / [0.0400]
pH = 3.95When the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer pH is = 3.77 + log₁₀ [0.0600] / [0.0400] = 3.95
What is Formic Acid?It is possible to Computation the pH of a buffer using H-H equation.
Then pH is = pka + log₁₀ [HCOONa] / [HCOOH]
Then If concentration of [HCOONa] is = [HCOOH] then = 0.50M and pH = 3.77:
3.77 is = pka + log₁₀ [0.50] / [0.50]
After that, 3.77 = pka
Then, Knowing pKa, the NaOH reacts with HCOOH, thus:
After that,[tex]HCOOH + NaOH \rightarrow HCOONa + H2O[/tex]
Now, That means the NaOH you add reacts with HCOOH producing more HCOONa.
Then, Initial moles of 100.0mL = 0.1000L:
After that, [HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH
Then, [HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa
After that, when the reaction, moles of each species is:
Then, 0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH
Now, 0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa
Then, With these moles of the buffer, you can calculate pH:
pH = 3.77 + log₁₀ [0.0600] / [0.0400]
Therefore, pH = 3.95
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When silver nitrate is added to an aqueous solution of magnesium chloride, a precipitation reaction occurs that produces silver chloride and magnesium nitrate. When enough AgNO3 is added so that 34.3 g of MgCl2 react, what mass of the AgCl precipitate should form
Answer:
103.62 g of AgCl.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
2AgNO3 + MgCl2 —> 2AgCl + Mg(NO3)2
Step 2:
Determination of the mass of MgCl2 that reacted and the mass of AgCl produced from the balanced equation.
This is illustrated below:
Molar mass of MgCl2 = 24 + (2x35.5) = 95 g/mol
Mass of MgCl2 from the balanced equation = 1 x 95 = 95 g
Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol
Mass of AgCl from the balanced equation = 2 x 143.5 = 287 g
Thus, from the balanced equation above,
95 g of MgCl2 reacted to produce 287 g of AgCl.
Step 3:
Determination of the mass of AgCl produced from the reaction of 34.3 g of MgCl2.
The mass of AgCl produced from the reaction can be obtained as follow:
Form the balanced equation above,
95 g of MgCl2 reacted to produce 287 g of AgCl.
Therefore, 34.3 g of MgCl2 will react to produce = (34.3 x 287)/95 = 103.62 g of AgCl.
Therefore, 103.62 g of AgCl were produced from the reaction.
how salt solution can be determined by using hydrometer
Answer:
Salt solution may be calculated by measuring the specific gravity of a sample of water using a hydrometer.
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