Answer:
Chemical reactions, kinetics, organic chemistry
Explanation:
You might study the chemical reaction, learn about the differences between products and reactants, about delta H and exothermic and endothermic reactions. You may also study Kinetics by studying the rates of reactions with certain chemicals in a maple's enzymatic processes.
Another thing that you might learn about is organic chemistry. The glucose molecules, carbohydrates, lipids, nucleic acids, all have a structure based on the Carbon atom. You can learn about the specific structures of some chemicals that are involved in photosynthesis and simple hydrocarbons that are involved in photosynthetic/bio-synthetic pathways.
There's probably a lot more - but these are the most basic things I could think of.
Compounds A and BB are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 gg of sulfur with 5.99 gg of oxygen, and compound BB results from combining 8.60 gg of sulfur with 12.88 gg of oxygen. Show that the mass ratios in the two compounds are simple multiples of each other.
Answer:
Mass ratio of sulfur and oxygen in compounds A and B is 3:2 which confirms that the mass ratios in the two compounds are simple multiples of each other
Explanation:
This question seeks to establish/confirm the law of multiple proportions which posits that elements combine to form different substances which are whole number multiples of each other. Best example of this plays out in the formation of several oxides of the same element. Looking at the ratio in which the elements combine in each of the oxides, we can assume that these ratios are simple whole number multiples of each other.
Now back to the question.
In substance A, we have 6 g of sulfur combining with 5.99 g of oxygen
Now, lest us calculate the ratio of the mass of sulfur to that of oxygen = 6g/5.99g = 1
Now let us calculate the mass ratio of sulfur to oxygen in the second compound = 8.6/12.88 = 0.668
Now the ratios in both compounds are 1 to 0.668. 0.668 to fraction is approximately 1/1.5.
So therefore, the ratio we are having would be 1:1/1.5 or 1:0.668
This is same as 1/(1÷1.5) which is 1.5/1 or simply 3/2
This gives a ratio of approximately 1.5 to 1 or 3 to 2
The ratio 3 to 2 indicates that the mass ratios in both com pounds are simple multiples of each other
In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for electrons. II. the oxidized form has a lower affinity for electrons. III. the reduced form has a higher affinity for electrons. IV. the greater the tendency for the oxidized form to accept electrons.
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
the reaction between aluminum and iron(iii) oxide can generate temperatures approaching 3000c and is used in welding metals. In one process, 268g of Al are reacted with 501g of Fe2O3. identify the limiting reagent and calculate the theoretical mass
Answer:
- Iron (III) oxide is the limiting reactant.
- [tex]m_{Al_2O_3}=319.9gAl_2O_3[/tex]
- [tex]m_{Fe}=350.4gFe[/tex]
Explanation:
Hello,
In this case, we consider the following reaction:
[tex]2Al + Fe_2O_3 \rightarrow Al_2O_3 +2Fe[/tex]
Thus, for identifying the limiting reactant we should compute the available moles of aluminium in 268 g:
[tex]n_{Al}=268gAl*\frac{1molAl}{26.98gAl} =9.93molAl[/tex]
Next, we compute the moles of aluminium that are consumed by 501 grams of iron (III) oxide via their 2:1 molar ratio:
[tex]n_{Al}^{consumed}=501gFe_2O_3*\frac{1molFe_2O_3}{159.69gFe_2O_30}*\frac{2molAl}{1molFe_2O_3}=6.27molAl[/tex]
Thus, we notice there are less consumed moles of aluminium than available, for that reason, it is in excess; therefore, the iron (III) oxide is the limiting reactant.
Moreover, the theoretical mass of aluminium oxide is:
[tex]m_{Al_2O_3}=6.27molAl*\frac{1molAl_2O_3}{2molAl} *\frac{101.96gAl_2O_3}{1molAl_2O_3} =319.9gAl_2O_3[/tex]
And the theoretical mass of iron is:
[tex]m_{Fe}=6.27molAl*\frac{2molFe}{2molAl} *\frac{55.845 gFe}{1molFe} =350.4gFe[/tex]
Best regards.
An unknown compound, B, has the molecular formula C7H12. On catalytic hydrogenation 1 mol of B absorbs 2 mol of hydrogen and yields 2-methylhexane. B has significant IR absorption band at about 3300 and 2200 cm-1. Which compound best represents B?
a. 5-methyl-1,3-hexadiene
b. 5-methyl-1-hexyne
c. 3-methyl-1-hexyne
d. 5-methyl-2-hexyne
e. 2-methyl-1,5-hexadiene
Answer:
B and D
Explanation:
If we use the info given we have a band a 3300 cm-1 and 2200 cm-1 this indicates that we have an alkyne functional group. Additionally, the hydrogenation of the unknown molecule will consume two moles of hydrogens this fits with the 2 pi bonds in the alkyne functional group. So, we can discard "a" and "e". The product of this hydrogenation is 2-methylhexane therefore we can discard c because the methyl group is placed on carbon 3. Structures b and d can work.
See figure 1
I hope it helps!
Acetic acid and ethanol react to form ethyl acetate and water.If 94.0 mmol of C2H5CO2CH3 are removed from a flask containing a mixture of HCH3CO2, C2H5OH, C2H5CO2CH3 and H2O at equilibrium, then following questions are to be answered. 1. What is the rate of the reverse reaction before any C2H5CO2CH3 has been removed from the flask?2. What is the rate of the reverse reaction just after the C2H5CO2CH3 has been removed from the flask?3. What is the rate of the reverse reaction when the system has again achieved equilibrium?4. How much less C2H5CO2CH3 is in the flask when the system has again reached equilibrium?
Answer:
Explanation:
The equation for the react between Acetic acid and ethanol to form ethyl acetate and water is :
[tex]HCH_3CO_2_{(aq)}+C_2H_5OH_{(aq)} \to C_2H_5CO_2CH_3_{(aq)} + H_2O_{(l)}[/tex]
Imagine if 94.0 mmol of [tex]C_2H_5CO_2CH_3[/tex] are removed from a flask; Then:
We are to answer the following questions:
1. What is the rate of the reverse reaction before any [tex]C_2H_5CO_2CH_3[/tex] has been removed from the flask?
The reaction above is called an esterification reaction;
So the rate of reverse reaction before any [tex]C_2H_5CO_2CH_3[/tex] is removed is greater than zero and equal to forward reaction rate.
2. What is the rate of the reverse reaction just after the [tex]C_2H_5CO_2CH_3[/tex] has been removed from the flask?
Just after the [tex]C_2H_5CO_2CH_3[/tex] has been removed from the flask, the rate of the reverse reaction is greater than zero but less than forward reaction rate.
3. What is the rate of the reverse reaction when the system has again achieved equilibrium?
When the system has again achieved equilibrium, the rate of the reverse reaction is greater than zero and equal to forward reaction rate because we it has achieved the equilibrium, hence, the reaction tends to proceed in the forward direction.
4. How much less [tex]C_2H_5CO_2CH_3[/tex] is in the flask when the system has again reached equilibrium?
The [tex]C_2H_5CO_2CH_3[/tex] in the flask when the system has again reached equilibrium is lesser by 94.0 mmol as given right from the question
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.
Is chemical engineering suits for a person who gets bored fast and needed to learn new things?
Answer:
yes
Explanation:
because it will keep them entertained and will learn new things
If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later?
Answer:
below
Explanation:
28th 10;24 am
If the Moon rises at 7 P.M. on a particular day, then approximately what time will it rise six days later at 12A.M.
How much time changes between Moon rises from one day to the next?This movement is from the Moon's orbit, which takes 27 days, 7 hours and 43 minutes to go full circle. It causes the Moon to move 12–13 degrees east every day. This shift means Earth has to rotate a little longer to bring the Moon into view, which is why moonrise is about 50 minutes later each day.
So knowing that moonrise is about 50 minutes later each day, we have:
[tex]7+50 minutes = 7:50\\8:40\\9:30\\10:20\\11:10\\12:00 A.M[/tex]
See more about moon at brainly.com/question/13538936
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Use bond energies to calculate ΔHrxn Δ H r x n for the reaction. 2H2(g)+O2(g)→2H2O(g) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )
Answer:
[tex]\large \boxed{\text{-486 kJ}}[/tex]
Explanation:
You calculate the energy required to break all the bonds in the reactants.
Then you subtract the energy needed to break all the bonds in the products.
2H₂ + O₂ ⟶ 2H-O-H
Bonds: 2H-H 1O=O 4H-O
D/kJ·mol⁻¹: 436 498 464
[tex]\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.[/tex]
Hydrocarbon X has the formula C6H12. X reacts with one molar equivalent of hydrogen in the presence of a palladium catalyst to form a product having 12 primary hydrogens. Treatment of X with ozone followed by zinc in aqueous acid gives a mixture two aldehydes. What is the structure of X
Answer:
Explanation:
X( C₆H₁₂ )= (CH₃)₃-C-CH=CH₂
(CH₃)₃-C-CH=CH₂ + H₂ = (CH₃)₃-C-CH₂-CH₃ ( 12 primary hydrogen bonds )
(CH₃)₃-C-CH=CH₂ + O₃ = (CH₃)₃-C-CH= O + HCHO
Nitrogen is a group 15 element. What does being in this group imply about the structure of the nitrogen atom?
O A. Nitrogen has 15 valence electrons.
OB.
Nitrogen has 15 neutrons.
OC. Nitrogen has 5 valence electrons.
D.
Nitrogen has 5 neutrons.
Answer:
D. Nitrogen has 5 valence electrons.
Explanation:
Nitrogen is an element in group 5A of the periodic table. Elements in group 5A all contain just 5 valence electrons. (Electrons in the outer shell).
**Elements are organized into these groups in a periodic table based on the number of valence electrons which determines their charge. (Does not apply to transition metals)
If complications arise after cataract surgery, the ophthalmologist will use a Nd:YAG laser to perform a posterior capsulotomy. If the wavelength of the laser used is 1064 nm (infrared), and the pulse duration is 2.00 x 10–6 s whose energy is 0.245 J per pulse, how many photons are produced in each pulse?
Answer: 1.311 × 10^18 photons are produced in each pulse
Explanation: Please see the attachments below
Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.13 m FeCl3 A. Highest boiling point 2. 0.19 m Mg(CH3COO)2 B. Second highest boiling point 3. 0.30 m KI C. Third highest boiling point 4. 0.53 m Glucose(nonelectrolyte) D. Lowest boiling point An error has been detected in your answer. Check for typos,
Answer:0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point-C
0.13 m FeCl3---- Lowest boiling point-D
Explanation:
Using the boilng point elevation formula
ΔTb=m* kb *i
where m= molality
kb= elevated boiling point constant( here kb values will be same for all soluton)
i= vant hoff factor = number of ions present in a solution
Using the number of ions and molarity present in a solution as a collagative property, since kb is constant, we can determine which of the species has the highest boiling point.
1.) 0.13 m FeCl3= Fe³⁻ + Cl⁻
i=4
ΔTb=m* kb* i= molarity x number of ionsx Kb= 0.13 x 4= 0.52kb
2) 0.19 m Mg(CH3COO)2 = Mg²⁺ + CH₃COO⁻
i= 3
ΔTb=m* kb* i= molarity x number of ions= 0.19 x 3= 0.57kb
3. 0.30 m KI = K⁺ + I⁻
i= 2
ΔTb=m *kb *= imolarity x number of ions xKb= 0.30x 2= 0.60kb
4. 0.53 m Glucose(nonelectrolyte) =
i= 1 for nonelectroytes
ΔTb=m* kb* i = molarity x number of ionsx Kb= 0.53 x 1= 0.53Kb
therefore,
0.30 m KI ---- A. Highest boiling point
0.19 m Mg(CH3COO)2 ---- B. Second highest boiling point
0.53 m Glucose(nonelectrolyte) ---- Third highest boiling point
0.13 m FeCl3---- Lowest boiling point