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
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
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]
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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
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.