Answers
Answer:
See figure 1
Explanation:
For this question, we have to remember that definition of an "isomer" in an isomer we have the same condensed formula (in this case [tex]C_6H_14[/tex]) and different structures. The first structure is a linear structure of 6 carbons (hexane). Then we can have a 5 carbon linear structure in which we have to add a methyl group. This methyl group can be attached to carbon 2 or carbon 3 (2-methylpentane and 3-methylpentane). Finally, we can have a 4 carbon linear structure in which we have to add 2 methyl groups. We can do this addition in carbon 2 (2,2-dimethylbutane) or we can do this addition in carbon 2 and carbon 3 (2,3-dimethylbutane).
See figure 1
I hope it helps!
The first component is a six-carbon linear structure. Then we'll need to add a methyl group to a five-carbon linear structure.
Constitutional isomers:This methyl group can be linked to either carbon 2 or carbon 3 in the carbon chain. Finally, we can make a four-carbon linear structure by adding two methyl groups. This addition can be done with carbon 2 alone or with carbon 2 and carbon 3 together.
The constitutional isomers have the same chemical formula, but their atom connections are different. Chain isomers, position isomers, and functional group isomers are the three forms of constitutional isomers.
Find out more information about 'Constitutional isomers'.
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Related Questions
Which of the possible compounds has a mass of 163 grams when
6.13 x 1024 molecules of the compound is put on a scale?
Answers
Answer:
CH4
Explanation:
In solving this problem, we must remember that one mole of a compound contains Avogadro's number of elementary entities. These elementary entities include atoms, molecules, ions etc. Recall that one mole of a substance is the amount of substance that contains the same number of elementary entities as 12g of carbon-12. The Avogadro's number is 6.02 × 10^23.
Hence we can now say;
If 163 g of the compound contains 6.13 ×10^24 molecules
x g will contain 6.02 × 10^23 molecules
x= 163 × 6.02 × 10^23 / 6.13 × 10^24
x= 981.26 × 10^23/ 6.13 ×10^24
x= 160.1 × 10^-1 g
x= 16.01 g
x= 16 g(approximately)
16 g is the molecular mass of methane hence x must be methane (CH4)
Phosphorus pentachloride decomposes to phosphorus trichloride at high temperatures according to the equation:
PCl5(g) = PCl3(g) + Cl2(gu)
at 250 Co , 0.125M PCl5 is added to the flask , if Kc=1.80, what are the equilibrium concentrations of each gas?
A) [PCl5]=0.0625M,[PCl3]=0.0335M,and [Cl2]=0.335M
B) [PCl5]=1.80M,[PCl3]=1.80M,and [Cl2]=1.80M
C) [PCl5]=0 M,[PCl3]=1.34M,and [Cl2]=1.34M
D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M
E) [PCl5]=3.96M,[PCl3]=3.83M,and [Cl2]=3.83M
Answers
Answer:
D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117M
Explanation:
Based on the reaction:
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
And knowing:
Kc = [PCl₃] [Cl₂] / [PCl₅] = 1.80
When you add PCl₅ into a flask, this gas will react producing PCl₃ and Cl₂ until [PCl₃] [Cl₂] / [PCl₅] = 1.80
This could be written as:
[PCl₃] = X
[Cl₂] = X
[PCl₅] = 0.125M - X
Where X represents the moles of PCl₅ that react, reaction coordinate.
Replacing in Kc expression:
[PCl₃] [Cl₂] / [PCl₅] = 1.80
[X [X] / [0.125 - X] = 1.80
X² = 0.225 - 1.80X
0 = -X² -1.80X + 0.225
Solving for X:
X = -1.9M → False solution, there is no negative concentrations
X = 0.11735M → Right solution.
Replacing, concentrations in equilibrium are:
[PCl₃] = X
[Cl₂] = X
[PCl₅] = 0.125M - X
[PCl₃] = 0.117M
[Cl₂] = 0.117M
[PCl₅] = 0.00765M
And right option is:
D) [PCl5]=0.00765M,[PCl3]=0.117M,and [Cl2]=0.0.117MHow many moles are there in 60.5 g of mercury(II) chloride?
Answers
Hey there!
Mercury (II) chloride has a molar mass of 271.52 g/mol.
So, one mole of mercury (II) chloride has a mass of 271.52 g/mol.
We have 60.5g of mercury (II) chloride.
60.5g ÷ 271.52g/mol = 0.223 mol
There are 0.223 moles in 60.5g of mercury (II) chloride.
Hope this helps!
Answer:
Mercury (II) chloride has a molar mass of 271.52 g/mol.
So, one mole of mercury (II) chloride has a mass of 271.52 g/mol.
We have 60.5g of mercury (II) chloride.
60.5g ÷ 271.52g/mol = 0.223 mol
There are 0.223 moles in 60.5g of mercury (II) chloride.
Hope this helps!
Answer the question NOW Mistystar4881
I hope this was helpful HATERZ like some name above me lol JK about LOL. LOL
A mixture with H2 and He exerts a total pressure of 0.48 atm. If there is 1.0 g of H2 and 1.0 g of He in the mixture, what is the partial pressure (in atmospheres) of hydrogen?
Answers
Answer:
Partial pressure of hydrogen H₂ = 0.32 atm
Explanation:
Given:
Total pressure = 0.48 atm
Find:
Partial pressure of hydrogen
Computation:
Number of mole of H₂ = 1 / 2 = 0.5 moles
Number of mole of He = 1 / 4 = 0.25 moles
Total moles = 0.5 + 0.25 = 0.75
Partial pressure of hydrogen H₂ = [moles / total moles] Total pressure
Partial pressure of hydrogen H₂ = [0.50 / 0.75]0.48 atm
Partial pressure of hydrogen H₂ = 0.32 atm
Answer: 0.32 atm
Explanation:
First convert the mass of H2 to moles using the molar mass.
(1.0 gram H2 ⋅ (1.0 mol H2 / 2.016 g H2)) ≈ 0.50 mol H2
Next, convert the mass of helium He to moles using the atomic mass.
(1.0 gram He ⋅ (1.0 mol He / 4.003 g He)) ≈ 0.25mol He
The total number of moles is about 0.75 moles . The partial pressure of a component of a gas mixture can be found by multiplying the mole fraction by the total pressure.
PH2 = XH2 × Ptotal
PH2 = (0.50 mol / 0.75 mol)(0.48 atm) = 0.32 atm
solution solution solution
Answers
Answer:
Oxygen present in food items makes then rancid due to the presence of oils and fats. If the food is flushed with nitrogen, it prevents it from being oxidised (the nitrogen acts as an antioxidant).
Hope it helps ! :)
The graph below shows the half life values of parent isotopes
Based on the graph, it can be concluded that the isotope which is most likely to be found with 12.5% of its original amount in 42 billion years is
Answers
Answer:
Thorium-235
Explanation:
Half-life is defined as the time taken for a radioactive material to reduce to half of its original amount. If the original amount of a radioactive substance N is 100%, then;
1st half-life- 50% of N is left
2nd half life - 25% of N is left
3rd half-life- 12.5% of N is left
The half-life of Thorium-235 is 15 billion years, hence three half lives will take place in 45 billion years. Hence 12.5% of the original amount of Thorium-235 present will remain after about 42 billion years.
the reation between sulfur dioxide and oxygen is a dynamic equilibrium. what happens when the pressure of the system is
increased ?
Answers
Answer:
In the above reaction, sulfur dioxide and oxygen react together to form sulfur trioxide. This means that an increase in pressure would move the equilibrium to the right and result in more sulfur trioxide being formed. Pressure can only affect the position of equilibrium if there is a change in the total gas volume.
A 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. A buffer region was found around a pH of 3.5. The unknown compound is
Answers
Answer:
The unknown compound is a weak acid.
Explanation:
Given that :
a 25 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution.
A buffer region was found around a pH of 3.5. We know that a pH of 3.5 is a weak acid. So, it is likely to be an organic acid
Let assume the solution of the unknown sample to be CH₃COOH
Now :
25 mL of CH₃COOH reacted with 0.115 M of NaOH
The equation for the reaction will be :
CH₃COOH + NaOH -----> CH₃COONa + H₂O
at x mole of 0.115y M of
CH₃COOH NaOH is present
If NaOH was added in excess;
CH₃COOH + NaOH -----> CH₃COONa , NaOH will be lost then CH₃COOH and CH₃COONa will be present
Therefore;
At equilibrium : Only CH₃COONa will be present but if it is above equilibrium NaOH will be present because the pH will increase due to the presence of the strong base
Grams of cl in 38g of cf3cl
Answers
Answer:
114 grams
Explanation:
3chlorines per compound*38grams=114
Platinum can be found in which of the following formations
A.Sedimentary Bands
B. Aggregates
C. Chromite Bands
D. Hydrothermal Veins
Answers
Answer:
Explanation:
D
If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of the reaction?
Answers
Answer:
The correct answer is - 0.570 grams
Explanation:
The balanced chemical reaction is given by
Cu(NO3)2(aq) + 2NaOH(aq) --------> Cu(OH)2(s) + 2NaNO3(aq)
1.0 mole 2.0 mole 1.0 mole 2.0 mole
number of mol of Cu(OH)2,
n = Molarity * Volume
= [tex]35.0*0.167 = 5.845[/tex] millimoles
As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2
Mass of Cu(OH)2 = number of mol * molar mass
= [tex]97.5*5.845*10^-3[/tex]
= 0.570 grams
Thus, the correct answer is - 0.570 grams
0.5moles of monoalkanoic acid weighs 44g. determine the molecular formular
Answers
Answer:
C4H8O2
Explanation:
Butanoic acid
C4H8O2
In the naturally occurring D series of ketoses, the carbonyl group is found on carbon number ___________, whereas in aldoses, the carbonyl group is found on carbon number _________.
Answers
Answer:
second carbon atom from the end
end carbon atom
Explanation:
Carbohydrates are naturally occurring organic compounds containing carbon, hydrogen and oxygen. The general molecular formula of Carbohydrates is [tex]C_x(H_2O)_y[/tex].
Carbohydrates can be classified based on structures,
Carbohydrates with the structure of alkanals (-CHO) are known as aldose while those of the structure of alkanones (C=O) are known as ketose.
In stereochemistry , D series is a kind of configurational arrangement where the hydroxyl group attaches itself to the right hand side.
Thus; in naturally occurring D series of ketoses, the carbonyl group is found on carbon number second carbon atom from the end whereas in aldoses, the carbonyl group is found on carbon number end carbon atom.
Select the correct classification for the reaction.
4NH3(g) + 502(g) – 4NO(g) + 6H20(9)
Answers
ΔGrxn = 958 kJ
(answer on Edge)
To solve such this we must know the concept of redox reaction. Redox reaction is the one in which oxidation and reduction takes place simultaneously. The classification for the given reaction is redox reaction.
What is chemical reaction?
Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.
The balanced equation is
4NH[tex]_3[/tex](g) + 50[tex]_2[/tex](g) – 4NO(g) + 6H[tex]_2[/tex]0(9)
In the given reaction, Nitrogen in NH[tex]_3[/tex] is getting reduced, Oxygen in 0[tex]_2[/tex] is getting oxidized.
Therefore, the classification for the given reaction is redox reaction.
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Calculate the energy of a photon of electromagnetic radiation at each of the following wavelengths.
a. 632.8 nm (wavelength of red lightfrom helium-neon laser)
b. 503 nm (wavelength of maximumsolar radiation)
c. 0.0520 nm (a wavelength contained inmedical X-rays)
Answers
Answer:
A)566×10^-16J
B)711.99×10^-16J
C)688.69×10^-12J
Explanation:
A)
But we know that
1 nm = 10^-9 m
Then we can Convert from nanometer to metre which is SI unit
632.8 nm = 632.8×10^-9 m = 6.328×10^-7 m
The speed of light is c = 2.998×10^8 m/s
λ = 6.328×10^-7 m
But the frequency can be calculated as;
ƒ = c / λ
ƒ = (2.998×10^8 m/s) / (6.328×10^-7 m)
Then
ƒ = 4.738×10^14 s-1
To calculate Energy we use
Energy= hf
Where h is plank constant= 6.626× 10^-34
Energy= 6.626× 10^-34 × 4.738×10^14=566×10^-16J
B)
But we know that
1 nm = 10^-9 m
Then we can Convert from nanometer to metre which is SI unit
Then 503 nm = 503×10^-9 m = 5.03×10^-7 m
c = 2.998×10^8 m/s
λ = 5.03×10^-7 m
But the frequency can be calculated as;
ƒ = c / λ
ƒ = (2.998×10^8 m/s) / (5.03×10^-7 m)
ƒ = 5.960×10^14 s-1
Energy= 6.626× 10^-34 × 5.960×10^14 s-1= 711.99×10^-16J
C)
1 nm = 10^-9 m
0.0520 nm = 0.0520×10^-9 m = 5.20×10^-11 m
Where the speed of light is
c = 2.998×10^8 m/s
λ = 5.20×10^-11 m
But the frequency can be calculated as;
ƒ = c / λ
ƒ = (2.998×10^8 m/s) / (5.20×10^-11 m)
ƒ = 5.765×10^18 s-1
Energy= 6.626× 10^-34 × 5.765×10^18 s-1= 688.69×10^-12J
2. A student has a centrifuge tube containing 14.0 g of t-butanol and is asked to make a 1.2 m solution of ethanol/t-butanol. How much ethanol would the student need to add in mL and in g? Show your calculations. Show your calculations. (6 pts)
Answers
Answer:
0.774g of ethanol
0.970mL of ethanol
Explanation:
Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.
In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).
14.0g of butanol are 0.014kg and as you want to prepare the 1.2m solution, you need to add:
0.014kg × (1.2moles / kg) = 0.0168 moles of solute = Moles of ethanol
To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:
0.0168moles Ethanol ₓ (46.07g / mol) =
0.774g of ethanolAnd to convert mass in g to mL you require density of the substance (Density of ethanol = 0.798g/mL):
0.774g ₓ (1mL / 0.798g) =
0.970mL of ehtanolWhen aqueous solutions of NH4OH(aq) and CuCl2(aq) are mixed, the products are NH4Cl(aq) and Cu(OH)2(s). What is the net ionic equation for this reaction
Answers
Answer:
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
Explanation:
The net ionic equation usually shows the main ionic reaction that goes in the system. The other ions that do not participate in this net ionic equation are called spectator ions. Spectator ions do not participate in the main reaction occurring in the system.
The net ionic equation quite often result in the formation of a solid precipitate in the system such as Cu(OH)2.
The net ionic equation for this reaction is;
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
For a given sample of C3H8(g), the enthalpy change during the reaction is -784kJ. How many grams of CO2 are produced? Group of answer choices g
Answers
Answer:
[tex]m_{CO_2}=46.6gCO_2[/tex]
Explanation:
Hello,
In this case, considering the combustion of propane:
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)\ \ \ \Delta _CH=-2220.0 kJ/mol[/tex]
We can compute the burnt moles of propane as shown below:
[tex]n=\frac{-784kJ}{-2220.0 kJ/mol} =0.353molC_3H_8[/tex]
Then, by noticing propane and carbon dioxide are in a 1:3 molar ratio, we can compute the grams carbon dioxide by using the shown below stoichiometric procedure:
[tex]m_{CO_2}=0.353molC_3H_8*\frac{3molCO_2}{1molC_3H_8} *\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=46.6gCO_2[/tex]
Best regards.
the reaction between acid and metal can be considered as
Answers
Answer:
A single displacement reaction.
Explanation:
Acid + Metal => Salt + Hydrogen
Since, Acids usually have an H+ ion so the Anion part of the acid combines with the Metal (Which is also a cation). So, Anion part of acid is singly displaced by the Metal.
A sample containing 1.20 moles of Ne gas has an initial volume of 7.50 L. What is the final volume of the gas, in liters, when each of the following changes occur in the quantity of the gas at constant pressure and temperature?
1) a leak allows one half of the Ne atoms to escape
2) a sample of 3.10 moles of Ne is added to the 1.20 moles of Ne gas in the container
3) a sample of 35.0 g of Ne is added to the 1.20 moles of Ne gas in the container.
Answers
Answer:
1 = 3.75L
2 = 26.91L
3 = 18.37L
Explanation:
Hello,
The question above can be solved when we know Avogadro's law which states that the volume of a fixed mass of gas is directly proportional the number of moles present provided temperature and pressure are kept constant.
Mathematically,
N = kV k = n/v
N1/V1 = N2/V2 = N3/V3 =......=Nn/Vn
N1 = 1.20 moles
V1 = 7.50L
1) if half of the Ne atoms escaped what would be the final volume.
1 mole of Ne = 6.022×10²³ atoms
½(1.20) moles of Ne = ?
0.6 moles of Ne = ?
1 mole = 6.022×10²³ atom
1.2 moles = ?
x = 1.2 × 6.022×10²³ atoms
x = 7.23×10²³atoms
If ½ of 7.23×10²³ atoms escaped, how many would be left
½ × 7.23×10²³ atoms = 3.61×10²³atoms
Now we have to find the number of moles and then use our equation.
1 mole = 6.022×10²³ atoms
y mole = 3.61×10²³ atom
y = 0.6 mole
N2 = 0.6 mole
N1 / V1 = N2 / V2
Make V2 the subject of formula,
V2 = (N2 × V1) / N1
V2 = (0.6 × 7.50) / 1.20
V2 = 3.75L
The volume after half of the Ne atoms escaped is 3.75L
2)
When a sample of 3.10 mole is added to 1.20 moles present
N1 = 1.20
V1 = 7.51
N2 = (1.20 + 3.10) = 4.30L
V2 = ?
N1 / V1 = N2 / V2
V2 = (N2 × V1) / N1
V2 = (4.30 × 7.51) / 1.20
V2 = 26.91L
The volume of Ne gas if 3.10 moles is added to it is 26.91L
3)
A sample of 35g is added to the 1.20 mole Ne in the container.
We need to convert the mass (35g) to moles. This can be done using mass-molarmass relationships
Number of moles = mass / molar mass
Molar mass of Ne = 20.17g/mol
Number of moles = 35 / 20.17
Number of moles = 1.735 moles
N2 = 1.20 moles + 1.735 moles
N2 = 2.935 moles
N1 / V1 = N2 / V2
V2 = (N2 × V1) / N1
V2 = (2.935 × 7.51) / 1.20
V2 = 18.37L
On addition of 35g of Ne gas to the container, the volume is increased to 18.37L
(4Ga + 3S2 → 2Ga2S3)
1. How many moles of Sulfur are needed to react with 12.5 moles of Gallium
Answers
Answer:
9.375
Explanation:
According to the chemical equation, for every 4 moles of gallium, 3 moles of sodium are needed to react. Set up a ratio using this relationship to solve.
4/3 = 12.5/x
4x = 37.5
x = 9.375
You need 9.375 moles of sulfur.
A 2.0 g sample of hydrocarbon was burned in the calorimeter. The temperature rose from 29°c to 32°c and heat and combustion is 11. Kj/g. Thr heat capacity of the calorimeter is
Answers
Answer:
THE HEAT CAPACITY OF THE CALORIMETER IS 3666.67 J/C
Explanation:
Mass = 2 g
Temperature difference = 32 C - 29 C = 3 C
Heat of combustion = 11 kJ/g
Heat capacity of the calorimeter = unknown
It is important to note that the heat of combustion of the reaction is the heat absorbed by the calorimeter in raising the mixture by 3 C
So therefore,
Heat = heat capacity * temperature difference
Heat capacity = Heat / temperature difference
Heat capacoty = 11 000 J / 3 C
Heat capacity = 3666.67 J/ C
According to valence bond theory, which orbitals overlap in the formation of the bond in HCl?
a) 1s on H and 3p on Cl
b) 1s on H and 4s on Cl
c) 1s on H and 2p on Cl
d) 2s on H and 3p on Cl
e) 2s on H and 2p on Cl
Answers
In HCl, the H atom has only one valance electron. Each share an electron an therefore a single covalent bond is formed between the two. The bond in HCl is therefore a result of an overlap between 1s orbital and ONLY ONE of the lobes of the 3p orbital of Chlorine.
Which aqueous solution will have the highest boiling point temperature? A. 0.100 molal NiBr2(aq) B. 0.250 molal CH3OH(aq) C. 0.100 molal MgSO4(aq) D. 0.150 molal Na2SO4(aq) E. 0.150 molal NH4NO3(aq)
Answers
Answer: 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.
Explanation:
Formula used for Elevation in boiling point :
[tex]\Delta T_b=i\times k_b\times m[/tex]
where
[tex]\Delta T_b=T_b-T^o_b[/tex]= elevation in boiling point
[tex[k_b[/tex] = boiling point constant
m = molality
i = Van't Hoff factor
A) 0.100 m [tex]NiBr_2[/tex]
i = 3 as [tex]NiBr_2\rightarrrow Ni^{2+}+2Br^-[/tex]
concentration will be [tex]3\times 0.100=0.300[/tex]
B) 0.250 m [tex]CH_3OH[/tex]
i = 1 as [tex]CH_3OH[/tex] is a non electrolyte
concentration will be [tex]1\times 0.250=0.250[/tex]
C) 0.100 molal [tex]MgSO_4(aq)[/tex]
i = 2 as [tex]MgSO_4\rightarrrow Mg^{2+}+SO_4^{2-}[/tex]
concentration will be [tex]2\times 0.100=0.200[/tex]
D. 0.150 molal [tex]Na_2SO_4(aq)[/tex]
i = 3 as [tex]Na_2SO_4\rightarrrow 2Na^{+}+SO_4^{2-}[/tex]
concentration will be [tex]3\times 0.150=0.450[/tex]
E. 0.150 molal [tex]NH_4NO_3(aq)[/tex]
i = 2 as [tex]NH_4NO_3\rightarrrow NH_4^{+}+NO_3^{-}[/tex]
concentration will be [tex]2\times 0.150=0.300[/tex]
The solution having the highest concentration of ions will have the highest boiling point and thus 0.150 m [tex]Na_2SO_4(aq)[/tex] will have highest boiling point.
The aqueous solution that would have the highest temperature at boiling point would be:
D). 0.150 molal Na2SO4(aq)
What is a boiling point?The boiling point is described as the temperature at which the solution starts boiling or the vapor pressure becomes equivalent to the provided external/outer pressure.
To determine the elevation in boiling point, we will use:
Δ[tex]T_{b}[/tex] [tex]= i[/tex] × [tex]k_{b}[/tex] × [tex]m[/tex]
with
[tex]T_{b}[/tex] [tex]= T_{b} - T^{0}_{b}[/tex]
[tex]k_b[/tex] [tex]=[/tex] constant of boiling point
Using this formula,
0.150 molal Na2SO4(aq)
Given,
[tex]i = 3[/tex]
[tex]Na2So4[/tex] will have
[tex]2Na^{+}[/tex] [tex]+[/tex] [tex]SO^{2-}_{4}[/tex]
So,
Concentration [tex]= 3[/tex] × [tex]0.15[/tex][tex]0[/tex]
[tex]= 0.45[/tex][tex]0[/tex]
∵ 0.150 molal [tex]Na2SO4[/tex]Na2SO4(aq) has the maximum concentration.
Thus, option D is the correct answer.
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Describe what happens when two substances at different temperatures cine into contact. Describe how the law of conservation of energy applies to this system
Answers
Answer:
The substance with the highest heat gives heat to the lowest temperature, equating both temperatures,
In this situation there is talk of giving up heat but not matter, it is here that the law of conservation of energy comes into play.
Explanation:
The law of conservation of energy talks about that energy is transformed and never lost between two substances or two bodies that interact with each other, these interactions can be heat exchanges, as in this example.
Calculate the amount of heat required to convert 10.0 grams of ice at –20.°C to steam at 120.°C. (Sp. heat of H2O(s) = 2.09 J/g•°C, Sp. heat of H2O(l) = 4.18 J/g•°C, Sp. heat of H2O(gas) = 2.03 J/g•°C; heat of fusion of H2O(solid) = 333 J/g, heat of vaporization of H2O(liquid) = 2260 J/g).
Answers
Answer:
THE AMOUNT OF HEAT REQUIRED TO CONVERT ICE FROM -20 C TO STEAM AT 120 C IS 30 946 J OR 30.946 KJ OF HEAT.
Explanation:
Mass = 10 g
To convert 10 g of ice at -20°C to steam at 120°C, the heat involved is:
1. Heat involved in converting the ice from -20 °c to ice at 0 °C:
Heat = mass * specific heat of water solid * change in temperature
heat = 10g * 2.09 J/g°C * ( 0- (-20))
Heat = 10 * 2.09 * 20
heat = 418 J
2. Heat required to convert the ice from 0°C to water at 0°C:
Heat = mass * specific heat of fusion of water solid
Heat = 10 * 333
Heat = 3330 J
3. Heat required to convert water at 0 C to water at 100 C:
Heat = mass * specific heat of water * change in teperature
Heat = 10 * 4.18 * (100 -0)
Heat = 4180 J
4. Heat required to convert water at 100 C to steam at 100 C:
Heat = mass * specific heat of vaporization
Heat = 10 * 2260
Heat = 22600 J
5. Heat required to convert steam from 100 C to steam at 120 C:
Heat = mass * specific heat of water * change in temperature
Heat = 10 * 2.09 * (120 -100)
Heat = 10 * 2.09 * 20
Heat = 418 J
T
he heat required to convert 10 g of ice at -20 C to steam at 120 C is therefore the total of the individual heat of reactions
Total amount of heat = ( 418 J + 3330 J + 4180 J + 22600 J + 418 J)
Total heat = 30946 J
Which of the following aqueous solutions are good buffer systems? . a. 0.12 M calcium hydroxide + 0.29 M calcium bromide . b. 0.25 M perchloric acid + 0.16 M sodium perchlorate . c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide .
Answers
Answer:
c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide .
Explanation:
A buffer is defined as the aqueous mixture of a weak acid with its conjugate base or vice versa. Based on the systems:
a. 0.12 M calcium hydroxide + 0.29 M calcium bromide. IS NOT A GOOD BUFFER SYSTEM because Ca(OH)₂ is a strong base.
b. 0.25 M perchloric acid + 0.16 M sodium perchlorate. IS NOT A GOOD BUFFER SYSTEM because perchloric acid is a strong acid
c. 0.34 M hydrocyanic acid + 0.27 M sodium cyanide. IS A GOOD BUFFER SYSTEM because HCN is a weak acid, and its conjugate base, CN⁻, is obtained in the dissolution of NaCN as Na⁺ and CN⁻ ions.
A 32.3-gram sample of gas is found to have a volume of 1.9 liters at 301 K and 1.21 atm. What is the molar mass of this gas? Show all of the work used to solve this problem.
Answers
Answer:
351.1g/mol
Explanation:
you can find the answer using The ideal gas equation
n= PV/RT
n=(1.21*1.9/0.082*301)mol
n=0.092 mol
molar mass=Mass/mole
m=32.3g/0.092mol
m=351.1g/mol
To distinguish smokers from nonsmokers, an assay for the determination of 2,5-dimethylfuran concentration in whole blood is performed. The concentration of 2,5-dimethylfuran in blood is directly proportional to the number of cigarettes smoked per day. Individuals who have a 2,5-dimethylfuran concentration above 0.024 parts per billion (ppb), or 24 parts per trillion (ppt), are likely to identify themselves as smokers. Draw the compound 2,5-dimethylfuran.
Answers
Answer:
See explanation
Explanation:
When we have to write an organic molecule, we have to remember the IUPAC rules. In this case, we have: "2,5-dimethylfuran". We can divide the name into two parts:
a) "2,5-dimethyl"
b) "Furan"
For part a) we have the indications of the "subgroups" bonded to the main structure, in this case, we have 2 methyl groups ([tex]CH_3[/tex]) bonded to carbons 2 and 5.
For part b) we have the main, structure. In this case furan, this compound is a cyclic structure of 5 members in which we have in the middle of the molecule an oxygen atom and 2 double bonds (see figure 1).
Finally, when we add the methyl groups to the furan we will get the 2,5-dimethylfuran. (see figure 2).
I hope it helps!
If the reaction consumes methane gas ( CH4 ) at a rate of 2.08 M/s, what is the rate of formation of H2 ? the balanced equation is CH4 + N2Cl4 = CCl4 + N2 + 2 H2
Answers
Answer:
4.16M/s
Explanation:
Based on the reaction:
CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂
1 mole of methane, CH₄, produce 2 moles of H₂.
That means whereas 1 mole of methane is consumed, 2 moles of H₂ are formed
Having this in mind, if you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.
Thus, rate of formation of H₂ is:
2.08M/s ₓ 2 =
4.16M/s
The rate of formation of H2 is 4.16M/s
The calculation is as follows:Based on the reaction:
CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂
here
1 mole of methane, CH₄, produce 2 moles of H₂.
In the case when you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.
Thus, rate of formation of H₂ is:
2.08M/s ( 2) = 4.16M/s
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