Answer:
it should be N/m or newton per meter.
Consider the following equilibrium:
2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.
What is Kp?
Answer:
Consider the following equilibrium:
2H2(g)+S2(g)⇌2H2S(g)Kc=1.08×107 at 700 ∘C.
What is Kp?
Explanation:
Given,
[tex]Kc=1.08 * 10^7[/tex]
The relation between Kp and Kc is:
[tex]Kp=Kc * (RT)^d^e^l^t^a^(^n^)[/tex]
Where delta n represents the change in the number of moles.
For the given equation,
The Delta n = Number of moles of products - number of moles of reactants
(2-(2+1))
=-1.
Hence,
Kp=Kc/RT.
Thus,
[tex]Kp=1.08 * 10^7 / 8.314 J.K6-1.mol^-^1 x 973 K\\Kp=1335.06[/tex]
The answer is Kp=1335.06
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
Explanation:
The relation between [tex]K_p \& K_c[/tex] is given by:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
Where:
[tex]K_c[/tex] = The equilibrium constant of reaction in terms of concentration
[tex]K_p[/tex] = The equilibrium constant of reaction in terms of partial pressure
R= The universal gas constant
T = The temperature of the equilibrium
[tex]n_g[/tex]= Change in gaseus moles
Given:
An equilibrium reaction, 700°C:
[tex]2H_2(g)+S_2(g)\rightleftharpoons 2H_2S(g),K_c=1.08\times 10^7[/tex]
To find:
The equilibrium constant in terms of partial pressure, [tex]K_p[/tex].
Solution:
The equilibrium constant of reaction in terms of concentration= [tex]K_c[/tex]
[tex]K_c=1.08\times 10^7[/tex]
The equilibrium constant of reaction in terms of partial pressure =[tex]K_p=?[/tex]
The gaseous moles of reactant side = [tex]n_r= 3[/tex]
The gaseous moles of product side = [tex]n_p= 2[/tex]
The temperature at which equilibrium is given = T
[tex]T = 700^oC+273.15 K=973.15K[/tex]
The change in gaseous mole = [tex]n_g=n_p-n_r=2-3 = -1[/tex]
[tex]K_p=1.08\times 10^7\times (0.0821 atm L/mol K\times 973.15 K)^{-1}\\K_p=1.35\times 10^5[/tex]
The value of [tex]K_p[/tex] is [tex]1.35\times 10^5[/tex].
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Identify a process that is NOT reversible. Identify a process that is NOT reversible. melting of snow baking of bread deposition of carbon dioxide freezing water melting of aluminum
Answer:
Identify a process that is NOT reversible.
Melting of snow
baking of bread
deposition of carbon dioxide
freezing water
melting of aluminum
Explanation:
A physical change is the one in which there is a change only in its physical state, color, the appearance of the substance. But the chemical composition of the substance remains unchanged.
It is a temporary change and can be reversed easily.
For example:
melting, freezing, deposition etc.
Baking is a permanent change and the chemical composition of the substance changes.
Hence, among the given options, baking of bread is not a reversible change.
Out of the following all are physical changes except baking of bread and physical changes are reversible so the process which is not reversible is baking of bread.
What are physical changes?Physical changes are defined as changes which affect only the form of a substance but not it's chemical composition. They are used to separate mixtures in to chemical components but cannot be used to separate compounds to simpler compounds.
Physical changes are always reversible using physical means and involve a change in the physical properties.Examples of physical changes include melting,boiling , change in texture, size,color,volume and density.Magnetism, crystallization, formation of alloys are all reversible and hence physical changes.
They involve only rearrangement of atoms and are often characterized to be changes which are reversible.
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Where do reactions in a solid
generally take place?
A. All throughout the solid
B. At the center of the solid.
C. Only on opposite sides of the solid
due to repelling forces.
D. On the surface of the solid.
Answer:
A
Explanation:
because solid stick together like magnetic and can feel every vibration
Answer:D
Explanation: on the surface of a solid
Methyl orange can change color by transitioning from one chromophore to another. When added to a clear solution and the solution turns red, it is determined to be a(n) __________ in its __________ stable form.
Answer:
acidic titration in its stable form
Explanation:
Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.
Which is the primary type of radiation from the sun that is absorbed by the ozone layer?
A. infrared radiatin
B. UV-B
C. X-rays
D. UV-C
E. UV-A
the answer to the question is B.UV-B
can anybody pls help me with chemistry how to balance an equation of class 7 ??? pls
really I don't like chemistry..
I need only the step(s)..
Answer:
[tex] This\:may\: help[/tex]
Answer:
(1) Write down the chemical reaction in the form of word equation,keeping reactants on left hand side and products on right hand side.
(2) Write symbol and formula of all reactants and products in word equation. (3) Balance the equation by multiplying the symbols and formula by smallest possible figures.
when a polar bond is formed between 2 atoms which atom receives a partial positive charge
Answer:
The more electronegative atom in a covalent bond
One of the purposes of this lab is to determine the order of the reaction with respect to the Allura Red dye by creating first and second-order graphs for all four trials. The correct order of the reaction is the one where the slopes of the graphs for the four trials are roughly the same. Why is this important when choosing the order of the reaction
Answer: Hello the options related to your question are attached below
The slope is related to the rate constant so all four trials should have the same slope since the reactions are all the same ( Option C )
Explanation:
It is important when choosing the order of the reaction because the concentration of the bleaches used in the four trials are in excess hence their slopes have to be roughly the same and also because the reactions are similar and they where done at the same temperature, hence the slope of the first and second-order graphs will be the same.
Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O). Suppose 5.7 g of hydrobromic acid is mixed with 0.980 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.
Answer:
The maximum mass of water that could be produced by the chemical reaction=0.441g
Explanation:
We are given that
Given mass of HBr=5.7 g
Given mass of sodium hydroxide=0.980 g
Molar mass of HBr=80.9 g/ Mole
Molar mass of NaOH=40 g/mole
Molar mass of H2O=18 g/mole
Reaction
[tex]HBr+NaOH\rightarrow H_2O+NaBr[/tex]
Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]
Using the formula
Number of moles of HBr=[tex]\frac{5.7}{80.9}=0.0705 moles[/tex]
Number of moles of NaOH=[tex]\frac{0.980}{40}=0.0245moles[/tex]
Hydrogen bromide is in a great excess and the amount of water produced.
Therefore,
Number of moles of water, n(H2O)=Number of moles of NaOH=0.0245moles
Now,
Mass of water=[tex]n(H_2O)\times Molar\;mass\;of\;water[/tex]
Mass of water=[tex]0.0245moles\times 18=0.441g[/tex]
Hence, the maximum mass of water that could be produced by the chemical reaction=0.441g
Which Group is in the second column of the periodic table?
O A. Noble gases
O B. Alkaline earth metals
O C. Alkali metals
O D. Halogens
Answer:
O B. Alkaline earth metals
Explanation:
Noble gases → 8th column.
Alkali metal → first column.
Halogen → 7th
Answer:
B. Alkaline earth metals
Explanation:
Alkaline-earth metals: The alkaline-earth metals make up Group 2 of the periodic table, from beryllium (Be) through radium (Ra). Each of these elements has two electrons in its outermost energy level, which makes the alkaline earths reactive enough that they're rarely found alone in nature. But they're not as reactive as the alkali metals. Their chemical reactions typically occur more slowly and produce less heat compared to the alkali metals.
If a hydrogen atom and a helium atom have the same kinetic energy:________
a. the wavelength of the hydrogen atom will be about 4 times longer than the wavelength of the helium atom.
b. the wavelength of the hydrogen atom will be about 2 times longer than the wavelength of the helium.
c. the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
d. the wavelength of the helium atom will be about 2 times longer than the wavelength of the hydrogen atom.
e. the wavelength of the helium atom will be about 4 times longer than the wavelength of the hydrogen atom.
Answer: If a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
Explanation:
The relation between energy and wavelength is as follows.
[tex]E = \frac{hc}{\lambda}\\[/tex]
This means that energy is inversely proportional to wavelength.
As it is given that energy of a hydrogen atom and a helium atom is same.
Let us assume that [tex]E_{hydrogen} = E_{helium} = E'[/tex]. Hence, relation between their wavelengths will be calculated as follows.
[tex]E_{hydrogen} = \frac{hc}{\lambda_{hydrogen}}[/tex] ... (1)
[tex]E_{helium} = \frac{hc}{\lambda_{helium}}[/tex] ... (2)
Equating the equations (1) and (2) as follows.
[tex]E_{hydrogen} = E_{helium} = E'\\\frac{hc}{\lambda_{hydrogen}} = \frac{hc}{\lambda_{helium}} = E'\\\lambda_{helium} = \lambda_{hydrogen} = E'[/tex]
Thus, we can conclude that if a hydrogen atom and a helium atom have the same kinetic energy then the wavelength of the hydrogen atom will be roughly equal to the wavelength of the helium atom.
a. Draw 2,3-dichloro octane.
b. Write the lewis structure for H20 molecule.
Answer:
a.draw 2,3 dicholoro octane
Explanation:
mag isip ka kung paano hehe
1. Show that heat flows spontaneously from high temperature to low temperature in any isolated system (hint: use entropy change that occurs during the process for your proof).
2. Work out the entropy change for the decomposition of mercuric oxide using mathematical and graphical arguments.
Answer:
1 ) Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
2) ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Explanation:
1) To show that heat flows spontaneously from high temperature to low temperature
example :
Pick two(2) solid metal blocks with varying temperatures ( i.e. one solid block is hot and the other solid block is cold )
Place both blocks for time (t ) in an insulated system to reduce heat loss or gain to or from the environment
Check the temperature of both blocks after time ( t ) it will be observed that both blocks will have same temperature after time t ( first law of thermodynamics )
Δs ( entropy change for hot block ) = - Q / th ( -ve shows heat lost to cold block )
Δs ( entropy change for cold block ) = Q / tc
∴ Total Δs = ΔSc + ΔSh
= Q/tc - Q/th
2) Entropy change for Decomposition of mercuric oxide
2HgO (s) → 2Hg(l) + O₂ (g)
Δs = positive
there is transition from solid to liquid and the melting point of mercury ( the point at which reaction will take place ) = 500⁰C
hence ΔSdecomposition = S⁻ Hg - S⁻ HgO =
Δh of reaction = 181.6 KJ
Temp = 500 + 273 = 773 k
hence ΔSdecomposition = Δh / Temp = ( 181.6 * 10^3 / 773 ) = 234.928 J/k
Balance the chemical equation below using the smallest possible whole number stoichiometric coefficients.
P4(s) + NaOH(aq) + H2O(l) -> Ph3(g) + Na2HPO3(aq)
Answer:
I used a,b c, d in the equation as substituted coefficients to find the unknown for each element of P, Na, O, H, and I got
P4(s) + 4NaOH(aq) + 2H20(l)---->2Ph3 +2Na2HPO3(aq).
which I think should be the answer.
Choose the correct statement. A) The cathode is the electrode where the oxidation takes place. B) The cathode is the electrode where the reduction takes place. C) Both oxidation and reduction may take place at the cathode, depending on the cell. D) The cathode is always positive
determine the budget for tge fitness event.consider the attendance (should tou ask for registration fee.give shirts etc)
Silver has two naturally occurring isotopes with the following isotopic masses: 10747Ag 10947Ag 106.90509 108.9047 The average atomic mass of silver is 107.8682 amu. The fractional abundance of the lighter of the two isotopes is ________.
Answer: The fractional abundance of lighter isotope is 0.518
Explanation:
Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex] ......(1)
Let the fractional abundance of Ag-107 isotope be 'x'
For Ag-107 isotope:Atomic mass = 106.90509 amu
Fractional abundance = x
For Ag-109 isotope:Atomic mass = 108.9047 amu
Fractional abundance = (1 - x)
Average atomic mass of silver = 107.8682 amu
Plugging values in equation 1:
[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]
Fractional abundance of Ag-107 isotope (lighter) = x = 0.518
Hence, the fractional abundance of lighter isotope is 0.518
1. What is the equivalent pressure measurement in mmHg of 2.50 atm?
Answer:
Atmosphere to mmHg Conversion Example. Task: Convert 8 atmospheres to mmHg (show work) Formula: atm x 760 = mmHg Calculations: 8 atm x 760 = 6,080 mmHg Result: 8 atm is equal to 6,080 mmHg.
Explanation:
This answer is helpfull for you I now20.1 cm3 of metal hydroxide (MOH) containing 4.8 g/dm3 react Completely with 23.0 cm3 of the HCL and Containing 3.8g/dm3. Identify the metal of M of MOH
Answer:
A metal M readily forms water-soluble sulphate MSO4, water-insoluble hydroxide M(OH)2 and oxide MO. The oxide and hydroxide are soluble in NaOH. The M is:
why might the melting point of the crystals obtained in this experiment be close to but below one of the reference melting points and melt slowly over several degrees
Answer: hello the experiment related to your question is missing but I will provide a more general answer within the scope of your question
answer :
presence of Impurities
Explanation:
The melting point of the crystals as obtained in the experiment will be close to but below reference melting points and will also melt slower because of the presence of impurities in the compound
Impurities alter the melting and freezing points from ideal freezing and melting points of compounds
Zn-64 = 48.63%
Zn-66 = 27.90%
Zn-67 = 4.10%
Zn-68 = 18.75%
Zn-70 = .62%
Calculate the average atomic mass/given their percent abundance
Answer:
A = 65.46 u
Explanation:
Given that,
The composition of zinc is as follows :
Zn-64 = 48.63%
Zn-66 = 27.90%
Zn-67 = 4.10%
Zn-68 = 18.75%
Zn-70 = .62%
We need to find the average atomic mass of the given element. It can be solved as follows :
[tex]A=\dfrac{48.63\times 64+27.90\times 66+4.1\times 67+18.75\times 68+0.62\times 70}{100}\\A=65.46\ u[/tex]
So, the average atomic mass of zinc is 65.46 u.
Combustion of 29.26 gg of a compound containing only carbon, hydrogen, and oxygen produces 33.86 gCO2gCO2 and 13.86 gH2OgH2O. Part A What is the empirical formula of the compound
Answer:
C2H4O3
Explanation:
Empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.
To solve this question we need to find the moles of carbon = Moles of CO2, the moles of hydrogen (Using moles of H2O) and the moles of oxygen (Finding the mass of the mass of each atom) as follows:
Moles Carbon -Molar mass CO2: 44.01g/mol-:
33.86g CO2 * (1mol/44.01g) = 0.769 moles CO2 = Moles C * (12g/mol) =
9.23g C
Moles Hydrogen -Molar mass H2O: 18.01g/mol-
13.86g H2O * (1mol/18.01g) = 0.770 moles H2O * (2mol H / 1mol H2O) = 1.54 moles H * (1g/mol) = 1.54g H
Moles Oxygen:
Mass: 29.26g - 9.23g C - 1.54g H = 18.49g O * (1mol/16g) = 1.156 moles O
Dividing each number of moles in the moles of C (Lowe number of moles):
C = 0.769 moles C / 0.769 moles C = 1
H = 1.54 moles H / 0.769 moles C = 2
O = 1.156 moles O / 0.769 moles C = 1.5
As the number must be a whole number each ratio twice:
C = 2
H = 4
O = 3
Empirical formula is:
C2H4O36ooo kg into quintal
Answer:
60 q
Explanation:
The conversion factor is 100; so 1 quintal = 100 kilograms. In other words, the value in q multiply by 100 to get a value in kg.
Which of the following was NOT explained by Dalton's atomic theory?
ANSWER:
A. the Law of Multiple Proportions
B. the difference between elements and compounds
C.?the difference between isotopes of an element
D. the Law of Conservation of Mass
Answer:
A. the Law of Multiple Properties
Answer:
A. the law of multiple proportions
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation
MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)
How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL Cl2(g) at 25 °C and 745 Torr ?
mass of MnO2:
Answer:
0.605 g
Explanation:
MnO₂(s) + 4HCl(aq) ⟶ MnCl₂(aq) + 2H₂O(l) + Cl₂(g)First we calculate how many Cl₂ moles need to be produced, using the PV=nRT formula:
P = 745 Torr ⇒ 745 / 760 = 0.980 atmV = 185 mL ⇒ 185 / 1000 = 0.185 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 25 °C ⇒ 25 + 273.16 = 298.16 KInputting the data:
0.980 atm * 0.185 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 Kn = 0.00696 molThen we convert 0.00696 moles of Cl₂ to MnO₂ moles:
0.00696 mol Cl₂ * [tex]\frac{1molMnO_2}{1molCl_2}[/tex] = 0.00696 mol MnO₂Finally we convert 0.00696 moles of MnO₂ to grams, using its molar mass:
0.00696 mol MnO₂ * 86.94 g/mol = 0.605 gCalculate the displacement (the total volume of the cylinder through which the piston move) of a 5.70L automobile engine in cubic inches, (1inch=2.54cm)
Answer:
348 inches³
Explanation:
From our previous knowledge of units conversion:
We know that 1000 cm³ makes 1 Liter.
Thus, for a 5.70 L automobile engine in cubic meters will be:
= 5.70 × 1000 cm³
= 5700 cm³
Now, the displacement of the automobile in cubic inches provided that 1 inch = 2.534 cm is:
⇒ 5700× (1/ (2.54)³) in³
= 5700×0.0610 in³
= 347.7 in³
≅ 348 inches³
Question 16(Multiple Choice Worth 5 points)
(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?
O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.
Answer:
It is equal to the total mass of the products.
Explanation:
Hope this helps :)
Acetoacetyl-ACP is formed as an intermediate during fatty acid biosynthesis. The CO2 used to synthesize malonyl-CoA is lost. Would this help make the reaction more or less energetically favorable
Answer:
This would make the reaction more energetically favourable
Explanation:
Fatty acid synthesis starts with the carboxylation of acetyl CoA to malonyl CoA. This irreversible reaction is the committed step in fatty acid synthesis. The synthesis of malonyl CoA is catalyzed by acetyl CoA carboxylase, which contains a biotin prosthetic group.
What would be the specific mathematical effect on the reaction rate if you carried out the sodium iodide-in-acetone reactions on the alkyl halides using an iodide solution half as concentrated? ("Slower" or "faster" is not specific enough.)
Answer:
Slower
Explanation:
The reaction between alkyl halides and sodium iodide-in-acetone is an SN2 reaction. The rate of reaction depends on the concentration of the alkyl halide as well as the concentration of the sodium iodide. It is a bimolecular reaction.
This means that if the concentration of any of the reactants is halved, the rate of reaction decreases accordingly.
Therefore, if the iodide solution is half as concentrated, the reaction is observed to be slower in accordance with the rate law;
Rate = k[alkyl halide] [iodide]
The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?
Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
Explanation:
Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]
Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g
Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.
[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]
Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
Explanation:
Given:
Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]
To find:
The number of atoms of tantalum in 37.1 milligrams.
Solution:
Mass of tantalum = 37.1 mg
[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]
The number of atoms in 0.0371 grams of tantalum = N
Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]
Then a mass of N atoms of tantalum will be:
[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
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