The answer would be 4.
The agencies are:
Department of Agriculture's Forest Service (AFS)
Department of the Interior's Bureau of Land Management (BLM)
Fish and Wildlife Service (FWS)
National Park Service (NPS)
These four agencies cover around 95% of the land and are the only federal agencies.
I learned about these agencies in school.
Which is true of protons and neutrons?
1. They have approximately the same mass and the same charge.
2) They have approximately the same mass but different charge.
by The have different mass and different charge.
O sette
4) They have different mass but the same charge.
Answer:
[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
Determine the molarity for each of the following Solution: 98.0 of phosphoric acids H3PO4 in 1.00L of Solution.
The molarity of the solution is 0.01.
Brainliest?
plate tectonics-1.gif
What is the above image a representation of?
Answer:
Subduction Process where the oceanic plate subducted under the continental plates because it denser than the Continental plate.
Dissolution of KOH, ΔHsoln:
KOH(s) → KOH(aq) (10.1)
Neutralization of solid KOH, ΔHneut:
KOH(s) + HCl(aq) → H2O(l) + KCl(aq) (10.2)
1) Using Hess's law, show how to combine Reaction 10.1 and Reaction 10.2 to give
KOH(aq) + HCl(aq) → H2O(l) + KCl(aq) (10.3)
2)How should ΔHsoln and ΔHneut be combined (mathematically) to give the change in enthalpy for Reaction 10.3, ΔH?
Using Hess's law we found:
1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)
2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).
The reactions of dissolution (10.1) and neutralization (10.2) are:
KOH(s) → KOH(aq) ΔHsoln (10.1)
KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)
1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.
Hence, to get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
We need to add reaction 10.2 to the reverse of reaction 10.1
KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)
Canceling the KOH(s) from both sides, we get reaction 10.3:
KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)
2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:
[tex] \Delta H = \Delta H_{soln} + \Delta H_{neut} [/tex]
The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:
[tex] \Delta H = \Delta H_{neut} - \Delta H_{soln} [/tex]
Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.
Learn more here:
https://brainly.com/question/2082986?referrer=searchResultshttps://brainly.com/question/1657608?referrer=searchResultsI hope it helps you!
A 0.48-mole sample of helium gas occupies a volume of 11.7 L . What is the volume of 0.72 mol of helium gas under the same conditions
AnswerExplanation:I finsished d
Answer:
17.55L
Explanation:
0.48mol : 11.7
0.72mol : x
0.48x = 8.424
x = 17.55
What question would a student need to ask to form a compound with Group 16 nonmetals
A.
Will group 16 elements lose electrons to bond with group 2 in an XY format?
B.
Will group 16 elements gain electrons to bond with group 1 in an XY2 format?
C.
Will group 16 elements gain electrons to bond with group 2 in an XY format?
D.
Will group 16 elements lose electrons to bond with group 1 in an X2Y format?
Since nonmetals gain electrons, the correct question to ask about group 16 elements is; "Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Group 16 elements are divalent and they form divalent negative ions. The periodic table is arranged in groups and periods. The elements in the same group have the same number of valence electrons. All elements in group 2 have six valence electrons.
If a wants to form a compound with the non metals of group 16, the correct question to ask is;"Will group 16 elements gain electrons to bond with group 2 in an XY format?"
Learn more: https://brainly.com/question/14281129
Question 2
Which of the following models the arrangement of atoms in a gas?
O None of these
PLEASE HELP ASAP
Answer:
The option where the atoms are furthest apart.
Explanation:
Gasses are the most energetic of the three basic states of matter, their atoms have more kinetic energy than either solids or liquids and will therefore also have the most spread out atoms.
if 2.4l of chlorine at 400 mm hg are compressed to 725 mm hg at a constant temperature. what is the new volume?
Answer: 1.324L
Explanation: use Boyles law, sorry so late!
answer this please thank you
Answer:
6am
Explanation:
answer this please thank you
The melting point of H₂O(s) is 0 °C. Would you expect the melting point of H₂S(s) to be 85 °C, 0 °C or -85 °C.? Justify your choice
Answer:
-85 °C
Explanation:
O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.
H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter
molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.
HELP HELP HELP ASAPPPP!
So the question is:
Relate the electron configuration pattern to the general trends in atomic properties in the periodic table including definitions
a. Atomic radii (size)
b. Electronegativity
c. Ionization energy
Answer:
b. Electronegativity
Explanation:
electronic configuration
Select the container from the figure (Figure 1) that represents the dilution of a 4 % (m/v) KCl solution to each of the following: Figure1 of 1 There is a diagram showing several containers. One container is filled with 4 percent of mass to volume solution of NaCl. Container 1 is filled with a solution in which volume is two times less than the volume of NaCl solution. Container 2 is filled with a solution of a volume two times larger compared to the NaCl solution. Container 3 is filled with a solution of a volume two times larger than the volume of the solution in container 2. Part A 2 % (m/v) KCl
From the dilution formula, we have that at constant value of the solute, the volume of the solution is inversely proportional to the concentration.
The correct responses are;
Part A: Container 2Part B: Container 3Reasons:
The given parameters are;
The concentration of the KCl solution = 4% m/v
Taking the solution as solution of KCl
The volume of the solution in container 1 = Two times less than the volume of KCl solution.
[tex]V_{container \, 1} = \displaystyle \mathbf{ \frac{1}{2} \cdot V_{4\% \, solution}}[/tex]
The volume of the solution in container 2 = Two times larger compared to the volume of KCl solution.
[tex]V_{container \, 2} = \mathbf{\displaystyle 2 \times V_{4\% \, solution}}[/tex]
The volume of the solution in container 3 = Two times larger than the container two solution volume.
[tex]V_{container \, 3} = \displaystyle \mathbf{ 2 \times V_{container \, 2}}[/tex]
Therefore;
[tex]V_{container \, 3} = \displaystyle 2 \times 2 \times V_{4\% \, solution} = \mathbf{4 \times V_{4\% \, solution }}[/tex]
Part A Required:
a. To select the container that represent the dilution of the 4% solution to 2%
Solution:
The dilution formula is; C₁·V₁ = C₂·V₂
Therefore;
[tex]\displaystyle V_1 = \mathbf{\frac{C_1 \cdot V_1}{C_2}}[/tex]
C₁ =4%, C₂ = 2%, we get;
[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{2} = 2 \cdot V_1[/tex]
The volume of the container that represents a 2% dilution is container 2
which is filled with a solution of a volume two times larger compared to the
KCl solution.
Part B:
Required:
The container diluted to a 1% m/v KCl solution.
Solution;
Using the dilution formula, we have;
C₁ = 4%, C₂ = 1%
Therefore;
[tex]\displaystyle V_1 = \frac{C_1 \cdot V_1}{C_2}[/tex]
[tex]\displaystyle V_1 = \frac{4 \cdot V_1}{1} = \mathbf{4 \cdot V_1}[/tex]
The volume of the solution is four times the volume of the 4% KCl solution, which is equivalent to the volume in container 3.
Possible parts of the question are;
Select the container that represents the dilution of the 4% (m/v) KCl solution to obtain the solutions that follows;
Part A: a 2% (m/v) KCl solution
Part B: a solution that is a 1% (m/v) KCl solution
Please see attached drawings
Learn more here:
https://brainly.com/question/11493179
Oxidation Unit Test
Use the galvanic cell reaction to answer the question.
2Cr(s) + 3Cu2+(aq) → 2Cr3+ (aq) + 3Cu(s)
Which half reaction occurs at the cathode?
(1 point)
The reduction half equation of the galvanic cell is is; 3Cu2+(aq) + 3e ------> 3Cu(s).
A galvanic cell is a cell that produces chemical energy by a spontaneous chemical reaction. The equation of the reaction is; 2Cr(s) + 3Cu2+(aq) → 2Cr3+ (aq) + 3Cu(s). We can see that chromium was oxidized and copper was reduced.
Since reduction occurs at the cathode, it follows that the reduction half equation is; 3Cu2+(aq) + 3e ------> 3Cu(s).
Learn more about galvanic cell; https://brainly.com/question/25606438
HELP ME OUT PLEASE!!!!
Which statement correctly describes one of the changes?
A) Picture I shows a chemical change, because a new substance is formed.
B) Picture Il shows a chemical change, because a new substance is formed.
C) Picture I shows a chemical change, because the same substance changes form,
D) Picture Il shows a chemical change, because the same substance changes form.
Answer:
Answer D. Picture II shows a chemical change, because the same substance changes form
Explanation:
This is the temperature that water molecules slow down enough to stick to each other and form a solid crystal
Answer:
D Im 90% Sure
Explanation:
If its not right i owe you one I did this one before
What does IUPAC stand for?
Answer: International Union of Pure and Applied Chemistry
The International Union of Pure and Applied Chemistry (IUPAC), established in 1919, is the international body that represents chemistry and related sciences and technologies.
Who wants to simp for me??
Answer:
qrtyuioplkjhgfdssssssazxcvbn
The octahedral complex ion [MnCl6] 3- has more unpaired spins than the octahedral complex ion [Mn(CN)6] 3- . How many unpaired electrons are present in each species
[MnCl6] 3- is high spin and has five unpaired electrons while [Mn(CN)6] 3- has only two unpaired electrons.
A complex may be low spin or high spin depending on the kind of ligand attached to the central metal atom/ion. If the ligand is a weak field ligand, the complex may be high spin (maximum number of unpaired electrons). If the complex is low spin, there are few unpaired electrons (minimum number of unpaired electrons). In that case, the ligand is a strong field ligand.
In the octahedral geometry, [MnCl6] 3- is high spin and has five unpaired electrons since the chloride ion is a weak field ligand. On the other hand [Mn(CN)6] 3- has only two unpaired electrons because the cyanide ion is a strong field ligand.
Learn more: https://brainly.com/question/6111443
PLS HELP THIS IS TO HARD PLS
Br2(l) + 2Nal(aq) — 12(s) + 2NaBr(aq)
Which elements are oxidized and reduced in the reaction?
(1 point)
O Sodium (Na) is oxidized, and bromine (Br) is reduced.
O Bromine (Br) is oxidized, and iodine (1) is reduced.
O Bromine (Br) is oxidized, and sodium (Na) is reduced.
Olodine (I) is oxidized, and bromine (Br) is reduced.
Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.
To determine which elements are oxidized and reduced,
First, we will define the terms Oxidation and Reduction
Oxidation is simply defined as the loss of electrons. It can also be defined as increase in oxidation number.
Reduction is defined as the gain of electrons. It can also be defined as decrease in oxidation number.
The given chemical equation is
Br₂(l) + 2NaI(aq) → I₂(s) + 2NaBr(aq)
Oxidation number of Bromine decreased from 0 to -1.
Therefore, Bromine is reduced.
Oxidation number of Iodine increased from -1 to 0.
Therefore, Iodine is oxidized.
Oxidation number of sodium did not change.
Therefore, Sodium is neither oxidized nor reduced.
Hence, Iodine (I) is oxidized, and bromine (Br) is reduced. The correct option is the last option - lodine (I) is oxidized, and bromine (Br) is reduced.
Learn more here: https://brainly.com/question/12913997
The pH of an acidic solution is 4.83. What is [H"]?
[tex]pH = -\log[H^{+}] \\\\\implies \log[H^{+}] = -pH\\\\\implies [H^{+}] = 10^{-pH}\\\\\implies [H^{+}] = 10^{-4.83} = 0.000015[/tex]
In experiment 9, in one operation, we heat up the alcohol with acid and do a concurrent distillation. What was the purpose of doing this
Answer:
we heat up because the component with lower boiling evaporates first,
leaving the other behind
Which spheres include the water cycle? (Select all that apply.)
biosphere
geosphere
hydrosphere
atmosphere
Answer:
A, B, C, D
Explanation:
The water cycle is included in the atmosphere (water vapor in the air), the geosphere (water collection), the hydrosphere (water on earth: lakes, oceans, rivers), and the biosphere (where water interacts with all living things).
In an ecosystem, the water cycle is included in all the spheres that is biosphere,geosphere,hydrosphere and atmosphere.
What is an ecosystem?Ecosystem is defined as a system which consists of all living organisms and the physical components with which the living beings interact. The abiotic and biotic components are linked to each other through nutrient cycles and flow of energy.
Energy enters the system through the process of photosynthesis .Animals play an important role in transfer of energy as they feed on each other.As a result of this transfer of matter and energy takes place through the system .Living organisms also influence the quantity of biomass present.By decomposition of dead plants and animals by microbes nutrients are released back in to the soil.
Learn more about ecosystem,here:
https://brainly.com/question/13979184
#SPJ2
radium-223 decays with a half-life of 11.4 days, how long will it take for a 0.240-mol sample of radiuim to decay to 1.88 x 10-3 mol
The time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
We'll begin by calculating the number of half-lives that has elapsed.
Original amount (N₀) = 0.240 mole
Amount remaining (N) = 1.88×10¯³ mole
Number of half-lives (n) =?N = 1/2ⁿ × N₀
1.88×10¯³ = 1/2ⁿ × 0.240
Cross multiply
1.88×10¯³ × 2ⁿ = 0.240
Divide both side by 1.88×10¯³
2ⁿ = 0.240 / 1.88×10¯³
2ⁿ = 128
2ⁿ = 2⁷
n = 7Thus, 7 half-lives has elapsed
Finally, we shall determine the time.Number of half-lives (n) = 7
Half-life (t½) = 11.4 days
Time (t) =?t = n × t½
t = 7 × 11.4
t = 79.8 daysTherefore, the time taken for 0.240 mole sample of radiuim to decay to 1.88×10¯³ mole is 79.8 days
Learn more: https://brainly.com/question/13266270
hormones that are essential for normal body growth and maturation include all the following except?
a. thyroid hormone
b. growth hormone
c. Ghrelin
d. insulin
Hello can someone please help me. Me and my mom went to the doctor and the doctor told me I can’t have sex till 19 because of my health problems. But they didn’t tell me what health problems! Can someone tell us there any health things or a name for that?
Answer:
bladder or bowel diseases, or arthritis. These might be your answer.
What will happen to the temperature of an object if the kinetic energy of the particles increases?
The temperature of an object will increase if the kinetic energy of the particle increases.
Kinetic energy is an energy that is said to be in motion. According to the kinetic molecular theory of ideal gas, the particles of the gas are usually moving in constant random motion and they exert no force on each other.
Also, the temperature varies directly proportional to the average kinetic energy of the gas particles. As a result, when the kinetic energy of the particles increases, the temperature will also increase.
Learn more about the kinetic theory of gas here:
https://brainly.com/question/9949658?referrer=searchResults
When the pressure of an equilibrium mixture of SO2, O2, and SO3 is doubled at constant temperature, what the effect on Kp
When the pressure of the equilibrium mixture of SO2, O2, and SO3 is doubled at constant temperature, the Kp is also doubled.
The equation of the reaction is given by; 2SO2 + O2 → 2SO3. The Kp of the reaction is obtained from;
Kp = pSO3^2/pSO2^2 . pO2
Since the Kp depends on the individual partial pressures of each of SO2, O2, and SO3 at equilibrium, if the pressure of the equilibrium mixture is doubled, the Kp is also doubled.
Learn more: https://brainly.com/question/11324711
The table below shows the dimensions of two colored cubes.
Dimensions of Cubes
Cube Side (cm) Mass
(g)
Yellow 3 135
Black 2 48
Answer: The correct answer is black because the product of its side and mass is lower.
Explanation: The density of a substance is defined as the amount of matter that can be stored in a given volume.
Hence, the black cube will be denser because the product of its side and mass is lower.
Answer: it was wrong on my test
Explanation:
literally dont believe them
hydrodistillation explain ????
Answer:
Explanation:
Hydrodistillation is a traditional method for the extraction of bioactive compounds from plants. In this method, plant materials are packed in a still compartment then water is added in sufficient amount and brought to a boil. ... The vapor mixture of water and oil is condensed by indirect cooling with water.
What would be the freezing point of a solution that has a molality of 1.506 m which was prepared by dissolving biphenyl (C12H10) into naphthalene
The freezing point of a solution prepared by dissolving biphenyl (C12H10) into naphthalene is 67.99oC.
Freezing point can be obtained from;
ΔT = K m i
ΔT = freezing point depression
m = molality of the solution = 1.506 m
i = Van't Hoff factor = 1 for molecular substances
K = Freezing constant of naphthalene = 8.15 oC/m
ΔT = 8.15 oC/m × 1.506 m × 1 = 12.27oC
Freezing point of pure naphthalene = 80.26 °C
ΔT = Freezing point of pure naphthalene - Freezing point of solution
Freezing point of solution = Freezing point of pure naphthalene - ΔT
Freezing point of solution = 80.26 °C - 12.27oC = 67.99oC
Learn more: