During which phase does earth see more than half of the moons surface?
Answer:
waxing crescent phase
Answer:
Waxing crescent
Explanation:
because i got the question right
How many molecules is 1.50 moles of CO2
Answer:
The answer to your question is 9.034 x 10²³ molecules
Explanation:
Data
number of molecules = ?
number of moles = 1.50
Process
To solve this problem, use Avogadro's number. This number relates 1 mol of a substance to 6.023 x 10²³ molecules.
1 mol of CO₂ --------------- 6.023 x 10 ²³ molecules
1.5 moles of CO₂ ---------- x
x = (1.5 x 6.023 x 10²³) / 1
x = 9.034 x 10²³ molecules
Briefly describe the solvation of sodium chloride to form aqueous solution.
(-5)+(+7)-(-4) + (+12)
A 8
B 10
C 18
d 20
Answer:
C-18
Explanation:
Step one follow order of operations
Add and subtract from left to right(-5)+(7)=2-(-4)+(12)
STEP 2
Apply negative Rule -(-4)=+4=2+4+(12)
then add 2+4+12=18
a force of 20 N acts upon 5 kg block caculate the accerleration of the object
Answer:
[tex]\boxed{\text{4 m $\cdot$ s$^{-2}$}}[/tex]
Explanation:
F = ma
[tex]a = \dfrac{F}{m} = \dfrac{\text{20 N}}{\text{5 kg}} \times \dfrac{\text{1 kg$\cdot$ m $\cdot$ s$^{-2}$}}{\text{1 N}} = \text{4 m $\cdot$ s$^{-2}$}\\\\\text{The acceleration of the object is } \boxed{\textbf{4 m $\cdot$ s$^{\mathbf{-2}}$}}[/tex]
If 455 J of heat is transferred to 25.0g of water at 45.0 degrees Celsius, what is the final temperature of the water?
Answer:49.3
Explanation:4.1j/g c * 25g * (t2-45c)=455j
T2-45c = 455j/4.1j/g c * 25g
455/104.6
45+4.3= 49.3 celsius
The final temperature of water if 455 J of heat is transferred to 25.0g of water at 45.0 degrees Celsius is 63.2°C.
How to calculate final temperature?The final temperature of water can be calculated using the following formula:
Q = m × c × ∆T
Where;
Q = quantity of heat absorbed or releasedm = mass of substancec = specific heat capacity∆T = change in temperatureAccording to this question, 455 J of heat is transferred to 25.0g of water at 45.0 degrees Celsius.
455 = 25 × (T - 45)
455 = 25T - 1125
25T = 1580
T = 63.2°C
Therefore, the final temperature of water if 455 J of heat is transferred to 25.0g of water at 45.0 degrees Celsius is 63.2°C.
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[tex]C_{12}H_{22}O_{11}(aq) + H_2O(l) \rightarrow 2C_6H_{12}O_6(aq)[/tex]
The chemical equation shown above represents the hydrolysis of sucrose. Under certain conditions, the rate is directly proportional to the concentration of sucrose. Which statement supports how a change in conditions can increase the rate of this reaction?
a. Increasing the amount of water in which the sugar is dissolved will increase the frequency of collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis.
b. Decreasing the temperature will increase the frequency of the collisions between the sucrose molecules and the water molecules resulting in an increase in the rate of hydrolysis.
c. Increasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.
d. Decreasing the concentration of sucrose will increase the rate of hydrolysis by increasing the frequency of the collisions between the sucrose and the water molecules.
Answer:
a
Explanation:
Michael drove 36 000 s at 65 km/hr. How far did Michael drive in meters? Show each step of the process and use correct significant digits and units in your final answer. You must write this out as one long calculation in order to get full marks. *
The Michael drove 648,000 m.
Explanation:
It is given that the time taken by Michael, that is t = 36000 s
Speed is nothing but distance traveled by the time taken in seconds. Its unit is m/s.
Here we have the speed in km/hr, we have to convert it to m/s as,
1 km = 1000 m
1 hr = 3600 s
So it is converted as,
[tex]$\frac{65 \times 1000}{3600}[/tex]
= 18 m/s
Now we have the speed in m/s.
18 m/s = [tex]$\frac{x m}{36000s}[/tex]
We can find the distance x as = 18 m/s ×36000 s = 648,000 m
So the Michael drive 648,000 m.
Please help me right now
If the surface area of a substance increases,the rate of dissolving will _____ ????
Answer:
decrease
Explanation:
If the surface area of a substance increases, the rate of dissolving will increase due to more frequent and effective collisions between the solvent molecules and solute particles.
This effect happens because dissolution is a surface phenomenon, where solvent molecules must collide with the outer surface of the solute. More surface area means more opportunities for these collisions. For instance, powdered CaCO₃ will react faster with HCI than a lump of CaCO₃ because the increased surface area allows more particles to be in contact with the acid, leading to more frequent collisions. Furthermore, an effervescent tablet will dissolve more rapidly when broken into smaller pieces since the available surface area for reaction is increased.
Additionally, the rate of dissolution is influenced by factors such as agitation and temperature. Stirring or shaking a solution can facilitate solute particles contact with solvent, thus increasing dissolution rate. Similarly, dissolving a substance in a warmer solvent can lead to faster dissolution due to increased molecule's kinetic energy.
3) Tarnished silver contains Ag2S. The tarnish can be removed by placing silverware in an aluminum pan containing an inert electrolyte Such as NaCl. Explain the electrochemical principle for this procedure. The standard reduction potential for the half-cell reaction: Ag2S (s) + 2e- 2Ag (s) + S-2 (aq) is -0.71 V
Tarnished silver ware is cleaned by placing it in an aluminium pan as E°cell is positive.
Explanation:
The tarnish silver due to silver sulphate is removed by placing the tarnished ware in an aluminium pan. Certainly, aluminium is a stronger reducing agent than silver. Hence, it reduces the silver ions(Ag+) to metallic silver. The silvery white white colour will slowly get deposited on the walls of silwer ware and then the tarnish will be removed.
Al(s)+ 3Ag+ (aq)→ Al3+ (aq)+ 3Ag(s)
E° cell = E ° cathode - E° anode
= (-0.71)- (-1.66) = 0.95 V is positive.
Hence, the reaction is feasible and the tarnish will be removed.
The electrochemical principle behind the procedure is aluminum is a strong reducing agent.
What is an electrochemical reaction?
Electrochemical reaction is the reaction in which the electrons flow between a solid electrode.
The reaction is [tex]Ag^2S (s) + 2e- 2Ag (s) + S-2 (aq) is -0.71 V[/tex]
E° cell = E ° cathode - E° anode
(-0.71)- (-1.66) = 0.95 V is positive.
Thus, the electrochemical principle behind the procedure is aluminum is a strong reducing agent.
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2. How do chemical reactions involve energy?
Answer:Chemical reactions often involve changes in energy due to the breaking and formation of bonds. Reactions in which energy is released are exothermic reactions, while those that take in heat energy are endothermic. exothermicA description of a chemical reaction that releases heat energy to its surroundings.
The normal boiling point of a substance is defined to be the temperature at which the liquid phase of the substance is in equilibrium with the gas phase at 1 atm pressure. The normal boiling point of methanol is 80oC and ∆H vap = 38 kJ/mol. What is the ∆Scrap value
Answer:
ΔSv = 0.1075 KJ/mol.K
Explanation:
Binary solution:
∴ a: solvent
∴ b: solute
in equilibrium:
μ*(g) = μ(l) = μ* +RTLnXa....chemical potential (μ)⇒ Ln (1 - Xb) = ΔG/RT
∴ ΔG = ΔHv - TΔSv
⇒ Ln(1 -Xb) = ΔHv/RT - ΔSv/R
∴ Xb → 0:
⇒ Ln(1) = ΔHv/RT - ΔSv/R
∴ T = T*b....normal boiling point
⇒ 0 = ΔHv/RT*b - ΔSv/R
⇒ ΔSv = (R)(ΔHv/RT*b)
⇒ ΔSv = ΔHv/T*b
∴ T*b = 80°C ≅ 353 K
⇒ ΔSv = (38 KJ/mol)/(353 K)
⇒ ΔSv = 0.1075 KJ/mol.K
Draw the structure of the expected major organic product if this compound was to undergo monobromination with br2/febr3.
Answer:
Explanation:
Bromine comes in handy with reactivity of organic compounds such as alkenes and arenes( aromatic hydrocarbon). In arenes , it is an electrophilic substitutional reaction where bromine becomes the electrophile to form a substituted aromatic ring. The use of bromine in monobromination (i.e substitution of one bromine atom) is due to the higher electronegative effect in bromine which makes it to be a strong electrophile.
In the diagram; the left hand side shows the arene compound and the right hand side shows the major product after it has undergo monobromination with Br₂/FeBr₃.
Which traits explain the differences between halogens and alkali metals? Select all that apply Question 5 options: Halogens are better oxidizing agents than their alkali metal counterparts within the same period. Halogens have a larger van der Waal's radii than their alkali metal counterparts within the same period. Halogens have a higher effective nuclear charge than their alkali metal counterparts within the same period. Halogens have a higher first ionization energy than their alkali metal counterparts within the same period. Halogens have a higher electron affinity than their alkali metal counterparts within the same period.
Answer:
Halogens are better oxidizing agents than their alkali metal counterparts within the same period
Halogens have a higher first ionization energy than their alkali metal counterparts within the same period.
Halogens have a higher electron affinity than their alkali metal counterparts within the same period.
Halogens have a higher effective nuclear charge than their alkali metal counterparts within the same period
Explanation:
Effective nuclear charge increases across a period and accounts for increase in ionization energy across a period. This explains why halogens in the same period with alkali metals have higher effective bucket charge as well as ionization energy than the alkali metals.
Similarly, electron affinity increased across a period, halogens have greater ekctron affinity than corresponding alkali metals in the same period.
The volume of a sample of N2 gas was decreased from 13.81 to 3.63 L. If the final pressure exerted by the N2 sample was 261.1 kPa, what pressure did the N2 exert before its volume was decreased?
Answer: 68.6 kPa
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=?\\V_1=13.81L\\P_2=261.1kPa\\V_2=3.63L[/tex]
Putting values in above equation, we get:
[tex]P_1\times 13.81L=261.1\times 3.63L\\\\P_1=68.6kPa[/tex]
Thus pressure before its volume was decreased was 68.6 kPa
How many moles of CH3OH are there in 43.7 mL of 0.400 M CH30
Answer:
0.0174 mol
Explanation:
Molarity is (moles of solute)/(liters of solvent). So, to find moles you need to multiply liters by molarity.
0.0437 L × 0.400 M = 0.0174 mol
The number of moles of CH₃OH present is 0.01748 mole
From the question,
We are to determine the number of moles of CH₃OH that are in 43.7 mL of 0.400 M CH₃OH
Using the formula
Number of moles = Concentration × Volume
From the question
Concentration of CH₃OH = 0.400 M
Volume of CH₃OH = 43.7 mL = 0.0437 L
∴ Number of moles of CH₃OH present = 0.400 × 0.0437
Number of moles of CH₃OH present = 0.01748 mole
Hence, the number of moles of CH₃OH present is 0.01748 mole
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Solid NaCl is added slowly to a solution containing 0.10M AgNO3 and 0.20M Pb(NO3)2. Ksp AgCl = 1.8 E-10 Ksp PbCl2 = 1.6 E-5 Write a net ionic equation and corresponding Ksp expression for the dissolution of solid i. Silver Chloride (AgCl) ii. lead (II) chloride (PbCl2) How do I do this?
Answer:
Ag+(aq) + Cl-(aq) <--->AgCl(s)
Ksp = [Ag+] [Cl-]
aii) Pb2+(aq) + 2Cl-(aq) <--->PbCl2(s)
Ksp = [Pb+] [Cl-]^2
Explanation:
In this question, we are to write the net ionic equation and corresponding Ksp expression for the salts in the question.
We proceed as follows;
Ag+(aq) + Cl-(aq) <--->AgCl(s)
Ksp = [Ag+] [Cl-]
aii) Pb+2(aq) + 2Cl-(aq) <--->PbCl2(s)
Ksp = [Pb+] [Cl-]^2
Final answer:
To write net ionic equations for AgCl and PbCl2, the equations are AgCl(s)
ightarrow Ag+(aq) + Cl-(aq) with Ksp = [Ag+][Cl-], and PbCl2(s)
ightarrow Pb2+(aq) + 2Cl-(aq) with Ksp = [Pb2+][Cl-]^2. These represent the dissolution of silver chloride and lead (II) chloride, respectively.
Explanation:
To write the net ionic equations and corresponding Ksp expressions for the dissolution of solid silver chloride (AgCl) and lead (II) chloride (PbCl2), you should consider both substances dissolving separately in water:
AgCl
Net ionic equation: AgCl(s)
ightarrow Ag+(aq) + Cl-(aq)
Ksp expression: Ksp = [Ag+][Cl-]
This represents the solubility product for silver chloride, which is given as 1.8 imes 10^-10.
PbCl2
Net ionic equation: PbCl2(s)
ightarrow Pb2+(aq) + 2Cl-(aq)
Ksp expression: Ksp = [Pb2+][Cl-]^2
This represents the solubility product for lead (II) chloride, which is given as 1.6 imes 10^-5. Note that because there are two chloride ions for every lead ion that dissolves, the concentration of chloride appears squared in the expression.
Due to the small and highly electronegative nature of fluorine, the oxyacids of the this element are much less common and less stable than those of the other halogens. Bonding theory, however, does allow one to propose structures for these acids and use formal charges for the evaluation of these structures. For a molecule of fluorous acid, the atoms are arranged as HOFO. (Note: In this oxyacid, the placement of fluorine is an exception to the rule of putting the more electronegative atom in a terminal position.) What is the formal charge on each of the atoms?
Answer:
HOFO = (0, 0, +1, -1)
Explanation:
The formal charge (FC) can be calculated using the following equation:
[tex] FC = V - N - \frac{1}{2}B [/tex]
Where:
V: are the valence electrons
N: are the nonbonding electrons
B: are the bonding electrons
The arrange of the atoms in the oxyacid is:
H - O₁ - F - O₂
Hence, the formal charge (FC) on each of the atoms is:
H: FC = 1 - 0 - 1/2*(2) = 0
O₁: FC = 6 - 4 - 1/2*(4) = 0
F: FC = 7 - 4 - 1/2*(4) = +1
O₂: FC = 6 - 6 - 1/2*(2) = -1
We can see that the negative charge is in the oxygen instead of the most electronegative element, which is the F. This oxyacid is atypical.
I hope it helps you!
How many grams of MgBr2 are in 4.3x10^23 atoms of MgBr2??
Answer:
The answer to your question is 131.43 grams of MgBr₂
Explanation:
Data
number of grams = ?
number of atoms = 4.3 x 10²³
Molecule = MgBr₂
Process
1.- Calculate the molar mass of MgBr₂
MgBr₂ = 24.3 + (79.9 x 2) = 24.3 + 159.8 = 184.1 g
2.- Use proportions to calculate the number of atoms
184.1 g of MgBr₂ -------------- 6.023 x 10²³ atoms
x -------------- 4.3 x 10²³ atoms
x = (4.3 x 10²³ x 184.1) / 6.023 x 10²³
x = 7.92 x 10²⁵ / 6.023 x 10²³
x = 131.43 grams
Solutions of sodium sulfate and silver nitrate are mixed together.
On a piece of scratch paper write a fully balanced double-replacement equation for this reaction.
Using the chemical equation you wrote; which answer correctly identifies the precipitate (if there is one) and the net ionic equation for the reaction that formed the precipitate?
a
Precipitate: silver sulfate
Net Ionic: 2Ag+ + SO42- --> Ag2SO4
b
There is no precipitate for the reaction; all products are soluble.
c
Precipitate: sodium nitrate
Net Ionic: Na+ + NO3- --> NaNO3
d
Precipitate: sodium nitrate
Net Ionic: Na2+ + 2NO3- --> Na(NO3)2
e
Precipitate: silver sulfate
Net Ionic: Ag+ + SO42- --> AgSO4
Final answer:
The correct option is (a), which states that the precipitate formed when sodium sulfate and silver nitrate solutions are mixed is silver sulfate, and it provides the correct net ionic equation for this precipitation reaction.
Explanation:
When solutions of sodium sulfate (Na2SO4) and silver nitrate (AgNO3) are mixed together, a double-replacement reaction occurs. The balanced chemical equation for this reaction is:
Na2SO4(aq) + 2AgNO3(aq) → Ag2SO4(s) + 2NaNO3(aq)
The precipitate formed in this reaction is silver sulfate (Ag2SO4). The net ionic equation for the formation of the precipitate is:
2Ag+(aq) + SO42-(aq) → Ag2SO4(s)
Answer (a) is the correct option, as it accurately identifies the precipitate and provides the correct net ionic equation for its formation.
During a volcanic eruption, large amounts of poisonous gases and particles are released into the atmosphere. How do some of these gases eventually reach the Earth's surface?
A) through rain
B) through wind
C) through lava
D) through trees
The gases eventually reach the Earth's surface through rain. Hence option A is correct.
What is volcanic eruption?Volcanic eruption is defined as when a volcano erupts, sometimes violently spewing lava and gas into the air. Volcanoes erupt with a powerfully destructive mixture of ash, lava, hot, toxic gases, and rock. Explosions from volcanoes have claimed lives. Volcanic eruptions can bring forth additional health risks such wildfires, floods, mudslides, electricity outages, and contaminated drinking water.
During a volcanic eruption, water vapor (H2O), carbon dioxide (CO2), and sulfur dioxide make up 90% of the gas molecules released (SO2). The remaining one percent is made up of trace amounts of hydrogen sulfide, carbon monoxide, hydrogen chloride, hydrogen fluoride, and other small gas species.
Thus, the gases eventually reach the Earth's surface through rain. Hence option A is correct.
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1 Copy and complete using the words below:
electrons, full, Group, lose, noble, unreactive
The elements in____ 0 of the Periodic Table are called the____gases. They are generally ____. because they have a____outer shell of electrons. So they do not need to gain____or share ___ with other atoms.
Answer:
The elements in__Group_ 0 of the Periodic Table are called the_noble__gases. They are generally __unreactive_. because they have a__full_outer shell of electrons. So they do not need to gain__lose_or share _electrons_ with other atoms.
How many moles are in 36.0g of H20
Answer:
The answer to your question is 2 moles
Explanation:
Data
mass of H₂O = 36 g
moles of H₂O = ?
Process
1.- Calculate the molar mass of water (H₂O)
H₂O = (1 x 2) + (16 x 1) = 2 + 16 = 18 g
2.- Use proportions and cross multiplication to find the answer.
18 g of H₂O ---------------- 1 mol
36 g of H₂O --------------- x
x = (36 x 1) / 18
x = 36/18
x = 2 moles
At high concentrations, inorganic fluoride inhibits enolase. In an anaerobic system that is metabolizing glucose as a substrate, which of the following compounds would you expect to immediately increase in concentration following the addition of fluoride?
A. 2-phosphoglycerate.
B. Glucose.
C. 3-phosphoglycerate
D. Phosphoenolpyruvate
E. Pyruvate
Answer:
A. 2-phosphoglycerate
Explanation:
Glycolysis is the process of breakdown of glucose into two 3-carbon molecules called pyruvate. The energy released during glycolysis is used to make ATP.
Enolase is the enzyme which plays very important role in glycolysis. In the 9th step of glycolysis, Enolase converts 2-phosphoglycerate into phosphoenolpyruvate.
This reaction of conversion of 2-phosphoglycerate to phosphoenolpyruvate is a reversible dehydration reaction.
Fluoride inhibits enolase, so when enolase is become non-functional then there is no convertion of 2-phosphoglycerate to phosphoenolpyruvate, so the concentraion of 2-phosphoglycerate is increases by the addition of fluoride.
A sample of a gas is contained in a closed rigid cylinder. According to the kinetic molecular theory, what occurs when the gas inside the cylinder is heated?
Answer:
When heating, the temperature is increased, therefore, the average kinetic energy is increased so the molecules will start moving more rapidly.
Explanation:
Hello,
Kinetic molecular theory is based on a series of specific statements:
- Gases are constituted by a raft of particles that are considered as both hard and spherical bodies under a state of constant and random movement.
- The particles are constantly moving in a straight line until they collide to each other or against walls of the container.
- There is no force of attraction nor repulsion among gas particles or among the particles and the walls of the container.
- Collisions are said to be perfectly elastic.
- The average kinetic energy of a raft of gas particles is temperature-dependent only.
- The higher the temperature the higher the average kinetic energy of the gas and the other way around.
In such a way, considering the last two statements, when heating, the temperature is increased, therefore, the average kinetic energy is increased so the molecules will start moving more rapidly in comparison to the initial state.
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A piece of iron with a mass of 56.2 grams is heated and placed into a calorimeter containing 93.64 grams of water at 7.0 degrees Celsius. The final temperature of the water and the iron is 90.5 degrees Celsius. Assuming no heat is lost to the surroundings, how much heat (in Joules) does the iron release? Round your answer to the nearest 0.1 Joules.
Answer:
[tex]Q_{iron} = -32730.083\,J[/tex]
Explanation:
By the First Law of Thermodynamics, the piece of iron and water reach thermal equilibrium when both have the same temperature. The heat released by the piece of iron is received by the water. That is to say:
[tex]Q_{iron} = -Q_{w}[/tex]
[tex]Q_{iron} = (93.64\,g)\cdot \left(4.186\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (7\,^{\textdegree}C-90.5\,^{\textdegree}C)[/tex]
[tex]Q_{iron} = -32730.083\,J[/tex]
For a reaction in which a diatomic molecule dis- sociates into atoms, what are the signs of ΔS and ΔH? Explain your answer. You may use the reaction N2(g) ⎯→ 2N(g) as an example of such a reaction.
Answer:
For the reaction N2 (g) and 2N (g) the energy is required to break the bond in N2 to from N atoms. This means AH is positive, is an endothermic reaction.
Explanation:
Entropy increases that is positive because atoms are now free to move.
*multiple choice*
What is the molarity of a 45.3g sample of KNO3 (101g) dissolved in enough water to make a 0.225L solution?
a) 0.45
b) 2
c) option 3
d) none of the above
1.95 or 2 is the molarity of a 45.3g sample of KNO3 (101g) dissolved in enough water to make a 0.225L solution.
The correct answer is option b
Explanation:
Data given:
mass of KN[tex]O_{3}[/tex] = 45.3 grams
volume = 0.225 litre
molarity =?
atomic mass of KNO3 = 101 grams/mole
molarity is calculated by using the formula:
molarity = [tex]\frac{number of moles}{volume of the solution}[/tex]
first the number of moles present in the given mass is calculated as:
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
number of moles = [tex]\frac{45.3}{101}[/tex]
0.44 moles of KNO3
Putting the values in the equation of molarity:
molarity = [tex]\frac{0.44}{0.225}[/tex]
molarity = 1.95
It can be taken as 2.
The molarity of the potassium nitrate solution is 2.
According to the following reaction, how many grams of hydrogen peroxide (H2O2) are needed to form 32.0 grams of oxygen gas?
Answer:
The answer to your question is 34 g of H₂O₂
Explanation:
Data
mass of H₂O₂ = ?
mass of O₂ = 32 g
Process
1.- Write the balanced chemical reaction
H₂O₂ ⇒ H₂ + O₂
2.- Calculate the molar mass of H₂O₂ and O₂
Molar mass H₂O₂ = (1 x 2) + (16 x 2) = 34 g
Molar mass of O₂ = 2 x 16 = 32 g
3.- Use proportions to calculate the mass of O₂
34g of H₂O₂ -------------- 32g of O₂
x -------------- 32g of O₂
x = (32 x 34) / 32
x = 34 g of H₂O₂