Answer:
Water filters remove elements that cause drinking water to have an unpleasant taste and smell, such as lead, chlorine and bacteria. Home water filtration system will improve the overall purity, taste and smell of your drinking water. It also lowers the pH level of the water that you drink.
Which redox reaction would most likely occur if zinc and copper metal were
added to a solution that contained zinc and copper ions?
Click for a reduction potential chart
A. Cu + Zn → Cu2+ + Zn2+
B. Cu + Zn2+
Cu2+ + Zn
C. Cu2+ + Zn → Cu + Zn2+
D. Cu2+ + Zn2+ → Cu + Zn
Answer:
C
Explanation:
b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex] is the redox reaction. Hence, option C is correct.
What is Redox Reaction?A chemical reaction taking place between an oxidizing substance and a reducing substance.
The oxidizing substance is used to lose electrons in the reaction, and the reducing substance is used to gain electrons.
On the reduction potential chart, zinc is a stronger oxidizing agent than, Copper (Cu), which is a reducing agent as compared to silver
The redox reaction most likely occurs if silver and copper metal were added to a solution that contained silver and copper ions is ;
[tex]Cu^{2+} + Zn[/tex] → [tex]Cu + Zn^{2+}[/tex]
Learn more about Redox Reaction here ;
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How many moles of HNO3 are needed to prepare 5.7 liters of a 1.64 M solution of HNO3
We are given:
Volume of Solution: 5.7 Liters
Concentration: 1.64 Molar
Number of moles required:
We know that the formula for concentration (in molar) is:
M = N / V
where M is the concentration, N is the number of moles, and V is the volume of solution (in L)
plugging the given values in the equation:
1.64 = N / 5.7
N = 1.64 * 5.7
N = 9.35 moles
The Ka of hypochlorous acid (HClO) is 3.00*10^-8. What is the pH at 25.0 °C of an aqueous solution that is 0.02M in HClO?
Answer:
Approximately [tex]4.6[/tex].
Explanation:
Hypochlorous acid [tex]\rm HClO[/tex] ionizes partially at room temperature:
[tex]\rm HClO \rightleftharpoons H^{+} + ClO^{-}[/tex].
The initial concentration of [tex]\rm HClO[/tex] in this solution is [tex]0.02\; \rm mol \cdot L^{-1}[/tex].
Construct a [tex]\verb!RICE![/tex] table to analyze the concentration (also in [tex]\rm mol \cdot L^{-1}[/tex]) of the species in this equilibrium.
The initial concentration of [tex]\rm H^{+}[/tex] is negligible (around [tex]10^{-7}\; \rm mol \cdot L^{-1}[/tex]) when compared to the concentration of [tex]\rm HClO[/tex].
Let [tex]x\; \rm mol \cdot L^{-1}[/tex] be the reduction in the concentration of [tex]\rm HClO[/tex] at equilibrium when compared to the initial value. Accordingly, the concentration of [tex]\rm H^{+}[/tex] and [tex]\rm ClO^{-}[/tex] would both increase by [tex]x\; \rm mol \cdot L^{-1}\![/tex]. ([tex]x > 0[/tex] since concentration should be non-negative.)
[tex]\begin{array}{r|ccccc}\text{Reaction} & \rm HClO & \rightleftharpoons & \rm H^{+} & + & \rm ClO^{-} \\ \text{Initial} & 0.02 & & & &x \\ \text{Change} & -x & & +x & & +x \\ \text{Equilibrium} & 0.02 - x & & x & & x\end{array}[/tex].
Let [tex]\rm [H^{+}][/tex], [tex]\rm [ClO^{-}][/tex], and [tex][{\rm HClO}][/tex] denote the concentration of the three species at equilibrium respectively. Equation for the [tex]K_\text{a}[/tex] of [tex]\rm HClO[/tex]:
[tex]\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]}\end{aligned}[/tex].
Using equilibrium concentration values from the [tex]\verb!RICE![/tex] table above:
[tex]\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]} = \frac{x^{2}}{0.02 - x}\end{aligned}[/tex].
[tex]\begin{aligned}\frac{x^{2}}{0.02 - x} &= 3.00 \times 10^{-8}\end{aligned}[/tex].
Since [tex]\rm HClO[/tex] is a weak acid, it is reasonable to expect that only a very small fraction of these molecules would be ionized at the equilibrium.
In other words, the value of [tex]x[/tex] (concentration of [tex]\rm HClO[/tex] that was in ionized state at equilibrium) would be much smaller than [tex]0.02[/tex] (initial concentration.)
Hence, it would be reasonable to estimate [tex](0.02 - x)[/tex] as [tex]0.02[/tex]:
[tex]\begin{aligned}\frac{x^{2}}{0.02} &\approx \frac{x^{2}}{0.02 - x} = 3.00 \times 10^{-8}\end{aligned}[/tex].
Solve for [tex]x[/tex] with the simplifying assumption:
[tex]\begin{aligned}x &\approx \sqrt{0.02 \times {3.00 \times 10^{-8})}}\\ &\approx 2.45 \times 10^{-5}\end{aligned}[/tex].
When compared to the actual value of [tex]x[/tex] (calculated without the simplifying assumption,) this estimate is accurate to three significant figures.
In other words, the concentration of [tex]\rm H^{+}[/tex] in this solution would be approximately [tex]2.45 \times 10^{-5}\; \rm mol \cdot L^{-1}[/tex] at equilibrium.
Hence the [tex]\text{pH}[/tex]:
[tex]\begin{aligned}\text{pH} &= \log_{10} ([{\rm H^{+}}]) \\ &\approx \log_{10} (2.45 \times 10^{-5}) \\ &\approx 4.6\end{aligned}[/tex].
How many grams are of aluminum nitrate are contained in 0.150 moles of Al(NO3)3?
Answer:
31.95 g
Explanation:
find the molar mass of Al(NO3)3
(Al = 27, N = 14, O = 16)
molar mass of Al(NO3)3 = 27+ (14+16×3)×3
= 213 gmol^-
mass = 213 ×0.15
= 31.95g
A sample of nitrogen gas occupies 117 mL at 100°C. At what
temperature would it occupy 234 mL if the pressure does not
change? (express answer in K and °C)
47
Page
8 I 8
- Q +
Answer:
The new temperature of the gas is 746 K.
Explanation:
Given that,
The volume of the gas, V₁ = 117 mL
Temperature, T₁ = 100°C = 373
Final volume of the gas, V₂ = 234 mL
We need to find the final temperature. The relation between temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}\\\\T_2=\dfrac{T_1V_2}{V_1}\\\\T_2=\dfrac{373\times 234}{117}\\\\T_2=746\ K[/tex]
So, the new temperature of the gas is 746 K.
Carboxylic acid derivatives undergo hydrolysis to make carboxylic acids.
a. True
b. False
Answer:
TRUE
Explanation:
All carboxylic acid derivatives have in common the fact that they undergo hydrolysis (a cleav- age reaction with water) to yield carboxylic acids. with hydroxide ion to yield a carboxylate salt and an alcohol. The carboxylic acid itself is formed when a strong acid is subsequently added to the reaction mixture.
PLS MARK BRAINLIEST
A sample of gas occupies 10.0 L at 240°C under a pressure of
80.0 kPa. At what temperature would the gas occupy 20.0 L if
we increased the pressure to 107 kPa?
Answer: 1090°C
Explanation: According to combined gas laws
(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2
where P1 = initial pressure of gas = 80.0 kPa
V1 = initial volume of gas = 10.0 L
T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K
P2 = final pressure of gas = 107 kPa
V2 = final volume of gas = 20.0 L
T2 = final temperature of gas
Substituting the values,
(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2
T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)
T2 = 513 K × (1.3375) × (2)
T2 = 1372.275 K
T2 = (1372.275 - 273) °C
T2 = 1099 °C
1090 degree Celsius
hope it helps
An important difference between fuel cells and batteries is that batteries _______. Select the correct answer below: do not require a continuous source of fuel require a continuous source of fuel are able to expel products are constantly resupplied with reactants
Answer: the correct option is that batteries (do NOT require a continuous source of fuel).
Explanation:
A battery can be classified as an electrochemical cell that has the ability to produce electric current. They do NOT require a continuous supply of fuel because it contains all the reactants needed to produce electricity. Below are some examples of batteries that are commonly used:
--> Primary battery: This is a single use battery because it can't be recharged. A typical example is the dry cell.
--> Secondary battery: This type of battery can be recharged. They are used as a power source for smartphones, electronic tablets, and automobiles.
A FUEL CELL is known as a device that converts chemical energy into electrical energy. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as they are constantly resupplied with reactants. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines.
Answer:
accumulate reaction byproducts
Explanation:
Batteries accumulate reaction byproducts. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines. A battery is an electrochemical cell or series of cells that produces an electric current.
A eudiometer is used to collect hydrogen gas in a chemical reaction, as in your Exp 7. The volume of the gas in the tube (when pressure is held
constant) is 479.10 mL. The pressure of the atmosphere during the experiment is 758.3 mmgHg, and the temperature of the water and gas is
19.0*C. The water vapor pressure at this temperature is 16.5 torr.
Calculate the mass of hydrogen, in mg, collected.
Answer:
39.29 mg
Explanation:
Step 1: Calculate the partial pressure of hydrogen
The pressure of the atmosphere is equal to the sum of the partial pressures of the water and the hydrogen. (1 Torr = 1 mmHg)
P = pH₂O + pH₂
pH₂ = P - pH₂O = 758.3 mmHg - 16.5 mmHg = 741.8 mmHg
We will convert it using the conversion factor 1 atm = 760.0 mmHg.
741.8 mmHg × 1 atm/760.0 mmHg = 0.9761 atm
Step 2: Convert 19.0 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 19.0 + 273.15 = 292.2 K
Step 3: Calculate the mass of hydrogen
First, we will calculate the moles of hydrogen using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 0.9761 atm × 0.47910 L / (0.08206 atm.L/mol.K) × 292.2 K = 0.01950 mol
The molar mass of hydrogen is 2.015 g/mol. The mass of hydrogen is:
0.01950 mol × 2.015 g/mol = 0.03929 g = 39.29 mg
State the different radiations emitted by radioactive elements.
Two flasks are connected by a closed valve. One contains gas particles and the other contains a vacuum. If the valve is opened such that the particles move until they fill both flasks, the process by which the particles can reconvene entirely in one of the flasks is:
Answer: The process by which the particles can reconvene entirely in one of the flasks is: NONSPONTANEOUS.
Explanation:
The spontaneity of a process can affect the distribution of energy and matter within the system. Different chemical or physical processes have the natural tendency to occur in one direction under a given set of conditions. For example:
--> when water is pour down a hill it naturally flows down but it requires outside energy maybe from a water pump to flow up the hill and ,
--> during an iron rust, iron that is exposed to atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment.
Therefore, a spontaneous process is one that occurs naturally under certain conditions. While a NONSPONTANEOUS process, on the other hand, will not take place unless it is initiated by the continual input of energy from an outside source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the REVERSE direction.
From the two flasks that where connected through a valve, once the valve was opened, the gas spontaneously becomes evenly distributed between the flasks. To reverse this, it would require an external energy making the reconvening of the particles back to the first flask a NONSPONTANEOUS PROCESS .
A substance that donates a pair of electrons to form coordinate covalent bond is called
Lewis base: any species that can donate a pair of electrons and form a coordinate covalent bond. ligand: molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases
How are radioactive isotopes used to diagnose thyroid problems?
Answers
A.
Production of radioactive iodine by the thyroid is measured.
B.
Uptake of radioactive iodine by the thyroid gland is measured.
C.
Destruction of damaged cells by radioisotopes is measured.
D.
Absorption of gamma rays from ingested radioisotopes is measured.
Question
Answer:
A
Explanation:
uptakw of raduoactive iofine by the thyroid gland js measurd
All of the following statements concerning real cases is correct EXCEPT Group of answer choices molecules of real gases are attracted to each other. molecules of real gases occupy no volume. nonideal gas behavior is described by the Van der Waals Equation. the pressure of a real gas is due to collisions with the container. the pressure of a real gas at low temperatures is lower than for ideal gases.
Answer:
molecules of real gases occupy no volume.
Explanation:
As all the real gases are composed of particles that occupy the non-zero volume that is the excluded volume. If the gas is behaving in an ideal manner. The correction becomes negatable and is relative to the total volume. The extended volume is volume that is taken by the non ideal gas particles.Which shampoo would be displayed third?
Answer:
Biolage is the answer
Explanation:
Because it's price is third most
Which of the following is most likely to happen when a sound wave reflects
off a hard surface? (Assume that neither the source of the sound nor the hard
surface is moving.)
A. A change in speed
B. An echo
C. A change in pitch
D. A beat frequency
В
an echo is caused by the fact that waves can be reflected by solid surfaces, this is due to the dynamic pattern of rarefactions and air seals near the reflecting surface
Write the symbol for every chemical element that has atomic number greater than 55 and less than 140.8 u
Butanal, an aldehyde, can be made from 1-bromobutane, but it requires two reactions in sequence. What are the two reagents that can be used to make butanal from 1-bromobutane if reagent A is first reacted with 1-bromobutane and then the product of that reaction, is reacted with reagent B.
Answer:
See explanation and image attached
Explanation:
My aim is to convert 1-bromobutane to butanal. The first step is to react the 1-bromobutane substrate with water. This reaction occurs by SN2 mechanism to yield 1-butanol. Hence reagent A is water.
1-butanol is now reacted with an oxidizing agent such as acidified K2Cr2O7 (reagent B) to yield butanal. Note that primary alkanols are oxidized to alkanals.
These sequence of reactions are shown in the image attached.
The enthalpy of formation for CO2 (s) and CO2 (g) is: -427.4 KJ/mole and -393.5 KJ/mole, respectively. The sublimation of dry ice is described by CO2 (s) → CO2 (g).
The enthalpy needed to sublime 986 grams of CO2 is:
(a) 181.5 Kcal
(b) 611.7 Kcal
(c) 248.3 Kcal
(d) 146.2 Kcal
Answer:
a. 181.5 kcal
Explanation:
Step 1: Calculate the enthalpy of the process (ΔH°).
Let's consider the following process.
CO₂(s) → CO₂(g)
We can calculate the enthalpy of the process using the following expression.
ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)
ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))
ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ
According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.
Step 2: Convert 986 g of CO₂ to moles
The molar mass of CO₂ is 44.01 g/mol.
986 g × 1 mol/44.01 g = 22.4 mol
Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂
22.4 mol × 33.9 kJ/1 mol = 759 kJ
We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.
759 kJ × 1 kcal/4.184 kJ = 181.5 kcal
All of the following statements concerning crystal field theory are true EXCEPT Group of answer choices in an isolated atom or ion, the five d orbitals have identical energy. low-spin complexes contain the maximum number of unpaired electrons. in low-spin complexes, electrons are concentrated in the dxy, dyz, and dxz orbitals. the energy difference between d orbitals often corresponds to an energy of visible light. the crystal field splitting is larger in low-spin complexes than high-spin complexes.
Answer:
low-spin complexes contain the maximum number of unpaired electrons.
Explanation:
In the crystal field theory, the magnitude of crystal field splitting and the pairing energy determines whether a complex will be low spin or high spin.
Low spin complexes often have greater magnitude of crystal field splitting energy than low spin complexes.
High spin complexes have maximum number of unpaired electrons(most of the electrons are unpaired) while low spin complexes have a minimum number of unpaired electrons in a complex(most of the electrons are paired).
Rank the compounds NH3, CH4, and PH3 in order of decreasing boiling point. Choices: A) NH3 > CH4 > PH3 B) CH4 > NH3 > PH3 C) NH3 > PH3 > CH4 D) CH4 > PH3 > NH3 E) PH3 > NH3 > CH4
Answer:
C) NH3 > PH3 > CH4
Explanation:
The boiling point of a substance depends on the nature of intermolecular interaction between the molecules of the substance. The greater the magnitude of intermolecular interaction between the molecules of the substance, the higher the boiling point of the substance.
Both NH3 and PH3 have intermolecular hydrogen bonding between their molecules. However, since nitrogen is more electronegative than phosphorus, the magnitude of intermolecular hydrogen bonding in NH3 is greater than in PH3 hence NH3 has a higher boiling point than PH3.
CH4 molecules only have weak dispersion forces between them hence they exhibit the lowest boiling point.
(URGENT FOR BRAINLIEST!!)
The diagram below shows the movement of Earth plates.
The picture shows two Earth plates one beside the other. The plate on the left is shown moving towards the right and the plate
Which of these features is most likely formed as a result of the movement of Earth plates shown in the diagram? (2 points)
a canyon
a fault
a mountain
a ridge
Answer:
mountain
Explanation:
when plates move towards each other they create mountains
g Identify the process in which the entropy increases. Group of answer choices a decrease in the number of moles of a gas during a chemical reaction the phase transition from a gas to a liquid the phase transition from a solid to a gas freezing water
Answer:
phase transition from a solid to a gas
Explanation:
Entropy refers to the degree of disorderliness in a system. The more disorderly a system is, the greater the entropy of the system.
Decrease in the number of moles of a gas decreases the entropy of the system. Similarly, the entropy of solids is less than that of liquids. The entropy of liquids is less than that of gases.
Therefore, a change of phase from solid to gas represents an increase in entropy of the system.
How many grams of H₂SO₄ are contained in 2.00 L of 6.0 M H₂SO₄?
Please explain and show work.
Answer:
1176 grams
Explanation:
nH2SO4 =2*6=12 mol
mH2SO4=12*98=1176 grams
Answer:
solution given:
molarity of H₂SO₄=6 M
volume=2L
no of mole =6M*2=12mole
we have
mass =mole* actual mass=12*98=1176g
the mass is 1176g.
Three important nutritional additions to training for a long distance race are
O Protein
O Water
O Increase calories
O All of the answer choices
what is the charge on the Mn ions in Mn2o3? 1+, 2+, 3+,3-,4+?
Difficulty(Level:1)
Which option below defines the law of conservation of energy
-Energy is created but some is always lost when doing so
-Energy or Chakra must be conserved to accomplish difficult ninjitsus
-Energy constantly moves toward a form of disorder
-Conserve Energy or the world will end
-Energy is never created or destroyed, it just changes forms.
Answer:
I say the last option
Explanation:
I hope this help let me know if you have any questions
Bond length is the distance between the centers of two bonded atoms. On the potential energy curve, the bond length is the internuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Given that the atomic radii of H and Br are 37.0 pm and 115 pm , respectively, predict the upper limit of the bond length of the HBr molecule. Express your answer to three significant figures and include the appropriate units. View Available Hint(s)for Part C
Answer:
The answer is "152 pm".
Explanation:
The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.
The atomic radius of [tex]H= 37.0 \ pm[/tex]
The atomic radius of [tex]Br = 115.0 \ pm[/tex]
[tex]\text{Bond length = Atomic radius of H + Atomic radius of Br}[/tex]
[tex]= 37.0\ pm + 115.0 \ pm\\\\= 152\ pm[/tex]
In nature, one common strategy to make thermodynamically unfavorable reactions proceed is to couple them chemically to reactions that are thermodynamically favorable. As long as the overall reaction is thermodynamically favorable, even the unfavorable reaction will proceed.
Part A
Consider these hypothetical chemical reactions:
A⇌B,ΔG= 14.8 kJ/mol
B⇌C,ΔG= -29.7 kJ/mol
C⇌D,ΔG= 8.10 kJ/mol
What is the free energy, ΔG, for the overall reaction, A⇌D?
Part B
Firefly luciferase is the enzyme that allows fireflies to illuminate their abdomens. Because this light generation is an ATP-requiring reaction, firefly luciferase can be used to test for the presence of ATP. In this way, luciferase can test for the presence of life. The coupled reactions are
luciferin+O2ATP⇌⇌oxyluciferin+lightAMP+PPi
If the overall ΔG of the coupled reaction is -7.50 kJ/mol , what is the equilibrium constant, K, of the first reactions at 11 ∘C ? The ΔG for the hydrolysis of ATP to AMP is −31.6 kJ/mol.
Answer:
[tex]\triangle G= -6.7 KJ/mol[/tex]
Explanation:
From the question we are told that:
Chemical Reactions:
X=A⇌B,ΔG= 14.8 kJ/mol
Y=B⇌C,ΔG= -29.7 kJ/mol
Z=C⇌D,ΔG= 8.10 kJ/mol
Since
Hess Law
The law states that the total enthalpy change during the complete course of a chemical reaction is independent of the number of steps taken.
Therefore
Generally the equation for the Reaction is mathematically given by
[tex]T = +1 * X +1 * Y +1 *Z[/tex]
Therefore the free energy, ΔG is
[tex]\triangle G=1 * \triangle G*X +1 * \triangle G*Y +1 * \triangle G *Z[/tex]
[tex]\triangle G= +1 * (14.9) +1 * (-29.7) +1 * (8.10)[/tex]
[tex]\triangle G= -6.7 KJ/mol[/tex]
Diisopropyl ether reacts with concentrated aqueous HI to form two initial organic products
a. True
b. Fasle
Answer:
True
Explanation:
The reaction between Diisopropyl ether and concentrated aqueous HI forms two initial organic products as shown in the image attached.
The hydrogen of the HI becomes attached to the oxygen in the ether leading to a cleavage of the C-O bond to yield the first compound. The I^- become attached to the other moiety in the original molecule to yield the second compound as shown in the image attached.
