Using proper arrow formalism, write down a mechanism for the reaction between methyl pyruvate and benzylamine in order to produce a corresponding Schiff base. If acid is required in your mechanism, use HA as a generic acid.a. How would you expect the IR spectrum of methyl pyruvate to differ from the Schiff base product formed in Part 1? Be specific with the location of the different, distinct IR signals that would be present in each compound, and which functional groups they correspond to
Answers
a) Using arrow formalism, the mechanism for the reaction between methyl pyruvate and benzylamine to produce the corresponding Schiff base is as follows:
1) HA + benzylamine → HA-benzylamine
2) methyl pyruvate + HA-benzylamine → Schiff base + HA
b) The IR spectrum of methyl pyruvate is expected to differ from that of the Schiff base product formed in Part 1 in several ways. For methyl pyruvate, there will be a strong peak at 1722 cm-1 corresponding to the carbonyl functional group in the carboxylic acid, and at 1449 cm-1 for the alkene. For the Schiff base product, there will be a strong peak at 1632 cm-1 corresponding to the C=N double bond, and at 3450 cm-1 for the NH2 group.
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Related Questions
You have a 400-mL container containing 55.0% He and 45.0% Ar by mass at 25°C and 1.5 atm total pressure. You heat the container to 100°C.Calculate the total pressure.
Answers
First, we need to find the number of moles, using the following formula:
\(pV=\text{nRT}\begin{cases}p=\text{pressure} \\ V=\text{volume} \\ n=\text{moles} \\ R=\text{constant of ideal gas law}=0.082\frac{L\cdot atm}{mol\cdot K} \\ \text{T=temperature}\end{cases},\)We're going to clean "n" and replace the initial data which is 40.0 mL, 25 °C, and 1.5 atm. Remember that the volume must be in L (liters) and temperature in K (kelvin).
*To calculate the volume from mL to L, you just divided the number by 1000, and to calculate the temperature in kelvin, you must sum 273 to the normal temperature.
\(n=\frac{pV}{RT}=\frac{1.5atm\cdot0.4\text{ L}}{0.082\frac{L\cdot\text{atm}}{\text{mol}\cdot K}\cdot298\text{ K}}=0.0245\text{ mol},\)Using this data, we're going to use the same formula with the final data which is 40.0mL (volume is constant), 100° C, and 0.025 mol using the correct units, clearing "P", like this:
\(p=\frac{nRT}{V}=\frac{0.0245mol\cdot0.082\frac{L\cdot atm}{mol\cdot K}\cdot373K}{0.4\text{ L}}=1.87\text{ atm.}\)Then, the final pressure will be 1.87 atm.
The density of water is 1.00 g/mL. Which substance below can float on water?
Answers
Answer:
ice, air, ducks and fabric
Explanation:
its common sense
2 moles of NO, was placed in an empty I dm' bottle and allowed to reach equilibrium according to the equation:
At equilibrium, 1.2 moles of N,O, dissociated. Calculate the value of the equilibrium constant for the reaction at that
temperature.
Answers
2NO(g) ⇌ N2(g) + O2(g)
According to the problem statement, 2 moles of NO were placed in a 1 dm^3 bottle and allowed to reach equilibrium, and at equilibrium, 1.2 moles of NO had dissociated. This means that the initial concentration of NO was:
[NO]initial = 2 mol / 1 dm^3 = 2 M
And the concentration of NO at equilibrium is:
[NO]equilibrium = (2 - 1.2) mol / 1 dm^3 = 0.8 M
Since the stoichiometry of the balanced equation is 2:1:1 for NO, N2, and O2, respectively, the equilibrium concentrations of N2 and O2 will also be 0.6 M.
The equilibrium constant (Kc) can be calculated using the equilibrium concentrations of the reactants and products, raised to the power of their stoichiometric coefficients. Therefore:
Kc = ([N2][O2]) / ([NO]^2)
Substituting the equilibrium concentrations into the equation, we get:
Kc = (0.6 M x 0.6 M) / (0.8 M x 0.8 M)
Kc = 0.5625
Therefore, the value of the equilibrium constant for the reaction at that temperature is 0.5625. Note that the units of Kc depend on the stoichiometry of the balanced equation. Since the stoichiometric coefficients are all 1, the units of Kc in this case are M^-1
$4000 principal earning 6% compounded annually. what will the balance be after 6 years
Answers
Answer:
Compound Interest is calculated on the initial Principal for a given unit of time(yearly, half-yearly, quarterly), it also includes all of the accumulated interests of the previous period of a loan or deposit.
In this type of interest, the amount after the first unit time becomes the principal for the second unit and the amount after second unit time becomes the principal for the third unit and so on.
So the difference between the final amount and the principal is called Compound interest.
If the interest is compounded yearly:-
A=P(1+R100)n
Here A = final Amount
P = Principal
R = interest rate
n = time period
Compound interest = Final amount - principal
Given:-
Principal (P) = $4000
Interest Rate (R) = 6%
Time period (n) = 5 years
6. A piece of solid gold was heated from 274K to 314K. 35.73 of energy was needed to raise the temperature.
What mass of gold was present?
Answers
From Q = mcΔT, we can rearrange the equation to solve for mass, m = Q/cΔT. The specific heat capacity, c, of solid gold is 0.129 J/g °C. I'm assuming that the energy is given in joules, as it's not specified in the question as written.
m = Q/cΔT = (35.73 J)/(0.129 J/g °C)(40.85 °C - 0.85°C)
m = 6.92 g of gold was present
The heat of the substance depends on the mass, specific heat, and temperature change of the system. The mass of gold that was present was, 6.92 g.
What is mass?Mass of the substance in the thermodynamic system is the ratio of the heat to the product of the specific heat and the temperature change. It is given as,
\(\rm m = \rm \dfrac{Q}{c \Delta T}\)
Given,
Heat (Q) = 35.73 J
Specific heat (c) = 0.129 J/g °C
Temperature change = 40 degrees celsius
Mass of gold is calculated as:
35.73 ÷ (0.129) (40) = 6.92 gm
Therefore, 6.92 gms of gold is present.
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Give an IUPAC name for the substances below: wireframe labels Wireframe Labels (Abbreviate ortho (o), meta (m), and para (p). if you use these terms. It is not necessary to use italics in writing compound names.) 1 st structure 2nd structure Submit An Try Another Version 2 Item attempts remaining
Answers
1st structure- 1-chloro-3-ethylbenzene (meta-ethyl chlorobenzene)
2nd structure- 2-nitrobenzoic acid (ortho-nitrobenzoic acid)
What is IUPAC's name?
An chemical nomenclature, the IUPAC nomenclature of organic chemistry is a method of naming organic chemical compounds as recommended by the International Union of Pure and Applied Chemistry. It is published in the Nomenclature of Organic Chemistry.
Hence, the 1st structure- 1-chloro-3-ethylbenzene (meta-ethyl chlorobenzene)
2nd structure- 2-nitrobenzoic acid (ortho-nitrobenzoic acid)
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4. An organism that creates its own food is called
• Decomposer
• Producer
• Scavenger
Consumer
hurry please!!
Answers
Answer:
producer
Explanation:
just believe me
The electrolyte that is used to make Hypochlorous Acid is made from Sodium Chloride (NaCl). the bottle says that the electrolyte solution contains 25% sodium chloride. If 2.4 ounces of the solution is added to the half gallon to prepare for the reaction. How many total grams of sodium chloride are used in the reaction. If the reaction produces 1100 PPM of HOCl (hypocholorus acid) how many grams of Hypocholrous are made? Do you think all the NaCl was converted to HOCl? or we can brew one more time and get more HOCl. If all of the NaCl converts to HOCl what would be the maximum possible PPM of the disinfectant
Answers
Answer:
63.9 grams. Yes, the Nacl was converted. Maximum possible ppm is 540ppm.
Explanation:
I this is college level chemistry not regular high school chem.
Answer:2
Explanation:
it is it do not ask questions
a sample of gas held at constant volume is raised from 34.0 atm to 5.90 atm. if the final temperature of the sample is 340.0 K, what was its initial temperature
Answers
Answer
To answer this question we need to use the following equation
P1/T1 = P2/T2
We can rearrange to make T1 subject of the formula as shown
T1 = (P1 x T2)/ P2
= (3.40 x 340)/ 5.90
= 195.9 K
Part A Write the condensed electron configurations for the Ca atom. Express your answer in condensed form as a series of orbitals. For example. He 2x 2p should be entered as (He)222p2. Submit Rest Answer Part B Write the condensed electron configurations for the Sn atom. Express your answer in condensed form as a series of orbitals. For example. He 2:2 should be entered as [he]2s22p^2 Submit Request Answer Part Wnite the condensed electron configurations for the Ga atom. Express your answer in condensed form as a series of orbitals. For example, (He)2x92p should be entered as [he]25*22p2. Submit Request Answet Part D Write the condensed electron configurations for the Vatom Express your answer in condensed form as a series of orbitals. For example. He 23'2p should be entered as [he]2s22p2. Submit Request Answer Part E Wite the condensed electron configurations for the Yalom Express your answer in condensed form as a series of orbitals. For example, (He 2.2" should be entered as (e)28*22p^2. Submit Best Answer Part F Write the condensed electron configurations for the Lu atom Express your answer in condensed form as a series of orbitals. For example, He/2.2p should be entered as (he) 25*22p2. Submit Request Answer
Answers
Part A: Ca atom has an atomic number of 20. Its electron configuration is:1s22s22p63s23p64s2,this is the full electron configuration of Ca.
The condensed electron configuration for the Ca atom is: [Ar]4s2
Part B: Sn atom has an atomic number of 50. Its electron configuration is:1s22s22p63s23p63d104s24p2,this is the full electron configuration of Sn.
The condensed electron configuration for the Sn atom is: [Kr]5s24d105p2
Part C: Ga atom has an atomic number of 31. Its electron configuration is:1s22s22p63s23p64s23d104p1, this is the full electron configuration of Ga.
The condensed electron configuration for the Ga atom is: [Ar]3d104s24p1
Part D: V atom has an atomic number of 23. Its electron configuration is:1s22s22p63s23p63d34s2, this is the full electron configuration of V.
The condensed electron configuration for the V atom is: [Ar]3d34s2
Part E: Y atom has an atomic number of 39. Its electron configuration is:1s22s22p63s23p63d104s24p64d15s2 ,this is the full electron configuration of Y.
The condensed electron configuration for the Y atom is: [Kr]4d15s2
Part F: Lu atom has an atomic number of 71. Its electron configuration is:1s22s22p63s23p63d104s24p64d105s25p65d16s2, this is the full electron configuration of Lu.
The condensed electron configuration for the Lu atom is: [Xe]4f145d16s2
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The transfer of thermal energy from a warmer object to a cooler object is called
Answers
Answer:
heat transfer
If 15 g of CzHo reacts with 60.0 g of Oz, how many moles of water (H2O)
will be produced? (Hint: use what you know about limiting reactant) *
2 C2H6 + 7 O2 + 4 CO2 + 6 H2O
O 1.5 mol H20
O
1.6 mol H20
O
1.0 mol H20
O
1.1 mol H20
Answers
Answer:
1.5 mol H2O
Explanation:
Limiting reactant was C2H6 - 0.5 moles
Fomula became .5C2H6 + 1.75O2 = 1CO2 + 1.5H2O
Which of the following is the standard temperature?
O None of these are correct
O All of these are correct
O 98.6 degrees fahrenheit
O 310 kelvin
O 37 degrees celcius
Answers
Answer:
i think its c but im sorry if im wrong
Explanation:
express balanced equations for the production of nitric acid from nitrogen
Answers
There are several methods for producing nitric acid from nitrogen, but two common methods are:
1. Ostwald process:
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
2 NO(g) + O2(g) → 2 NO2(g)
3 NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq)
Overall equation:
4 NH3(g) + 3 O2(g) + H2O(l) → 2 HNO3(aq) + 2 NO(g)
2. Birkeland-Eyde process:
N2(g) + 3 H2(g) → 2 NH3(g)
2 NH3(g) + 3 O2(g) → 2 NO(g) + 3 H2O(g)
2 NO(g) + O2(g) → 2 NO2(g)
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
Overall equation:
N2(g) + 3 O2(g) + 3 H2(g) → 2 HNO3(aq)
If you had an object with a density of 1.12 g/cm2, would it sink or float in a container of salt water that has a density of 1.35 g/ml?
a -sink
b-float
Answers
The object will float on the salt water.
We have an object of density 1.12 g/cm³.
We have to determine whether it will sink or float in salt water of given density.
How many milliliter are their in 1 cm³ ?In 1 cm³ there are 1 ml
According to the question -
Density of object = ρ(O) = 1.12 g/cm³.
Density of salt water = ρ(S) = 1.35 g/ml = 1.35 g/cm³.
For the object to sink its density should be greater than that of the salt water. Mathematically -
ρ(O) > ρ(S) → Object will sink.
ρ(O) < ρ(S) → Object will float.
Now, it can be seen that -
ρ(O = 1.12) < ρ(S = 1.35)
Therefore, the object will float on the salt water.
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What is the amount of grams in a sample of 1000 ml of C12H22O11 in a 2.0 M solution?
Answers
There are roughly 684.6 grams in a sample of 1000 ml of C12H22O11 in a 2.0 M solution.
To calculate the amount of grams in a sample of 1000 ml of C12H22O11 in a 2.0 M solutionWe need to use the formula:
mass = moles × molar mass
where
moles are equal to molarity times volume (in liters)One mole of a substance has a mass known as a molar massC12H22O11 (sucrose) has a molar mass of about 342.3 g/mol.
The volume must first be changed from milliliters to liters:
1000 ml = 1000/1000 L = 1 L
Next, we can determine how many moles of C12H22O11 are present in the solution:
moles = 2.0 M × 1 L = 2.0 moles
Finally, we can figure out how much C12H22O11 is present in the solution:
mass = moles × molar mass = 2.0 moles × 342.3 g/mol ≈ 684.6 grams
Therefore, there are roughly 684.6 grams in a sample of 1000 ml of C12H22O11 in a 2.0 M solution.
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There are roughly 684.6 grams in a sample of 1000 ml of C12H22O11 in a 2.0 M solution.
To calculate the amount of grams in a sample of 1000 ml of C12H22O11 in a 2.0 M solutionWe need to use the formula:
mass = moles × molar mass
where
moles are equal to molarity times volume (in liters)One mole of a substance has a mass known as a molar massC12H22O11 (sucrose) has a molar mass of about 342.3 g/mol.
The volume must first be changed from milliliters to liters:
1000 ml = 1000/1000 L = 1 L
Next, we can determine how many moles of C12H22O11 are present in the solution:
moles = 2.0 M × 1 L = 2.0 moles
Finally, we can figure out how much C12H22O11 is present in the solution:
mass = moles × molar mass = 2.0 moles × 342.3 g/mol ≈ 684.6 grams
Therefore, there are roughly 684.6 grams in a sample of 1000 ml of C12H22O11 in a 2.0 M solution.
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A Geiger counter detects radiation. Each click corresponds to ____ nucleus/nuclei decaying.
A. two
B. four
C. one
D. three
Answers
Answer:
THREE
Explanation:
What conclusion can you draw about the ability of metals to hold on to and attract electrons, as
compared to nonmetals?
Answers
Answer:
Metals react by losing electrons. So, there is high reactivity due to lower attraction. Non-metals react by gaining electrons. So, there is high reactivity due to higher attraction.
Explanation:
Metals react by losing electrons. So, there is high reactivity due to lower attraction. Non-metals react by gaining electrons. So, there is high reactivity due to higher attraction. Also, electrons lost by metals transfer to the nonmetals. It is easier for the metals to lose their valance electrons and form cations rather than gaining electrons.
Metals do not hold on to or attract electrons while nonmetals hold on to or attract electrons.
In the periodic table, metals are found towards the left hand side of the table while nonmetals are found towards the right hand side of the table.
Electron affinity of elements increase from left to right across the period. Electron affinity refers to the ability of elements to attract or hold electrons. This ability increase steadily across the period.
Usually, the electron affinity values of nonmetals are very high showing that they easily hold on to and attract electrons while the electron affinity values of metals is very low showing that they do not easily hold on to and attract electrons.
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Blood rich in carbon dioxide was called venous blood, as it was believed that only the veins carried this type of blood. Based on what you know about the circulatory system, explain why this statement is incorrect.
Answers
Answer:
Yes is incorrect
Explanation:
not all veins carry blood rich in CO2 , the pulmonary vein carries blood with O2 which means that not all veins carry blood rich in CO2
What does it mean for a unit to be "derived"?
Answers
Answer:
you differentiate to find the derivative
Explanation:
Answer:
A derived unit is a unit that results from a mathematical combination of SI base units.
OR
A unit of measurement obtained by multiplication or division of the base units of a system without the introduction of numerical factors.
Explanation:
i dont know if that answered your question
A Virginia class nuclear submarine has an internal volume of 7.9 million
liters at a pressure of 1.0 atm. If a crewman were to open one of the
hatches to the outside ocean while it was underwater (pressure = 25 atm),
what be would the new volume of the air inside the submarine?
Answers
Answer:
1500
Explanation:
I believe this is correct but am not sure, I shall update when I submit my quiz...
Plz help ASAP.so yea I’m so lost it’s confusing me.
Answers
Answer:
2Al + 3FeO->Al2O3+3Fe
Explanation:
plug in those numbers to balence the equation and yes Al2O3 does not have a number in front
Answer:
2Al +3FeO = 1 Al2O3 + 3Fe
Explanation:
you need 2 al on the left to get al2 on the right, and you need 3 o on the left to get o3 on the right. there's still 3 fe left over, so that's what you put there.
what happens if you don't disconnect the wires in a series circuit?
Answers
Answer:
the series circuit has one current so if you don't disconnect the wires nothing happens
but if you disconnect the wires in series circuit, the current stop and the current doesn't go to the end
Explanation:
(≧▽≦)hope it helps (≧▽≦)
What is the chemical formula for lithium and phosphorus?.
Answers
naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of co2 and 11.30 mg of h20. determine the molecular formula of naphthalene
Answers
The molecular formula of naphthalene is C6H6 if it involves the complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of co2 and 11.30 mg of h20.
We have the complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of co2 and 11.30 mg of h20.
Mass of C in CO2 is,
= 12/44 x 69.00
= 18.82mg
Mass of H in H2O is,
= 2/18 x 11.30
= 1.256mg
We have to find Mole ratios.
C= 18.82/12= 1.568
H= 1.256/1= 1.256
Now we have to divide by the smaller mole ratio.
C= 1.568/1.256= 1.2=1
H= 1.256/1.256= 1
So the empirical formula is CH.
The empirical formula of a compound is defined as the formula which gives us the lowest whole number ratio of that particular compound.
A molecular formula is a way of presenting information about the chemical proportions of atoms that constitute a particular chemical compound or molecule.
Empirical formula weight=12+1=13
We can get the mass of the empirical formula can be computed by dividing the molar mass of the compound by it.
Molecular weight=78
n= Molecular weight / Empirical formula weight
= 78/ 13
=6
Molecular formula=(CH) n =(CH) 6 =C6H6
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plz help me i swear will mark you brainiest
Answers
Answer:
C is the correct answer 35% and 9%
1. Describe your own signal transduction system that utilizes a 1st, 2nd, 3rd, and 4th messenger (please feel free to be creative while also adhering to the underlying science of actual signal transduction messengers and their functions as we discussed these in class).
2. Describe chemical transmission of a nervous message across a synapse.
Answers
A creative signal transduction system that utilizes first messenger like hormone X, second messenger like calcium +2, third messenger like cAMP and fourth messenger like protein kinase A is as follows : 1) Hormone X was the first messenger.
Consider that the first messenger in this system is hormone X. A signaling substance called hormone X attaches to a particular receptor on the cell membrane. 2)Calcium (Ca2+) is a second messenger. Hormone X releases calcium ions (Ca2+) from intracellular reserves when it binds to its receptor.
The second messenger in this system is calcium. 3) cAMP (cyclic adenosine monophosphate) is the third messenger. Adenylyl cyclase, an enzyme, is activated by the elevated calcium levels and transforms ATP (adenosine triphosphate) into cAMP (cyclic adenosine monophosphate).
The third messenger in this route is cAMP. 4) the Protein Kinase A (PKA) fourth messengerProtein kinase A (PKA), an enzyme that phosphorylates target proteins, is triggered by the high amounts of cAMP. The fourth messenger in this signaling chain is PKA.
Let's now list the actions involved in this signal transduction system: The receptor for hormone X is located on the cell membrane. Hormone X binding triggers a signaling cascade, which causes calcium ions (Ca2+) to be released from intracellular storage.
Adenylyl cyclase is triggered by elevated calcium levels and turns ATP into cAMP. Protein kinase A (PKA) is activated by increased cAMP levels. Specific target proteins are phosphorylated by PKA, which causes a variety of physiological reactions and downstream effects.
Although this is a hypothetical example, it follows the general rules of signal transduction systems that are present in biological systems. actual signal transduction pathways in real organisms, a large variety of messengers and chemicals can be involved, making them complex.
2. A crucial aspect of neuronal communication is the chemical transport of signals across synapse. Here is a step-by-step explanation of what happens: a) Arrival of Action Potential: The presynaptic terminal of the neuron sending the message receives an action potential, an electrical signal.
When the neuron's membrane potential exceeds a certain level, this action potential is produced. b) Presynaptic terminal depolarization is a result of the action potential's arrival at the presynaptic terminal. The presynaptic membrane's voltage-gated calcium channels open.
c) Calcium Influx: Calcium ions (Ca2+) can enter the presynaptic terminal when voltage-gated calcium channels open. The cytoplasm of the presynaptic terminal receives calcium ions as they migrate down the gradient of their concentration from the extracellular environment.
d) Release of Neurotransmitters: Vesicles containing neurotransmitters fuse with the presynaptic membrane as a result of calcium influx. The synaptic cleft, which is the minuscule space between the presynaptic terminal and the postsynaptic membrane, is where the neurotransmitters are released as a result of this fusion.
e) Neurotransmitter Diffusion: Across the synaptic cleft, the released neurotransmitters spread out. They pass through the narrow opening to travel to the postsynaptic membrane, which is home to the following neuron or target cell.
After passing through the postsynaptic membrane, the neurotransmitters attach to particular receptors on the surface of the postsynaptic neuron or target cell. Typically, these receptors are proteins incorporated into the postsynaptic membrane.
f) Postsynaptic reaction: A reaction in the postsynaptic neuron or target cell is brought on by the binding of neurotransmitters to their receptors. This reaction may be either excitatory, resulting in depolarization and a higher probability of an action potential, or inhibitory.
g) Reuptake: After the neurotransmitters have had their impact, they can be eliminated from the synaptic cleft via reuptake or enzyme breakdown. Reuptake is a typical mechanism where the presynaptic terminal pulls the neurotransmitters back up for reuse.
h) Transmission: of the signal is terminated by the removal or deactivation of neurotransmitters in the synaptic cleft. When another action potential occurs, the postsynaptic neuron goes back to its resting state and the process is ready to continue.
Overall, chemical transmission across a synapse entails the release, diffusion, and binding of neurotransmitters to receptors, which results in a response in the postsynaptic neuron or target cell and, eventually, permits communication between neurons in the nervous system.
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Select the net ionic equation for the reaction that occurs when sodium hydroxide and sulfuric acid are mixed. o No reaction occurs o 2Na+ (aq) + 2OH(aq) + 2H(aq) + SO2 (aq) → 2Na+ (aq) + SO, (aq) + 2 H2O(g) o H(aq) + OH(aq) → H2O
o H+(aq) + HSO4(aq) + 2OH- (aq) → 2 H2O(l) + SO4- (aq) o 2 Na+ (aq) +20H(aq) + 2H+ (aq) + S0,2-(aq) → 2 Na+ (aq) + SO42+ (aq) + 2 H2O(l) o 2 Na+ (aq) + SO42-(aq) → Na2SO4 (g)
Answers
The correct net ionic equation for the reaction that occurs when sodium hydroxide and sulfuric acid are mixed is H⁺(aq) + OH⁻(aq) → H₂O(l).
When sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄) are mixed, they react to form water (H₂O) and sodium sulfate (Na₂SO₄). The balanced chemical equation for this reaction is:
2 NaOH(aq) + H₂SO₄(aq) → 2 H₂O(l) + Na₂SO₄(aq)
To find the net ionic equation, we need to separate all the aqueous species into their respective ions:
2 Na⁺(aq) + 2 OH⁻(aq) + 2 H⁻(aq) + SO₄²⁻(aq) → 2 H₂O(l) + 2 Na⁺(aq) + SO₄²⁻(aq)
Next, we can cancel out any spectator ions that appear on both sides of the equation. In this case, the sodium ions (Na⁺) and the sulfate ions (SO₄²⁻) are spectator ions. After canceling out the spectator ions, we are left with the net ionic equation:
2 OH⁻(aq) + 2 H⁻(aq) → 2 H₂O(l)
Simplifying the equation, we get:
H⁺(aq) + OH⁻(aq) → H₂O(l).
This is the correct net ionic equation for the reaction that occurs when sodium hydroxide and sulfuric acid are mixed.
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Water is a liquid at room temperature. this is due to?
Answers
Solution: Water is liquid at room temperature because the molecules of water are bound by H bonding so the water molecules are not able to move freely and behave as a liquid.
Solid, liquid and gases differ in their densities, their arrangement of atoms/ molecules and their bonding or force of attraction between the atoms/ molecules. Water is liquid because molecules of water have medium intermolecular space between them and weak force of attraction which allow the molecules of water to move within the limit so it can only flow. Another reason for the liquid nature of water is presence of H bonding which hold the water molecules and don’t let them move freely. The energy needed to break the H bond is moderate and not available at room temperature. Hence water is liquid at room temperature.
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Help please!! Use complete sentences to differentiate between acids and bases on the basis of their behavior when dissolved in water give an example of each type
Please put this in simple terms
Answers
Answer:
Bases and Acids are chemically opposite from each other,and there are multiple ways to distinguish how they react when dissolved in water.
Explanation: