Aldol condensation is an easy way of making carbon-carbon bond.
In the field of chemistry, we can describe aldol condensation as a type of bond that occurs due to electrophilic substitution at the alpha carbons of the enol.
Aldol condensation is known to be the easiest way for one carbon to form bonds with another carbon.
Such an aldol reaction that takes place between carbonyl compunds instead of aldehydes or enols is referred to as crossed aldol condensation.
Aldol reactions are known to be an important part of synthesizing organic compounds as carbon-carbon bonds are formed due to it.
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atmospheric pressure at sea level is equal to a column of 760 mm hg. oxygen makes up 21 percent of the atmosphere by volume. the partial pressure of oxygen (po2) in such conditions is .
The partial pressure of oxygen in such conditions is equal to 160 mm Hg.
As the percentage of oxygen in the air is 21 percent, the mole fraction of oxygen in the air can be calculated as follows;
mole fraction of oxygen in air = 21 ÷ 100 = 0.21
Now we can determine the partial pressure of oxygen by using the equation given by Raoult's law as follows;
partial pressure of oxygen = total pressure of air × mole fraction of oxygen
As the atmospheric pressure at sea level is stated to be 760 mm hg and the calculated value of the mole fraction of oxygen is 0.21, substituting these values in the equation as follows;
partial pressure of oxygen = 760 mm hg × 0.21
partial pressure of oxygen = 160 mm hg
Therefore, the partial pressure of oxygen is calculated to be 160 mm hg.
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Identify the element symbol
1s22s22p63s23p64s23d104p65s24d5
Answer:
Tc
Explanation:
You just have to follow the rows with the exponents. Just remember that when we get to d, the number in the front is a period lower. Hope this helps!
a sample of gas occupies a volume of 70.3 ml . as it expands, it does 140.6 j of work on its surroundings at a constant pressure of 783 torr . what is the final volume of the gas?
A sample of gas occupies a volume of 70.3 ml as it expands, it does 140.6 j of work on its surroundings at a constant pressure of 783 torr and the final volume of the gas is 11.12L
A gas is a sample of matter that conforms to the shape of a container in which it is held and acquires a uniform density inside the container and even in the presence of gravity and regardless of the amount of substance in the container
Here given data is 11.12L
V₁ = Initial volume = 70.3 ml = 0.0703ml = (1L=1000ml)
w = 140.6 J = 1.387 J = (1Latm=101.3J)
Pressure = 783 torr = 1.03 atm = (1atm=760 torr)
V₂ = final volume = ?
1.387Latm = 1.03 atm × (V₂ - 0.0703ml)
(V₂ - 0.0703ml) = 11.05
V₂ = 11.12L
The final volume of the gas is 11.12L
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two mole of ideal gas is compressed isothermal from 10 to 10,000 atmospheric pressure at 500 degree kelvin. calculate the entropy change in the process.
Two mole of ideal gas is compressed isothermal from 10 to 10,000 atmospheric pressure at 500 degree kelvin then the entropy change in the process is -114.86 J/K
Ideal gas is the hypothetical gas composed of molecule which follow a some rules in that ideal molecule do not attract or repel to each other and here given data is
Ideal gas = 2 mole
Pressure = P₁ = 10
Pressure = P₂ = 10,000
Temprature = 500 degree kelvin
We have to calculate entropy change in the process =?
We know the entropy change
ΔS = cv ln(T₂/T₁) +nR ln (P₂/P₁)
For isothermal compression temprature constant,
ΔS = nR ln (P₂/P₁)
ΔS = 2×8.314×ln(10/10,000)
ΔS = -114.86 J K⁻
ΔS = -114.86 J/K
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There are 120.0 mL of O2 at 700. 0 mmHg and 15⁰ C. What is the number of grams present?
Answer:
0.1498 g of O2.
Explanation:
The Behavior of Gases => Ideal Gas Law.
The ideal gas law is a single equation that relates the pressure, volume, temperature, and the number of moles of an ideal gas, which is:
[tex]PV=nRT,[/tex]where P is pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.082 L*atm/mol*K), and T is the temperature in the Kelvin scale.
So we have to convert pressure from 700.0 mmHg to atm, volume from 120.0 mL to L, and 15 °C to K.
Let's convert pressure taking into account that 1 atm equals 760 mmHg, like this:
[tex]700.0\text{ mmHg}\cdot\frac{1\text{ atm}}{760\text{ mmHg}}=0.9211\text{ atm.}[/tex]Remember that 1 L equals 1000 mL, so 120.0 mL would be equal:
[tex]120.0\text{ mL}\cdot\frac{1\text{ L}}{1000\text{ mL}}=0.1200\text{ L.}[/tex]And the conversion from °C to K is just sum °C with 273, so 15 °C in K is:
[tex]K=\degree C+273=15\degree C+273=288\text{ K.}[/tex]Finally, we can use the ideal gas formula, solving for 'n' (number of moles) and replacing the data that we have, as follows:
[tex]\begin{gathered} n=\frac{PV}{RT}, \\ \\ n=\frac{0.9211\text{ atm}\cdot0.1200\text{ L}}{0.082\frac{L\cdot atm}{mol\cdot K}\cdot288\text{ K}}, \\ \\ n=4.680\cdot10^{-3}\text{ moles.} \end{gathered}[/tex]Now, the final step is to convert 4.680 x 10⁻³ moles of O2 to grams using the molar mass of O2 that can be calculated using the periodic table, which is 32 g/mol. The conversion will look like this:
[tex]4.68\cdot10^{-3\text{ }}moles\text{ O}_2\cdot\frac{32\text{ g O}_2}{1\text{ mol O}_2}=0.1498\text{ g O}_2.[/tex]The answer would be that there are 0.1498 g of O2.
Which element has 2 valence electrons and 5 energy levels?
Answer:
its Boron i hope this helped.
Explanation:
Distillation is a process used to separate a mixture of liquids based on different
a. boiling points
b.
densities
freezing points
d. solubilities
Answer:
freezing points hope it helps u
Distillation is a process used to separate a mixture of liquids based on different freezing points. Therefore, option C is correct.
What is distillation ?Distillation is the process of turning a liquid into a vapor, which is then condensed back into a liquid state. The simplest illustration of it is when steam from a kettle condenses into drops of distilled water that are left on a cold surface.
In general, distillation is the process of vaporizing a liquid to its boiling point, then condensing it back into a liquid to separate it from impurities or other solutes.
Among the many industrial uses of distillation are the production of alcoholic drinks, water purification, and oil refining. Distillation is a physical procedure that removes desired pure compounds from an initial source using heat and other techniques.
Thus, option C is correct.
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you need to prepare 1 l of the acetic acid/acetate buffer. you decided to mix 300 ml of 0.45 m acetic acid acid and 100 ml of 0.65 m naoh plus water to a final volume of one liter. the pka of acetic acid is 4.75. 1. calculate the ph of your buffer solution. 2. naoh is not the conjugate base to acetic acid. explain why you can use naoh to make an effective acetate buffer.
The pH of buffer solution is 4.72.
1) Molarity of acetic acid = 0.45 M
The volume of acetic acid = 300 ml
Molarity of NaOH = 0.65 M
Volume of NaOH = 100 ml
An acidic buffer solution is calculated by using Henderson's equation:
pH = pKa + ㏒[salt] / [acid]
where pKa = 4.75
pH = 4.75 + ㏒[0.065M] / [0.07M]
pH = 4.75 + ㏒(0.929)
pH = 4.75 - 0.03
pH = 4.72
Therefore, the pH of the buffer solution is 4.72.
2) NaOH is not the conjugate base to acetic acid. As we know NaOH is a strong base and acetic acid is a weak acid. Thus, an acid-base reaction occurs between acetic acid and NaOH, thus forming CH₃COONa, which is the conjugate base of acetic acid.
Acidic buffer solution is formed by mixing of weak acid and its conjugate salt of strong base. Thus, NaOH acts as limiting reactant and only CH₃COOH and CH₃COONa is present in solution. Thus, CH₃COOH is weak acid and CH₃COONa is its conjugate salt of strong base forms buffer solution. In the reaction of NaOH and CH₃COOH, NaOH gets completely reacted and forms CH₃COONa. Thus, NaOH is added to make an effective acetate buffer.
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in a beaker, 0.57 mol of potassium iodide and 0.52 mol of lead (ii) nitrate are dissolved in water. a reaction occurs, and a white precipitate is observed. calculate the mass (in grams) of the precipitate formed
The white precipitate formed is lead iodide (PbI₂) of mass 131.38g
What is a precipitate reaction?The term "precipitation reaction" may be described as "a chemical response taking place in aqueous solution wherein ionic bonds integrate to shape an insoluble salt". These insoluble salts fashioned in the course of precipitation reactions are referred to as precipitates. Precipitation reactions are normally double displacement reactions with the formation of stable residues referred to as precipitates.
One of the first-class examples of precipitation reactions is the chemical response among potassium chloride and silver nitrate, which precipitates stable silver chloride. This is an insoluble salt fashioned as a made from precipitation reactions. The chemical components for this precipitation response is proven below.
AgNO3(aqueous) + KCl(aqueous) —–AgCl(precipitate) + KNO3(aqueous)
For the given case,
The balanced reaction is as follows:
Pb(NO₃)₂ (aq) + 2KI (aq) → 2KNO₃ (aq) + PbI₂ (s)
This tells us that for every mole of lead iodide (PbI₂) produced, it requires 2 moles of potassium iodide (KI).
If you have 0.57 mole of KI, they will produce: 0.57/2 = 0.285 mol of PbI₂
Number of moles = [tex]\frac{mass}{molar mass}[/tex]
mass of lead iodide (PbI₂) = Number of moles × molar mass
Mass of lead iodide = 0.285 × 461.01
Mass of lead iodide = 131.38g
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Which set of numbers will balance the following equations? 1's have been included for clarity.__Mn3N4 + __NaF --> __MnF4 + __Na3N a 1; 4; 1; 4 b 1; 4; 3; 2 c 1; 12; 3; 4 d 3; 2; 3; 2
ANSWER
Option C
EXPLANATION
Given that;
[tex]\text{ ----- Mn}_3N_4\text{ }+\text{ ---- NaF }\rightarrow\text{ ---- MnF}_4\text{ }+\text{ ---Na}_3N[/tex]In the reaction above, we have the following data
At the reactants side;
3 atoms of manganese
4 atoms of nitrogen
1 atom of sodium
1 atom of fluorine
At the products side
1 atom of manganese
4 atoms of fluorine
3 atoms of sodium
1 atom of nitrogen
To balance the above equation, apply the law of conservation mass
Law of conservation of mass states that matter can neither be created nor destroyed but can e transformed from one formato another.
To balance the equation, 1 mole of Mn3N4 reacts with 12 moles of Na Tto give 3 moles of MnF4 and 4 moles of Na3N
So, the new equation becomes
[tex]\text{ Mn}_3N_4\text{ }+\text{ 12NaF }\rightarrow\text{ 3MnF}_4\text{ }+\text{ 4Na}_3N[/tex]The following data can be deduced in the above equation
At the reactants side
3 atoms of Mn
4 atoms of N
12 atoms of Na
12 atoms of F
At the products side
3 atoms of Mn
12 atoms of F
12 atoms of Na
4 atoms of N
Looking atthe vabove data, the number of atoms of each element at the reactants side is equal to the number of atoms of same elements at the products side.
Hence, the correct answer is option Ce
u
The _______ the number of theoretical plates, the _______ the separation of a mixture of two liquids.
The more the number of theoretical plates, the more easier the separation of a mixture of two liquids.
A measure of column efficiency is the theoretical plates number (N). It describes the number of plates as defined by plate theory and can be used to calculate column efficiency using a computation where the sharper the peaks, the higher the theoretical plate number. The more theoretical plates there are, the more ideal equilibrium stages there are in the system, and the more effective the separation is.
From all of these factors, it is obvious that more theoretical plates suggest greater separation quality since more equilibriums between stationary and mobile phases can be reached. This can be understood by using the example of a set of stairs. The length of each step grows as the number of stairs decreases.
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I need to determine electron configuration of rubidium and rubidium ion.
The atomic number of rubidium is 37, it means that its electron configuration is as follows:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^1[/tex]Rubidium is an alkali metal. Its ion is formed when it loses an electron. It means that the rubidium ion has 36 electrons instead of 37, and its configuration is as follows:
[tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^6[/tex]what elements does KOH contain
Answer:
K- Potassium
O- Oxygen
H- Hydrogen
KOH is called as Potassium Hydroxide and it contains K (Potassium), O (Oxygen) and H (Hydrogen) elements.
Caustic potash is a common name for potassium hydroxide, an inorganic compound with the formula KOH. The majority of its industrial and niche applications capitalize on its caustic nature and acid reactivity. It important because it is the starting point for most liquid and soft soaps, as well as many chemicals that contain potassium.
It is a dangerously corrosive white solid. KOH has a high level of thermal stability. Melt-casting it into pellets or rods with low surface areas and easy handling properties is common due to its high stability and low melting point. Because KOH is hygroscopic, these pellets become tacky when exposed to air.
Its dissolution in water is strongly exothermic, with water and carbonates accounting for the remaining 90% purity. A common name for concentrated aqueous solutions is potassium lye. Solid KOH does not easily dehydrate even at high temperatures.
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Use the concepts of energy and stability as they relate to
orbital filling to write the electron configuration for iron. (Look at image)
The orbital diagram shows us the number of electrons that can be found in the atom.
How do you fill up an orbital?We know that the orbital is a region in pace where there is a high probability of finding the electron. We also know that the filling of the electrons into the respective orbitals where they fall in can be done by the use of the Aufbau principle.
According to the Aufbau principle, it is clear that in the filling of the electrons into the orbitals, the orbitals that are lesser in energy are first filled. The orbitals that are lesser in energy are always found quite close to the nucleus.
This implies that the relative energies of the orbitals would tend to increase as we move away from the nucleus. In other words, the nucleus is the part of the atom that would be at the lowest energy. The orbital diagram of iron by the use of the Aufbau principle have been shown in the image attached to this answer.
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selective analysis of an analyte and internal standard yielded detector responses of 3.47 and 5.23 respectively when each had a concentration of 1.00 ppm. five ml of an unknown analyte solution were mixed with 1 ml of 5.00 ppm internal standard and then diluted to a total of 10 ml. the detector responses for analyte and internal standard for this mixture were 4.26 and 2.43. calculate the concentration of the analyte in the unknown solution. report your answer in ppm.
By calculating the response factor for the internal standard first, we can calculate that the concentration of the unknown analyte sample is 2.64 ppm.
When solving problems like this, we need to calculate the appropriate response factor:
A(X) / C(X) = F * A(S) / C(S)
A(X) - response for the analyte (3.47)
C(X) - concentration of the analyte (1 ppm)
F - response factor
A(S) - response for the internal standard (5.23)
C(S) - concentration of the internal standard (1 ppm)
F = A(X) * C(S) / (C(X) * A(S)
F = 3.47 * 1 ppm / (5.23 * 1 ppm)
F = 0.663
Now we can use this in the second calculation to obtain the concentration of the analyte in the sample. One thing to note is that in second measurement 1 mL of a 5.00 ppm internal standard was diluted to 10 mL, so the concentration of the internal standard is actually 5.00 ppm / 10 = 0.500 ppm.
4.26 / C(X) = 0.663 * 2.43 / 0.500 ppm
C(X) = 4.26 * 0.500 ppm / (0.663 * 2.43)
C(X) = 1.32 ppm
Because 5 mL of the analyte solution were diluted to 10 mL, the concentration is actually 2 * 1.32 ppm = 2.64 ppm.
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Facts for Globalization?
Answer:
Globalization has many benefits, such as the increased flow of goods, services, capital, and people across borders. This has resulted in increased trade and investment, which has led to higher economic growth rates and improved standards of living around the world.
Explanation:
have a good day
Explain how to make one liter of a 1.25 N sodium hydroxide solution.
Answer:
1.25 moles(N) of NaOH are present in 1000 gm of water.
and 1000gm = 1L
hence we need to put 1.25 Moles (N) of NaOH in 1L H2Oduring a phase change, even though heat is entering or leaving the system, the overall temperature of the system does not change. a good example of this is the freezing of liquid water to form ice cubes (an exothermic reaction) or the melting of an ice cube to liquid water (an endothermic reaction). if the temperature of the system does not change, what happens to the heat (energy) that is being released from (exothermic) or put into the system (endothermic)?
There is no temperature increase during a phase change because the heat supplied to the substance is used to give the molecules in the substance energy to break free from the forces holding them together.
What is a phase change in a substance?A phase change in a substance refers to the process by which a substance changes from one physical state to another when heat is added or removed from it.
A phase change is also called a change of state.
The various types of phase changes are as follows;
melting - this is the process whereby a substance changes from solid to liquidvaporization - this is the process whereby a substance changes from liquid to gassolidification - this is the process whereby a substance changes from liquid to solidcondensation - this is the process whereby a substance changes from gas to liquidDuring a phase change, even though heat is entering or leaving the system, the overall temperature of the system does not change. This is because the heat energy supplied is used to overcome the forces between the molecules of the substance.
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a gas expands against a constant external pressure of 1.15 atm: the initial volume of the gas is 0.150 l and the final volume is 1.65 l. calculate wgas.
The work done by the gas that expands at constant external pressure ( isobaric expansion) is 174.8 J
What is the work done by the gas that expands at constant external pressure?The work done by the gas that expands at constant external pressure is work done under isobaric conditions.
The work done by an isobaric expansion is calculated with the formula below:
Work done = P (Vf − Vi)
where;
P is pressureVf is final volume of gasVi is initial volume of gasP = 1.15 atm
P is converted to N/m² and volume is converted to m³
1.15 atm = 1.15 * 101325 N/m²
P = 116523.75 N/m²
Vf = 1.65 * 10⁻³ m³
Vi = 0.15 * 10⁻³ m³
W = 116523.75 N/m² ( 1.65 * 10⁻³ m³ - 0.15 * 10⁻³ m³)
Work done = 174.8 J
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Which polymer are synthetic
Answer:
Examples of synthetic polymers include nylon, polyethylene, polyester, Teflon, and epoxy.
Which of the following is not a factor that affects temperature?
Longitude
Altitude
Ocean currents
Latitude
What is the boiling point of a solution of 0.250g of glycerol, C3H8O3, in 40g of H2O? Glycerol is a molecular compound, i = 1
Explanation
Given:
mass of glycol = 0.250g
mass of solvent (water) = 40g = 0.04 Kg
i = 1
Required: The boiling point of the solution
Solution
[tex]\Delta T\text{ = Kbm}[/tex]where Kb is the boiling point elevation constant
m is the molal concentration (molality) of all species
Step 1: Find the moles of glycerol
n = m/M
n = 0.250g/92,09382 g/mol
n = 2.71x10^-3 mol
Step 2: Find the molality
molality (m) = moles of solute/Kg of solvent
molality = 2.71x10^-3/0.04Kg
molality = 0.068 m
Step 3: Find the change in temperature
[tex]\Delta T\text{ = Kbm}[/tex]Kb of water = 0.512 cm^-1
DT = 0.512 x 0.068
DT = 0.035 'C
Step 4: Find the boiling point
Boiling point of water = 100 'C
DT = 0.035 'C
New boiling point =100 + 0.035 = 100.035 'C
Answer
100.035 'C
as a result of the sigmoidal shape of the o2-hb dissociation curve, there is very little change in hb saturation across a range of po2 values when po2 is high (e.g. between 80 and 100 mm hg in the text book figures). according to the reading, what is an advantage of the relatively flat shape of the curve here?
Red blood cells contain hemoglobin. When carbonic acid, which is found in tissues, is present, hemoglobin releases the bound oxygen. The oxygen that is attached to hemoglobin is released into the blood's plasma and taken up by the tissues in the capillaries, which are also where carbon dioxide is created.
Throughout each of these deep breathing cycles, how do the partial pressures change?If more air is exchanged, the partial pressures of the outside atmosphere will approach but never reach them due to the remaining volume.
Combustible Oxide : The dissociation curve shifts to the left as a result of one CO molecule binding to hemoglobin increasing the affinity of the other binding locations for oxygen.
Lower CO2, greater pH, and lower temperature all move oxygen dissociation to the left.
The pCO2 is significantly influenced by two variables. The person's breathing rate and depth are the first thing to consider. Someone who is hyperventilating will "blow out" more CO2, which will lower pCO2 levels. Holding one's breath causes CO2 to be retained, which raises pCO2 levels.
Limited pressures : There are two reasons why the alveolar oxygen partial pressure is lower than the ambient O2 partial pressure. First, the upper airway humidifies the air as it reaches the lungs, lowering the partial pressure of oxygen to around 150 mmHg while increasing the partial pressure of water vapour (47 mmHg).
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Help me to balance this equation please
Answer:
H202(aq)+2Fe2+(aq)+2H+=>2fe3+(aq)+2H20(l)
Explanation:
H202(aq)+2Fe2+(aq)+2H+=>2fe3+(aq)+2H20(l)
H202(aq)+2Fe2+(aq)+2H+=>2fe3+(aq)+2H20(l)
Now read the question sooner passes an electric current through a sample of clear odorless colorless liquid as the experiment continues bubbles form and the volume of liquid and decreases soon eclectic samples of two colorless odorless gases that bubble out of the liquid one of these gases burn neither of neither the original nor the other gas burns which is the best explanation of her results
The best explanation will be that the sample was broken down by the electric current and formed a new substance that could burn. Therefore, the original liquid is a compound.
The utilization of the electric current to break down substances is called electrolysis.
The case displayed is the same as the electrolysis of water: the electrical current breaks down the water (a compound) into the components that frame it: hydrogen (a gas that burns) and oxygen (a gas that does not burn).
The test was broken down by the electric current and shaped into an unused substance that seems to burn. Hence the initial fluid may be a compound.
Bypassing an electric current through a test of clear, colorless, and odorless fluid. As the trial proceeds, bubbles shape, and the volume of fluid diminishes.
As the test proceeds, bubbles shape, and the volume of fluid diminishes. Suna collects tests of two colorless, odorless gasses that bubble out of the fluid. One of the gasses burns. Not one or the other the first fluid nor the other gas burns.
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A chemical reaction combines the metal sodium (NA) and the gas chlorine (cl2) to form sodium chloride, commonly known as table salt (nacl) why does this produce have different properties from the reactants
When a chemical reaction happens, the atoms in the reactants are rearranged to create compounds with various chemical characteristics. Although identical atoms are present in both the reactants and the products, sodium chloride was produced by rearranging the atoms in the reactants. The reactants lacked sodium chloride.
Sodium is a soft, silver-colored metal that may be sliced with a knife. With water, pure sodium metal reacts violently (and even explosively), creating sodium hydroxide, hydrogen gas, and heat.
2Na(s) + 2H₂O(l) ⟶ 2NaOH(aq) + H₂(g)
Chlorine is a poisonous yellow-green gas with a pungent smell.
But when sodium and chlorine combine, they create sodium chloride, or table salt, which is known to practically everyone in the world.
2Na(s) + Cl₂(g) ⟶ 2NaCl(s)
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the number of atoms in 0.009567 mole of HBrO3 is
Avogadro's Number:
Avogadro's number, which is equal to 6.02214076 x [tex]10^{23}[/tex], is the quantity of units in one mole of any material. Hence, 1 mole of HBrO3 will contain
6.02214076 x [tex]10^{23}[/tex] atoms.
Then, 0.009567 mole of HBrO3 will contain
= 0.009567 moles x 6.02214076 x [tex]10^{23}[/tex] atoms
= 0.0576138 x [tex]10^{23}[/tex] atoms = 5.761 x [tex]10^{21}[/tex] atoms.
Hence, 0.009567 mole of HBrO3 will contain 5.761 x [tex]10^{21}[/tex] atoms.
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what mass of magnesium bromide is formed when one grams of magnesium reacts with 5 g of bromine
Taking into account the reaction stoichiometry, a mass of 7.57 grams of MgBr₂ is formed when one grams of magnesium reacts with 5 g of bromine.
Reaction stoichiometryIn first place, the balanced reaction is:
Mg + Br₂ → MgBr₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
Mg: 1 moleBr₂: 1 moleMgBr₂: 1 moleThe molar mass of the compounds is:
Mg: 24.31 g/moleBr₂: 159.8 g/moleMgBr₂: 184.11 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
Mg: 1 mole ×24.31 g/mole= 24.31 gramsBr₂: 1 mole ×159.8 g/mole= 159.8 gramsMgBr₂: 1 mole ×184.11 g/mole= 184.11 gramsLimiting reagentThe limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
Limiting reagent in this caseTo determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 24.31 grams of Mg reacts with 159.8 grams of Br₂, 1 grams of Mg reacts with how much mass of Br₂?
mass of Br₂= (1 grams of Mg× 159.8 grams of Br₂) ÷24.31 grams of Mg
mass of Br₂= 6.57 grams
But 6.57 grams of Br₂ are not available, 5 grams are available. Since you have less mass than you need to react with 1 grams of Mg, Br₂ will be the limiting reagent.
Mass of MgBr₂ formedConsidering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 24.31 grams of Mg form 184.11 grams of MgBr₂, 1 grams of Mg form how much mass of MgBr₂?
mass of MgBr₂= (1 grams of Mg× 184.11 grams of MgBr₂)÷ 24.31 grams of Mg
mass of MgBr₂= 7.57 grams
Then, a mass of 7.57 grams of MgBr₂ can be produced.
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Question 5(Multiple Choice Worth 2 points)
(14.01 LC)
What is the function of the outer covering in fish?
Answer:
Protects the fish from their environment.
Explanation:
Fish have a mucus on their outer covering that is very slimy and slippery; this serves as a benefit for the fish that will allow it to swim more easily in water.
It also works as a little suit of armor for the fish to protect themselves against predators.
To find the order of a reaction with respect to one reactant, you will monitor the __________ as the _________of ________ is changed.
To find the order of a reaction with respect to one reactant, you will monitor the Percentage as the reactant of product is changed.
To determine the reaction order from experimental data, either the differential rate law or the integrated rate law can be used. In the rate law, the exponents are frequently positive integers: 1 and 2 or even 0. As a result, the reactions in each reactant are of zeroth, first, or second order. Changes in the concentration of the reactants or products can be used to calculate the rate. changes in reactant or product mass or volume If the reaction is the result of multiple reaction steps with different orders, the order of the reaction can be changed by adjusting the reagent concentration. Because the order is often ruled by the slowest reaction, you can manipulate the relative reaction speed by varying the concentrations.
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Answer:
Explanation:
You will watch the Percentage when the reactant of the product is changed in order to determine the sequence of a reaction with respect to one reactant.
Either the differential rate law or the integrated rate law can be used to infer the reaction order from experimental data.
The exponents of the rate law are frequently positive numbers, such as 1 and 2 or even 0.
Because of this, each reactant undergoes reactions of the zeroth, first, or second order.
The rate can be calculated using changes in the concentration of the reactants or products.
A change in the reactant, product, or reactant's volume by changing the reagent concentration, the reaction order can be modified if it involves several steps that have been performed in a different sequence.
You can also control the relative reaction speed by changing the concentrations because the order is frequently determined by the slowest reaction.
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