Answer:
Paired electrons are the electrons in an atom that occur in an orbital as pairs.
⇒paired electrons always occur as a couple of electrons
unpaired electrons are the electrons in an atom that occur in an orbital alone.
⇒unpaired electrons occur as single electrons in the orbital.
Answer:
Paired electrons are the electrons in an atom that occur in an orbital as pairs whereas unpaired electrons are the electrons in an atom that occur in an orbital alone. Therefore, paired electrons always occur as a couple of electrons while unpaired electrons occur as single electrons in the orbital.
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at a blood bank there are five problems with a labortory technician's work performance the first year of work
The performance of a laboratory technician in a blood bank is crucial as it directly impacts the quality of the blood products and patient safety. If there are five problems with a technician's work performance in the first year of work, it can have serious consequences for the blood bank's operations.
Some of the potential problems that may arise include:
Improper labeling of blood products: This can result in confusion and incorrect transfusions.
Mishandling of blood products: This can lead to contamination, spoilage, or improper storage, which can affect the quality of the blood products.
Failure to follow standard operating procedures: This can result in errors, deviations from protocols, and potential safety hazards.
Poor communication skills: This can result in misunderstandings, delays, and errors in documentation.
Inadequate training or knowledge: This can lead to mistakes, misinterpretation of test results, and failure to recognize potential problems.
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What is the mass of carbon in grams found in one molecule compound C7H804
Answer:
The mass of carbon in grams found in one molecule of the compound C7H804 is approximately 2.80 grams.
7. Propane reacts with oxygen to produce carbon dioxide and water according to
the equation pictured below.
Which of the following statements is correct?
Cz Hq + 50₂
50₂ 3C0₂ + 4H₂O
→
2
*
For every 4 moles of water produced, 3 moles of propane react.
For each mole of oxygen that reacts, 3 moles of carbon dioxide are produced.
For every 3 moles of carbon dioxide produced, 5 moles of oxygen react.
For each mole of propane that reacts, 5 moles of oxygen are produced.
Propane is a hydrocarbon gas that can react with oxygen to produce carbon dioxide and water. This reaction is an example of combustion, where a fuel reacts with oxygen to release energy in the form of heat and light.
The balanced chemical equation for this reaction is:
C3H8 + 5O2 → 3CO2 + 4H2O
This equation tells us that for every mole of propane that reacts, we need 5 moles of oxygen. However, it also tells us that for every mole of oxygen that reacts, we produce 3 moles of carbon dioxide.
This means that the ratio of carbon dioxide to oxygen in this reaction is 3:1. So if we want to produce a certain amount of carbon dioxide, we need to make sure we have enough oxygen to react with the propane. For example, if we want to produce 6 moles of carbon dioxide, we would need 2 moles of oxygen (since 6/3 = 2).
Similarly, if we want to know how much oxygen we need to react with a certain amount of propane, we can use the ratio of propane to oxygen (1:5). For example, if we have 2 moles of propane, we would need 10 moles of oxygen (since 2 x 5 = 10).
Overall, the reaction between propane and oxygen is an important one in many industries, including heating and cooking. Understanding the stoichiometry of this reaction can help us predict how much of each reactant we need to produce a certain amount of product, and can help us optimize our processes for efficiency and safety.
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When a chemical hand warmer is activated, it becomes warm to the touch. In terms of energy, what is occurring?
Answer:
Exothermic processes.
Explanation:
Answer:
Exothermic processes.
Explanation:
Exothermic processes will feel warm to the touch and show an increase in temperature. For example, when an 'instant hand warmer' is exposed to air a chemical reaction takes place that releases heat. This exothermic chemical reaction can be used to warm a person's hands when they are cold.
4. The Haber process involves a synthesis reaction. The two reactants are H₂ (g) and N₂ (g), and the products are ammonia NH3 (g) and heat. The reaction takes place in a reaction vessel in which one of the factors that influence the reaction is a high temperature of about 600°C. The reactant atoms move freely about in the reaction vessel. Explain this reaction in terms of collision theory.
The reactant molecules must hit with each other in the proper orientation and with adequate energy, according to collision theory, for a reaction to take place.
The Haber process is a chemical reaction that produces heat and ammonia (NH₃) from hydrogen gas (H₂) and nitrogen gas (N₂). At a high temperature of roughly 600 °C and a pressure of roughly 200 atm, iron catalyses the process.
By taking into account the interactions between molecules, collision theory explains how chemical processes take place.
Molecules that are reactants may either bounce off one another when they collide without reacting, or they may react and produce products.
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Calculate the pH of a solution prepared by dissolving 1.30g of sodium acetate, CH3COONa, in 60.5 mL of .20 M acetic acid, CH3COOH(aq). Assume the volume change upon dissolving the sodium acetate is negligible. Ka of CH3COOH is 1.75*10^-5.
The pH of the solution prepared by dissolving 1.30g of sodium acetate in 60.5 mL of 0.20 M acetic acid is approximately 3.34.
To calculate the pH of the solution, we need to consider the equilibrium between the acetic acid (CH3COOH) and its conjugate base, acetate ion (CH3COO-), using the Ka value provided. We will also take into account the presence of sodium acetate (CH3COONa), which will provide additional acetate ions.
First, let's determine the moles of acetic acid present in the solution:
Moles of acetic acid = concentration x volume
Moles of acetic acid = 0.20 M x 60.5 mL x (1 L / 1000 mL)
Moles of acetic acid = 0.0121 mol
Next, let's determine the moles of sodium acetate:
Moles of sodium acetate = mass / molar mass
Moles of sodium acetate = 1.30 g / 82.03 g/mol
Moles of sodium acetate = 0.0158 mol
Since sodium acetate dissociates completely, the concentration of acetate ions (CH3COO-) will be equal to the moles of sodium acetate divided by the total volume of the solution:
Concentration of acetate ions = moles of sodium acetate / total volume
Concentration of acetate ions = 0.0158 mol / 60.5 mL x (1 L / 1000 mL)
Concentration of acetate ions = 0.260 M
Now, we can set up an ICE (Initial, Change, Equilibrium) table to calculate the concentrations of acetic acid and acetate ions at equilibrium:
CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
Initial: 0.0121 M 0 M 0 M
Change: -x +x +x
Equilibrium: 0.0121 - x x x
The Ka expression for acetic acid is:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Using the given Ka value, we can set up the equation:
1.75 x 10^-5 = (x)(x) / (0.0121 - x)
Since x is expected to be small compared to 0.0121, we can approximate the denominator as 0.0121:
1.75 x 10^-5 = x^2 / 0.0121
Simplifying the equation:
x^2 = 1.75 x 10^-5 x 0.0121
x^2 = 2.1175 x 10^-7
x ≈ 4.6 x 10^-4
Now, we can calculate the concentration of H3O+ ions:
[H3O+] = x = 4.6 x 10^-4 M
Finally, we can calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(4.6 x 10^-4)
pH ≈ 3.34
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4. An oxygen gas has a volume of 225 mL at 75.0° C and 175 mmHg. What will be its volume at a
temperature of 20.0° C and a pressure of 11000 mmHg?
An oxygen gas has a volume of 225 mL at 75.0° C and 175 mmHg the volume of the oxygen gas at the new conditions is 12 mL.
We can use the combined gas law to solve for the volume of the oxygen gas at the second set of conditions:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.
Converting the initial conditions to SI units:
V1 = 225 mL = 0.225 L
T1 = 75.0 + 273.15 = 348.15 K
P1 = 175 mmHg = 0.23 atm
Converting the final conditions to SI units:
T2 = 20.0 + 273.15 = 293.15 K
P2 = 11000 mmHg = 14.47 atm
Now we can solve for V2:
V2 = (P1 * V1 * T2) / (T1 * P2)
V2 = (0.23 * 0.225 * 293.15) / (348.15 * 14.47)
V2 = 0.012 L = 12 mL
Therefore, the volume of the oxygen gas at the new conditions is 12 mL.
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Calculate the values of pH and pOH, based on Kw and the concentration of all species present in a neutral solution of water.
The values of pH and pOH will be 7 in a neutral solution of water.
pH is defined as the negative logarithm of H⁺ ion concentration.
pH is a measure of how acidic or basic a substance is. In our everyday routine, we encounter and drink many liquids with different pH. Water is a neutral substance. Soda and coffee are often acidic.
The pH is an important property, since it affects how substances interact with one another and with our bodies. In our lakes and oceans, pH determines what creatures are able to survive in the water.
Kw = ionic product of water
Kw = 10⁻¹⁴
Kw = [H⁺] [OH⁻]
From the value of Kw,
[H⁺] = [OH⁻] = 10⁻⁷
pH =pOH = -log 10⁻⁷
pH = pOH = 7 for a neutral solution of water.
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Which solvent mixture would be expected to deviate most strongly from Raoult’s law?
a) Sulfuric acid (H2SO4) + H2O
b) Acetic acid (CH3COOH) + H2O
c) Butane (C4H10) + octanol (C8H17OH)
d) Methanol (CH3OH) + H2O
Option (c) is expected to deviate most strongly from the given law.
The law states that the partial pressure of each component in an ideal solution is proportional to its mole fraction in the solution. Deviations from Raoult's law occur when the intermolecular interactions between the molecules of the components in the solution are different from those between the molecules of the pure components. The solvent mixture that is expected to deviate most strongly from Raoult's law is the one that has the largest difference in intermolecular interactions between its components.
Among the given choices, option (c) is expected to deviate most strongly from the given law. Butane (C4H10) and octanol (C8H17OH) have very different intermolecular interactions, as butane is nonpolar and octanol is polar. The resulting solution is expected to have different intermolecular forces than either pure component, leading to significant deviations from Raoult's law. Options (a), (b), and (d) are expected to show smaller deviations from Raoult's law, as the intermolecular interactions between the components in each of these mixtures are more similar.
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2.380 gm of a metal on treatment with nitric acid and subsequent ignition gave 3.022 gm of the metallic oxide. Specific heat of metal is 0.055. Calculate the exact atomic weight.
10. Using Valence Bonding theory, draw a hybrid orbital picture showing the hybrid and atomic orbitals that make up the bonding scheme for each molecule. Hint: Start by drawing a Lewis dot structure. For each molecule, complete the following questions: a) Draw and label a molecular orbital diagram; include all electrons in both the atomic and molecular orbitals. b) From your diagram, write out the ground-state molecular orbital configuration; namely o 1s² * 1s² ... c) Determine the bond order for the molecule. d) Would the following molecules be attracted to a magnet? Briefly explain i. B2^2 ii. Cz iii. HCl
The Valence Bond Theory (VBT) examines how atoms interact to explain chemical bonding.
.Valence bond theory states that the overlapping of partially filled atomic orbitals is what results in bonds.
.Ground-state molecular orbital configuration=σ 1 s, σ * 1 s, σ 2 s, σ * 2 s, σ 2 p z, π 2 p x = π 2 p y, π * 2 p x = π * 2 p y, σ * 2 p z
.bond order =1/2(bonding electron- non bonding electron)
A chemical bonding theory which explains the chemical connection between two atoms is the valence bond (VB) theory. It uses the concepts of quantum mechanics to describe bonding, similar to molecular orbital (MO) theory.
.Valence bond theory states that the overlapping of partially filled atomic orbitals is what results in bonds.
.Ground-state molecular orbital configuration=σ 1 s, σ * 1 s, σ 2 s, σ * 2 s, σ 2 p z, π 2 p x = π 2 p y, π * 2 p x = π * 2 p y, σ * 2 p z
.bond order =1/2(bonding electron- non bonding electron)
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The formula NaF would represent what type of bond?
a. Ionic
b. Covalent
c. metallic
d. James
What type of reaction will occur if ΔH is positive and entropy decreases?
Responses
spontaneous reaction
endothermic reaction
Gibbs free reaction
exothermic reaction
The reaction would be an endothermic reaction. Option B
What type of reaction occurs?With a positive H, endothermic processes take in energy from their environment. It is less advantageous for the process if the entropy drops because it indicates that the reactants are getting more organized.
The likelihood of the response happening spontaneously is decreased as a result. The reaction can, however, continue if additional energy is introduced to the system from an outside source as shown.
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How many µg of mercury are contained in 27.2 mL of a 14.10 ppm solution?
Consider an ideal gas, initially in one half of the apparatus, expands into a vacuum to fill the other half. Initially the gas is in thermal equilibrium with the surroundings.
Which of the following is FALSE about this gas expansion process?
∆T=0
∆E=0
∆S(surroundings)=0
w = q
∆KE=0
∆S(system)=0
∆H=0
∆PE=0
The false statement about the gas expansion process is "∆S(system)=0".
When the gas expands into the vacuum, it undergoes an irreversible process. The gas molecules move from a region of high concentration to a region of low concentration, and as a result, there is an increase in entropy (∆S) of the gas.
According to the Second Law of Thermodynamics, the entropy of an isolated system will always increase for an irreversible process. Therefore, the entropy of the system, in this case, the gas, increases as it expands into the vacuum. The false statement about the gas expansion process is "∆S(system)=0". The other statements are true for this process. Since there is no temperature difference between the gas and the surroundings, the change in temperature is zero (∆T=0). The change in internal energy (∆E) is also zero since the gas is expanding against no external pressure. Finally, since there is no heat transfer between the system and the surroundings, the work done (∆PE + ∆KE) is equal to the heat transfer (w = q).
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How and why is C is the right answer?
The correct answer to the question is option C as that is the only option in which the change in entropy is positive.
Entropy is the degree of randomness in the system. The gaseous state has the highest entropy among the state of matter.
a) Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s)
Since the aqueous state has higher entropy than solids, the change in entropy is negative
b) H₂O (l) ----> H₂O (s)
Since the liquid state has higher entropy than solids, the change in entropy is negative.
c) 2SO₃ (g) ----> 2SO₂ (g) + O₂ (g)
Since the reactant has only 2 moles of gas and the product has 3 moles of gas, the entropy of the reactant increases with the reaction.
d) CaO (s) + CO₂ (g) -----> CaCO₃ (s)
Since the reactant has one mole of gas and the product has none thus the change in entropy is negative.
e) N₂ (g) + 3H₂ (g) ------> 2NH₃ (g)
Since the reactant has more moles of gases than the product which are 4 and 2 moles of gas respectively, the change in entropy is negative.
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Which statement correctly describes a catalyst?
Responses
It is consumed and slows down a reaction.
It is consumed and speeds up a reaction.
It is not consumed and speeds up a reaction.
It is not consumed and slows down a reaction.
Heat 2kg ice from -5c to 0c
In order to melt 2kg of ice the heat that is required is: 709.87 kJ
What is the specific heat energy that is required?The total heat required is gotten from the formula:
Total heat = Heat required to convert 2kg of ice to 2kg of water at 0℃ + Heat required to convert 2kg of water at 0℃ to 2kg of water at 5℃
Thus:
Heat = mh_fg + mCpΔt
We ae given that:
m(mass of ice) = 2kg
hfg (latent heat of fusion of ice) = 334kJ
Cp of water (specific heat) = 4.187 kJ/kg-k
Δt(temperature difference) = 5℃
Therefore,
Heat required = (2 × 334) + (2 × 4.187 × (5 − 0))
= 709.87 kJ
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1. In which situation will erosion likely to occur? Why?
gras
The situation where erosion is likely to occur is B. A bare hillside with loose, dry soil
Why would erosion likely occur ?Erosion is an ongoing process of soil displacement when airflows or water currents move through landforms. Examples occur on hillside areas which have no plants, grasses, shrubs, or anything else to stabilize the dry dirt underneath – making it easy for wind and rainfall to literally transport the surface materials in downward cascades.
Unfortunately, this can yield serious consequences - from impoverished fertility to extra sediments being added to nearby creeks and rivers, to demolishing the infrastructure located at the base of the hill.
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Options include;
Hello. Need help answering this question.
A sample of Zn(s) is reacted with HCl(aq) to form hydrogen gas. The H2 gas bubbles out of aqueous solution and is collected in a 670.0 mL container at 576.0 Torr and 25 C. How many grams of zinc reacted?
0.04 grams of zinc reacted
The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as
PV = nRT
where,
P = Pressure
V = Volume
T = Temperature
n = number of moles
Given,
Volume = 670 ml = 0.67L
Pressure = 576 torr
Temperature = 25 degree celsius
PV = nRT
576 × 0.67 = n × 62.36 × 298
n = 0.02 moles
mass = moles × molar mass
= 0.02 × 2 = 0.04g
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What is the oxidation number of Boron? (2.2.1)
2+
2-
3+
3-
b. what in tro duc - improvements were made in measuring system with the of standard units
The use of standard units has revolutionized measuring systems by providing a consistent framework, promoting accuracy and precision, enabling comparability, and fostering advancements in measurement technology.
The introduction and use of standard units have greatly improved the measuring systems in several ways.
1. Consistency and Comparability: Standard units provide a common reference point for measurements, ensuring consistency and comparability across different systems and locations. By using standardized units such as the International System of Units (SI), measurements can be accurately compared and communicated worldwide.
2. Precision and Accuracy: Standard units are based on well-defined and internationally accepted definitions, allowing for precise and accurate measurements. They provide clear guidelines for calibration and measurement techniques, reducing errors and uncertainties in the measuring process.
3. Interdisciplinary Compatibility: Standard units facilitate interdisciplinary compatibility by enabling seamless integration of measurements from various fields of science and engineering. Researchers and professionals from different disciplines can exchange and combine data without the need for complex unit conversions, enhancing collaboration and knowledge sharing.
4. Traceability: Standard units are traceable to national or international measurement standards, which ensures the accuracy and reliability of measurements. Traceability provides a clear chain of measurement comparisons, allowing for the establishment of confidence intervals and uncertainties in measured values.
5. Technological Advancements: The adoption of standard units has driven advancements in measurement technologies. The need for accurate and traceable measurements has spurred the development of more precise instruments, improved calibration techniques, and enhanced measurement methodologies.
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The duration between two events is called ......
Answer:
The duration between two events is called Interval.
Consider the solubilities of a particular solute at two different temperatures.
Solubility (g/100 g H₂O)
44.3
81.4
Temperature (C)
20.0
30.0
Suppose a saturated solution of this solute was made using 51.0 g H₂O at 20.0 °C. How much more solute can be added if the
temperature is increased to 30.0 °C?
9.64 g more solute can be added if the temperature is increased to 30°C
The maximum amount of solute that can dissolve in a known quantity of solvent at a certain temperature is its solubility.
A saturated solution is a solution that contains the maximum amount of solute that is capable of being dissolved. An unsaturated solution is a solution that contains less than the maximum amount of solute that is capable of being dissolved.
Mass of solute present in a saturated solution with 51.0 g H₂O at 20.0 °C is
(44.3 × 51) ÷ 100 = 22.6g
The ratio of solubilities at two different temperautures = 81.4 / 44.3 = 1.83
This means that there can be 1.83 times more solute when the temperature is 30 °C.
22.6 × 1.83 = 41.35 g
So, the mass of solute that can be added = 51 - 41.35 = 9.64g
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a blank can be seen when a substance change into a new substance
As per chemical changes, a color change can be seen when a substance change into a new substance.
Chemical changes are defined as changes which occur when a substance combines with another substance to form a complete new substance.Alternatively, when a substance breaks down or decomposes to give new substance it is also considered to be a chemical change.
There are several characteristics of chemical changes like those of change in color, change in state , change in odor and even change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.
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PLEASE PLEASE PLEASE HELP ME!!!!!!!!
Starting with water, toggle between Real and Model at the top of the screen. The real view shows the experimentally determined bond angle. The model view shows the angle predicted by the VSEPR model. Carefully observe the bond angle in each case.
Complete the table by entering the real bond angle for each molecule listed.
Type the correct answer in each box.
The real bond angles are;
1) 104 degrees
2) 180 degrees
3) 119 degrees
4) 120 degrees
5) 107 degrees
6) 109.5 degrees
What is the VSEPR theory?The VSEPR (Valence Shell Electron Pair Repulsion) theory is a model used in chemistry to predict the shape of individual molecules based on the repulsion between valence electrons pairs in the outermost electron shell of an atom.
We know that we can look at the presence or the absence of the lone pairs or symmetry in the compound as as to ascertain the real bond angles of the compounds listed in the question.
The bond angles can be obtained from chemical litearture.
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Answer:
1) 104.5
2) 180
3) 119
4) 120
5) 107.8
6) 109.5
Explanation:
3. John Dalton in 1801, established the relationship between the partial pressure of con gases and total pressure of the gas mixture, which is called Dalton's law of partial pressure
Define partial pressure of gas.
Show that partial pressure of a component gas is the product of total press mole fraction.
c) Air contains 20% of O₂ and 80% of N₂ by mass. Find the partial pressure N₂ at the sea level at 25°C. (2) OR
Dalton's law, which states that a gas mixture's overall pressure is equal to the sum of its component gases' partial pressures.
Thus, The pressure that each gas would produce if it occupied the same volume of the mixture by itself at the same temperature is known as the partial pressure.
The English chemist John Dalton stated this empirical relationship in 1801.
It derives from the kinetic theory of gases on the basis of an ideal (perfect) gas and presupposes no chemical interaction between the constituent gases. It roughly holds true for real gases when pressures are low enough and temperatures are high partial pressure.
Thus, Dalton's law, which states that a gas mixture's overall pressure is equal to the sum of its component gases' partial pressures.
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What is the approximate base of the cumulus clouds if the surface air temperature at 1000 feet MSL is 70°F and the dewpoint is 48°F
Answer:
Explanation:
Here's the answer.
To estimate the approximate base of the cumulus clouds, we first need to determine the lifting condensation level (LCL), which is the height at which the air reaches saturation and clouds begin to form as it rises.
To calculate the LCL, we can use the following formula:
LCL = (T - Td) x 400
Where:
T is the temperature in Fahrenheit at the surface
Td is the dew point temperature in Fahrenheit at the surface
In this case, T = 70°F and Td = 48°F:
LCL = (70 - 48) x 400
LCL = 8800 feet MSL
So, the base of the cumulus clouds would likely be around 8800 feet MSL. However, this is just an estimate and the actual height of the cloud base can vary due to other factors such as atmospheric stability, moisture content, and local topography.
A patient is given 0.050 mg of technetium-99 m
(where m means metastable—an unstable but long-lived state), a radioactive isotope with a half-life of about 6.0 hours. How long until the radioactive isotope decays to 3.1×10−3 mg?
It will take 22 hours until the radioactive isotope decays to 0.0031mg.
The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration.
The half-life of a radioactive isotope is the amount of time it takes for one-half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.
Given,
Initial mass = 0.05 mg
Final mass = 0.0031 mg
Half Life = 6 hours
N / N₀ = [tex]( 1 / 2)^n[/tex]
0.0031 / 0.05 = [tex]( 1/2)^n[/tex]
n = 3.65
t = n × half life
= 3.65 × 6
= 21.9 hours = 22 hours
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Which of the following will not be the product(s) of a decomposition reaction
Two compounds an element and a compound two elements a single compound
A decomposition reaction is a chemical reaction that involves the breakdown of a single compound into simpler substances. In this type of reaction, the reactant molecule is split into two or more products, and the products formed are usually different from the original reactant.
The reactant may break down into two compounds, an element and a compound, two elements, or a single compound.
Out of the four options mentioned, a single compound will not be the product of a decomposition reaction. This is because a decomposition reaction breaks down a single compound into simpler substances, which means that it cannot produce a single compound as a product.
For example, consider the decomposition of water (H2O) into hydrogen gas (H2) and oxygen gas (O2). The reaction can be represented as:
2H2O → 2H2 + O2
In this reaction, a single compound (H2O) is broken down into two simpler substances (H2 and O2). This is an example of a decomposition reaction that produces two elements as products.
In conclusion, a decomposition reaction can produce two compounds, an element and a compound, or two elements as products, but it cannot produce a single compound as a product.
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