The pH of a 0.25 M solution of hypobromous acid (HBrO) is 4.60. What is the Ka of hypobromous acid?
A. 1.8 x 10-10
B. 1.0 x 10-4
C. 5.0 x 10-5
D. 1.0 x 10-8
E. 2.5 x 10-9

Diatomic elements are written with a subscript of 2 when they exist as a pure diatomic molecule in nature.

Diatomic elements are elements that exist in nature as two atoms of the same element bonded together. These elements include hydrogen (H₂), nitrogen (N₂), oxygen (O₂), fluorine (F₂), chlorine (Cl₂), bromine (Br₂), and iodine (I₂). Diatomic molecules are important in chemistry because they represent the simplest possible form of a covalent bond. They have an even number of electrons in their bonding molecular orbital, which makes them stable and nonpolar.

When diatomic elements are written in a chemical equation, they are written with a subscript of 2. This is because the chemical formula for a diatomic molecule consists of two atoms of the same element. For example, the chemical formula for hydrogen gas is H₂, and the chemical formula for oxygen gas is O₂. This notation helps to distinguish between elements that exist as single atoms (such as helium, He) and elements that exist as diatomic molecules (such as hydrogen, H₂).

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Answer:

The speed of sound in air is 333.33 m/s

Explanation:

The parameters given are the time it takes for the sound to male it to the cliff and back = 30 times in 9 seconds

Distance to the cliff = 50 meters

Therefore, the sound covers a distance of 30 × 100 = 3000 meters in 9 seconds

Which give the speed of sound in air using the formula for speed = Distance/Time

Therefore, the speed of sound in air = 3000/9 = 333.33 m/s.

Answer:

It is neither a synthesis reaction nor a decomposition reaction because two reactants form two products.

Explanation:

This reaction would be considered a single replacement reaction.

The chemical reaction involves the products and the reactants. The image is neither synthesis nor decomposition as the two products are formed from two reactants. Thus, option A is correct.

A chemical reaction includes the chemical interaction between the reactants to yield the products. In a synthesis reaction, two reactants interact to produce a different molecule or compound.

In a decomposition reaction, a reactant undergoes splitting to produce products and energy. But in the image two reactants produces two products that neither of the reaction shows.

This type of reaction is seen as the single replacement reaction where the more reactive element displaces the less reactive from the compound to form the new molecule.

Therefore, option a. Maya's picture represents a single replacement reaction but not a synthesis or decomposition reaction.

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\(2032533 \sq22222rt[2]{?} \)

Answer:

Gabriella is slowing down at the same rate that Kendall is speeding up, and Franklin is not accelerating.

Gabriella is slowing down at the same rate that Kendall is speeding up, and Franklin is not accelerating. Hence option B is correct.

Acceleration is defined as the speed at which an object's velocity changes over time. The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is accelerating. Acceleration can be expressed as

Acceleration = Displacement / Time

Velocity is defined as the pace at which an item changes position while moving in one direction, as seen from a certain point of view and as measured by a specific unit of time. Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is travelling along a path.

Thus, Gabriella is slowing down at the same rate that Kendall is speeding up, and Franklin is not accelerating. Hence option B is correct.

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Answer:

Bromine fluoride

Answer:

it moves through the water

Explanation:

Answer:

C/3. It moves through the water.

Explanation:

All credit goes to the person above!!!!!!

Thank you so much for being helpful!!!

Based on the data and calculations obtained, it can be concluded that the hypothesis stating, "If the density of the object is 19.32 g/ml, then the object is gold" is not supported.

To support this conclusion, the data and calculations regarding the density of the object should be analyzed. If the calculated density of the object differs significantly from the expected value of 19.32 g/ml, it indicates that the object is not gold.

The accuracy and precision of the data can be assessed by comparing the calculated density with the expected density value. If the calculated density is close to the expected value, it suggests high accuracy. Additionally, if multiple measurements of density yield consistent results, it indicates high precision.

Further experimentation could involve additional tests to determine the identity of the object. These tests could include assessing other physical or chemical properties such as melting point, electrical conductivity, or reactivity with certain substances.

Possible sources of error in the experiment could include instrumental errors in measuring the mass or volume of the object, contamination of the object, or inaccuracies in the known density of gold used for comparison. These factors could contribute to deviations between the calculated and expected densities.

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Answer:

CO32-

Explanation:

decay rate of approximately 1.69x10^17 Bq (becquerels),

The decay rate of a radioactive sample is determined by the number  of radioactive atoms present and the decay constant, which represents the probability of decay per unit of time.

To calculate the decay rate, we multiply the number of atoms in the sample by the decay constant. In this case, the sample has 8.31x10^17 atoms and a decay constant of 0.203/s. Multiplying these values gives a decay rate of approximately 1.69x10^17 Bq (becquerels), which represents the number of decays per second in the sample.

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Answer:

b

Explanation:

Answer:

ita b

Explanation:

trust fam

The value of k for this aqueous reaction at 298 k a b↽−−⇀c  dδ°=10. 35 KJ/mol is 30.37 mol.

K is the song which is used to donate the equilibrium of a reaction and the equilibrium of the reaction can only be obtained when the concentration of reactant is equal to the concentration of the product.

K is also known as the equilibrium constant.

δG or dδ° = - R T In K

                     R = universal gas constant

                     T = 298 k or 24.4Celciuss

                      dδ°= 10. 35

substituting the value in the equation

10. 35 = - 8.313 ×  24.4 celcius  In k

 In k =  - 8.313 ×  24.4 celcius / 10. 35

 In k = 30.37 mol.

Therefore, the value of k for this aqueous reaction at 298 k a b↽−−⇀c  dδ°=10. 35 KJ/mol is 30.37 mol.

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The steady-state temperatures and conversions, we need to solve the energy and material balance equations for the adiabatic CSTR. To determine stability, we can analyze the eigenvalues of the linearized system.

There may be multiple steady states depending on the reaction rate constant and operating conditions.

To find the steady-state temperatures and conversions in the adiabatic CSTR, we need to solve the energy and material balance equations. The energy balance equation for an adiabatic CSTR is given by:

Q - ΔHr*V = 0,

where Q is the heat transfer rate, ΔHr is the enthalpy change of the reaction, and V is the volume of the reactor. Since the reactor is adiabatic (no heat transfer), the equation simplifies to:

ΔHr*V = 0.

The material balance equation for a liquid phase reaction is given by:

v*(CAO - CA) = -r*V,

where v is the volumetric flow rate, CAO is the initial concentration of A, CA is the concentration of A at steady state, and r is the reaction rate.

Using the first-order reaction rate equation, r = k*CA, where k is the rate constant, we can rewrite the material balance equation as:

v*(CAO - CA) = -k*CA*V.

Simplifying the equation, we get:

v*CA + k*V*CA = v*CAO.

Rearranging the equation and substituting the given values, we can solve for CA, which represents the concentration of A at a steady state. With the known concentration of A, we can calculate the conversion, X, using the formula:

X = (CAO - CA)/CAO.

By solving the material balance equation and calculating the conversion, we can find the steady-state temperatures and conversions.

In terms of the number of steady states, there may be multiple solutions depending on the reaction rate constant, k, and the operating conditions. The system can exhibit multiple steady states if there are multiple roots to the material balance equation.

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The equilibrium constant can be obtained using ΔG°rxn = -RT lnK.

On the basis of this, 2H+ + 2e- → H2(g) reduction.

Therefore E° = 0V

Pb2+ + 2e- ⇄ Pb(s) - 0.126

Mg2+ + 2e- ⇄ Mg(s) -2.372

For total E°cell potential,

E°cell = Ered - Eoxd

Eoxd = -Ered of the one being oxidised

E°cell = -2.372 - (-0.126) = -2.246

E°cell = -2.246 V

Thus, calculating dG:

dG = -n x F x E°cell

dG = -2*96500*(-2.246)

=433478 J/mol

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Answer:

c. 80 g

Explanation:

metal alloy:

wt% = (mass compound/mass alloy)×100

for antimony:

⇒ 16% = (g antimony / g alloy)×100

⇒ 0.16 = g antimony / g alloy

g alloy = 500 g

⇒ (0.16)×(500 g) = g antimony

⇒ g antimony = 80 g

Answer:

6.14x10^24

Explanation:

convert the given 47.0g to mols then to molecules.

\(\frac{47.0g}{} x\frac{1mol}{46.07g}x\frac{6.022x10^{23} }{1 mol}\)

46.07 is the molar mass of ethanol

6.022x10^23 is avogadros number

Answer:

the full viral genome is retained in the host cell, but its expression is dramatically restricted, such that few viral antigens and no viral particles are produced.

Explanation:

The values of the caring, professional nurse identified by the American association of colleges of nursing (AACN) include the following below:

This is referred to as a healthcare professional who specializes in caring for the sick and ensuring adequate recovery is achieved.

It is required for a qualified nurse to possess some values which will be important and vital in her execution of tasks in a health center. An example is having integrity which will ensure that all the patients are treated fairly and equally regardless of any differences such as ethnicity, gender etc.

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The electron geometry of the middle carbon atom in a molecule of acetone is trigonal planar.

When determining electron geometry (and molecule geometry), we use the VSEPR (Valence Shell Electron Pair Repulsion) model. According to it, regions of high electron density (such as bonds and lone electron pairs) will arrange around the atom in a way that minimizes repulsion between the negatively charged electrons.

To apply this to the central carbon in acetone, we must determine the number of regions of high electron density around it. That atom doesn't have any lone electron pairs, but it does have 3 bonds (2 single bonds with other carbons and 1 double bond with oxygen). The geometry that minimizes negative repulsive interactions between these is the trigonal planar geometry, with all the bonds lying on the same plane, under 120° angles to one another.

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Answer: The correct option is A. The moon's orbit around Earth

Explanation: Newtons first law of motion says that an object in motion will always remain in motion and an object in erst will always stay at rest untill an external force acts upon it. This law is also called law of inertia which is the property of a body to resist any change in its state of rest or motion. Since moon is also revolving around the earth so it is in motion always and it continues its motion.

Answer:

the moon’s orbit around Earth

Explanation:

Answer:

Argon Gas

Explanation:

:)

Answer:

65.21g

Explanation:

\((48.89 \div 100) \times 63.929 = 31.25 \\ (27.81 \div 100) \times 65.996 = 18.35 \\ (23.3 \div 100) \times 67.002 = 15.61 \\ 31.25 + 18.35 + 15.61 = 65.21\)

The third law of thermodynamics describes the entropy of a: solid.

The third law of thermodynamics states that the entropy of a pure crystalline substance approaches zero as the temperature approaches absolute zero (0 Kelvin or -273.15 degrees Celsius). This law implies that at absolute zero, a perfectly ordered and pure crystalline solid will have zero entropy.

The third law of thermodynamics is not specific to liquids or gases but applies to solids. In a solid, the molecules are highly ordered and have fixed positions in a regular lattice structure. As the temperature decreases towards absolute zero, the thermal motion of the molecules reduces, and the system becomes more ordered, resulting in a decrease in entropy.

In contrast, liquids and gases have higher entropy compared to solids at absolute zero because their molecules have more freedom of movement and are not as tightly arranged. Therefore, the third law of thermodynamics specifically addresses the entropy of solids and does not apply to liquids or gases.

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