calculate the mass of barium in 288mg of barium arsenate, ba3(aso4)2. give your answer in mg.

Answer:

Refraction is the bending of a wave when it enters a medium where its speed is different, like going from the air (gas) to a glass of water (liquid). Absorption is when light energy penetrates an object. Typically this energy is then converted to heat.

To find the half-life of sodium-24, we can use the formula for exponential decay:

N(t) = N₀ * (1/2)^(t / t₁/₂)

Where:

N(t) is the amount of the substance remaining at time t

N₀ is the initial amount of the substance

t is the time elapsed

t₁/₂ is the half-life of the substance

Given:

N₀ = 6.00 mg (initial amount)

N(t) = 0.750 mg (amount after 45 hours)

Plugging in these values, we can solve for t₁/₂:

0.750 mg = 6.00 mg * (1/2)^(45 / t₁/₂)

Dividing both sides of the equation by 6.00 mg:

(0.750 mg) / (6.00 mg) = (1/2)^(45 / t₁/₂)

0.125 = (1/2)^(45 / t₁/₂)

To eliminate the base of 1/2, we can take the logarithm of both sides:

log₂(0.125) = 45 / t₁/₂ * log₂(1/2)

Using the property log₂(a^b) = b * log₂(a):

log₂(0.125) = 45 * log₂(1/2) / t₁/₂

We know that log₂(1/2) = -1, so we can simplify the equation further:

log₂(0.125) = -45 / t₁/₂

Now, we can solve for t₁/₂:

t₁/₂ = -45 / log₂(0.125)

Using a calculator:

t₁/₂ ≈ -45 / (-3) ≈ 15

The half-life of sodium-24 is approximately 15 hours.

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Answer:

True

Explanation:

For the B₁-B₂ cell pair, B₂ is the cathode, and B₁ is the anode. For the A₁-B₂ cell pair, B₂ is the cathode, and A₁ is the anode. For the A₂-B₂ cell pair, B₂ is the cathode, and A₂ is the anode.

B₁-B₂:

The half-reaction at B₁ is given as A⁺(aq) + 2e⁻ → A(s) with a reduction potential of -0.78 V.

The half-reaction at B₂ is given as B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.

Since the reduction potential of B₂ (-0.45 V) is more positive than B₁ (-0.78 V), B2 will be the cathode, and B1 will be the anode.

A₁-B₂:

The half-reaction at A₁ is given as C⁺(aq) + 2e⁻ → C(s) with a reduction potential of -0.95 V.

The half-reaction at B₂ is the same as mentioned before B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.

Since the reduction potential of B₂ (-0.45 V) is more positive than A₁ (-0.95 V), B₂ will be the cathode, and A₁ will be the anode.

A₂-B₂:

The half-reaction at A₂ is given as D⁺(aq) + 2e⁻ → D(s) with a reduction potential of -0.70 V.

The half-reaction at B₂ is the same as mentioned before B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.

Since the reduction potential of B₂ (-0.45 V) is more positive than A₂ (-0.70 V), B₂ will be the cathode, and A₂ will be the anode.

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We can see here that the acid-catalyzed addition of alcohols to alkenes proceeds in a mechanism analogous to the acid-catalyzed addition of water to yield ethers. True.

Alcohol is a broad term that refers to a class of organic compounds characterized by the presence of a hydroxyl (-OH) functional group attached to a carbon atom.

The acid-catalyzed addition of alcohols to alkenes is a useful reaction for the synthesis of ethers. It is also a relatively mild reaction, which makes it a good choice for the synthesis of ethers from sensitive compounds.

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Answer:

it is b

Explanation:

mid ocean ridge diverges meaning it moving in two different direction horizontally - left to right

The mass of the ice melted by a complete transfer of thermal energy is 83.84 g.

The mass of the ice melted by a complete transfer of thermal energy is calculated by applying the following equation as follows;

Q = mcΔθ

where;

heat gained by the ice = heat loss by the steam

The mass of the ice melted by a complete transfer of thermal energy is calculated;

m x heat of fusion = 12.39g x heat of vaporization

m x 334 J/g = 12.39 g  x 2260 J/g

334m = 28,001.4

m = 28,001.4 / 334

m = 83.84 g

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Answer:

(40 g O) / (15.99943 g O/mol) x (1 mol CaCO3 / 3 mol O) x (100.0875 g CaCO3/mol) =  

83 g CaCO3

So answer D), although three significant digits are not justified.

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In accordance with the diagram, activation energy is the distance from the reactants to the top of the "hill"

The activation energy represented on the energy diagram is as given as

"Activation energy is the distance from the reactant to the top of the hill"

So, the correct option is A.

The minimum amount of energy required by the reactant to do the reaction is called activation energy.

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Answer:

5

Explanation:

Na, N, O, O, O

Answer:

VOLTIC CELL

Battery

Electrical to chemical

Endothermic

Electrolysis

Chemical to electrical

Anode

Anode to cathode

Battery

Endothermic

Negative

Enthomeric

Reduction

Oxidation

ELECTRYOTIC CELL

No

Non spontaneous

positive

negative

yes

Anode

Electrical to chemical

Anode to cathode

Exothermic

positive

oxidation

Electroplating

Chemical to electrical

At a pressure of 537.7 torr, pure water would boil at approximately 95.8°C.

To calculate the boiling point of water at a given pressure, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = (-ΔHvap/R) × (1/T2 - 1/T1)

where:

P1 and P2 are the initial and final pressures,

ΔHvap is the molar enthalpy of vaporization,

R is the ideal gas constant (8.314 J/(mol·K)),

T1 is the initial temperature,

T2 is the final temperature.

In this case, we want to find the boiling point (T2) of water at a pressure of 537.7 torr. The normal boiling point of water is 100°C (373.15 K) at atmospheric pressure (760 torr).

We can rearrange the equation to solve for T2:

T2 = (1/(((-ΔHvap/R) × (1/T1)) + (ln(P2/P1))))

Substituting the given values:

P1 = 760 torr

P2 = 537.7 torr

ΔHvap = 40.7 kJ/mol (convert to J/mol by multiplying by 1000)

R = 8.314 J/(mol·K)

T1 = 373.15 K

T2 = (1/(((-40.7 kJ/mol × 1000 J/kJ) / (8.314 J/(mol·K)) × (1/373.15 K)) + (ln(537.7 torr/760 torr))))

Calculating this expression gives us:

T2 ≈ 95.8°C

Therefore, at a pressure of 537.7 torr, pure water would boil at approximately 95.8°C.

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Answer:

A

Explanation:

[H+] = 10^-pH

[H+] = 1 x 10^-4 M

Answer:

C: Warm water is denser than cold water

Hydrogen peroxide and water both contain the same two elements. The chemical formulas of hydrogen peroxide and water are respectively H2O2 and H2O. The two compounds have distinct physical and chemical properties. Here is a long answer to the question posed.

Water is a colorless, odorless, and tasteless liquid that is vital for life. Its melting point is 0 degrees Celsius, while its boiling point is 100 degrees Celsius. Water is the most prevalent chemical substance on Earth, with about 71% of the planet's surface covered by it.

Hydrogen peroxide is a pale blue liquid that is slightly more viscous than water. It has a bitter taste, and its odor is similar to that of bleach. The boiling point of hydrogen peroxide is 150.2°C, while the melting point is -0.43°C. Hydrogen peroxide is a powerful oxidizer that is commonly used as a bleaching agent, disinfectant, and antiseptic.


Chemical formulas are utilized in a variety of scientific applications, including chemical reactions and product descriptions. An incorrect chemical formula can have significant ramifications, potentially resulting in the formation of a hazardous substance.

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The wavelength that corresponds to photons with greater energy is 320 nm. Hence, option a is correct.

The energy of a photon is inversely proportional to its wavelength according to the equation:

E = hc/λ

where E is the energy of the photon, h is Planck's constant (approximately 6.626 x 10^-34 J·s), c is the speed of light (approximately 3.00 x 10^8 m/s), and λ is the wavelength of the photon.

Since energy (E) is inversely proportional to wavelength (λ), shorter wavelengths correspond to photons with greater energy. Therefore, the photon with a wavelength of 320 nm has greater energy compared to the photon with a wavelength of 530 nm.

The wavelength that corresponds to photons with greater energy is 320 nm (option a).

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You can take physics as a hobby and also it is great that you like both maths and physics as well and also want to be an computer science engineer because physics and math

Answer:

11g/cm^3

Explanation:

The density of a substance can be found By

density= Mass/volume

From the question

mass = 22 g

volume = 2 cm³

22/2=11

We have the final answer as

11 g/cm³

Answer:

11g/cm^3

Explanation:

It is because the constant formula density is

Density=Mass/ Volume.

Therefore we have

Density=22g/2cm^3

Which is = 11g/cm^3

Answer:

Light waves are electromagnetic waves while sound waves are mechanical waves. light waves are transverse while sound waves are longitudinal. sound waves require a material medium to travel, and hence, cannot travel in a vacuum. the speed of light in a medium is constant.

Explanation:

Hope this helped Mark BRAINLIEST!!

Answer:

Explanation:

H m sarav

The fact that there are more possible combinations for amino acids than the number of amino acids themselves implies that each amino acid can be encoded by multiple codons.

A codon is a sequence of three nucleotides in DNA or RNA that corresponds to a specific amino acid.

There are 20 standard amino acids used to build proteins in living organisms. However, there are 64 possible three-letter combinations of nucleotides (4³) that can be used to form codons. This means that on average, there are more than three codons that can encode each amino acid.

To calculate the number of codons per amino acid on average, we divide the total number of codons (64) by the number of amino acids (20). Therefore, the average number of codons per amino acid is 64/20 = 3.2.

This indicates that there is a degeneracy or redundancy in the genetic code, where multiple codons can specify the same amino acid.

For example, the amino acid leucine is encoded by six different codons (UUA, UUG, CUU, CUC, CUA, CUG), while methionine and tryptophan are each encoded by a single codon.

In conclusion, the existence of more possible combinations for amino acids than amino acids themselves means that the genetic code is degenerate, allowing for redundancy and flexibility in protein synthesis.

This redundancy helps to protect against errors in DNA replication and transcription and allows for evolution to occur through the accumulation of genetic variations.

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The pressure equilibrium constant expression for the given reaction 2Li3N(s) ⇌ 6Li(s) + N2(g) is given below: P = (PN2 ) (PLi)6/ (PLi3N )2Where, PN2 is the partial pressure of nitrogen PLi is the partial pressure of lithiumPLi3N is the partial pressure of lithium nitride.

The above formula of pressure equilibrium constant expression gives the ratio of the product of partial pressures of reactants and products to the power of their stoichiometric coefficients. Here, the stoichiometric coefficients of Li3N, Li and N2 are 2, 6 and 1 respectively. The reaction 2Li3N(s) ⇌ 6Li(s) + N2(g) represents the dissociation of lithium nitride to lithium and nitrogen gases. The dissociation reaction is endothermic.

As the temperature increases, the dissociation of lithium nitride increases which results in the formation of more lithium and nitrogen gas molecules. Therefore, the value of the equilibrium constant for this reaction increases with increasing temperature. The above-pressure equilibrium constant expression helps to determine the equilibrium concentration of reactants and products when the reaction is in equilibrium.

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Answer:

uhm i believe its D hope i helped

Explanation:

Answer:

D. decomposers

The transition wavelength, λ is given as:λ = (hc)/Eλ = (6.626 x 10^-34 Js * 3 x 10^8 m/s)/(4.585 x 10^-20 J)λ = 4.325 x 10^-7 mλ = 432.5 nmTherefore, if the length of the box is doubled, the transition wavelength remains the same as 200 nm.

The length of the box, L = 1 nm.The transition wavelength, λ = 200 nm.Formula:For particle in a one-dimensional box,For the energy of a particle in a one-dimensional box,E = (n²h²)/(8mL²)where, n = quantum numberh = Planck's constanm = mass of the particleL = length of the box.1. For the lowest energy level,n = 1E₁ = (n²h²)/(8mL²)E₁ = (1²h²)/(8mL²)E₁ = (1*6.626 x 10^-34 Js)²/(8(9.11 x 10^-31 kg)(1 x 10^-9 m)²)E₁ = 9.170 x 10^-20 JFrom the above formula, the probability of finding the particle within 0.01 nm of the center of the box is given as:ψ(x) = sqrt(2/L)sin(nπx/L)Probability, P = ∫ψ²dxFrom the above formula, the probability of finding the particle within 0.01 nm of the center of the box for the lowest-energy level is given as:P = ∫[sqrt(2/L)sin(πx/L)]²dxP = ∫2/L sin²(πx/L)dxP = 2/L * L/2P = 1Hence, the probability of finding the particle within 0.01 nm of the center of the box for the lowest-energy level is 1.2. For the first excited state,n = 2E₂ = (n²h²)/(8mL²)E₂ = (2²h²)/(8mL²)E₂ = (4*6.626 x 10^-34 Js)²/(8(9.11 x 10^-31 kg)(1 x 10^-9 m)²)E₂ = 3.668 x 10^-19 JFrom the above formula, the probability of finding the particle within 0.01 nm of the center of the box for the first excited state is given as:P = ∫[sqrt(2/L)sin(2πx/L)]²dxP = ∫2/L sin²(2πx/L)dxP = 2/L * L/2P = 0.5Hence, the probability of finding the particle within 0.01 nm of the center of the box for the first excited state is 0.5.3. (i) If the mass of the particle is doubled:For the energy of a particle in a one-dimensional box,E = (n²h²)/(8mL²)The energy of a particle is inversely proportional to the mass of the particle.Hence, if the mass of the particle is doubled, the energy of the particle is halved.E = 0.5E₁ = 0.5(9.170 x 10^-20 J)E = 4.585 x 10^-20 JFrom the above formula, the transition wavelength, λ is given as:λ = (hc)/Eλ = (6.626 x 10^-34 Js * 3 x 10^8 m/s)/(4.585 x 10^-20 J)λ = 4.325 x 10^-7 mλ = 432.5 nm(ii) If the charge of the particle is doubled:The charge of the particle does not have any effect on the energy of the particle.Therefore, the transition wavelength, λ remains the same as 200 nm.(iii) If the length of the box is doubled:For the energy of a particle in a one-dimensional box,E = (n²h²)/(8mL²)The energy of a particle is inversely proportional to the length of the box.Hence, if the length of the box is doubled, the energy of the particle is halved.E = 0.5E₁ = 0.5(9.170 x 10^-20 J)E = 4.585 x 10^-20 JFrom the above formula, the transition wavelength, λ is given as:λ = (hc)/Eλ = (6.626 x 10^-34 Js * 3 x 10^8 m/s)/(4.585 x 10^-20 J)λ = 4.325 x 10^-7 mλ = 432.5 nmTherefore, if the length of the box is doubled, the transition wavelength remains the same as 200 nm.

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Answer:

[tex]d=1.78\times 10^{-7}\ Dg/\mu m^3[/tex]

Explanation:

Given,

The density of Helium is [tex]1.78\times 10^4\ g/cm^3[/tex]

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

[tex]d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3[/tex]

So, the density of Helium is equal to [tex]1.78\times 10^{-7}\ Dg/\mu m^3[/tex].