If there are 400 radioactive atoms in a sample, how many will remain after four half-lives have passed?
Enter the exact number as your answer. Do not spell out the number.

Answer:

1.02 × 10⁻³ g

Explanation:

Step 1: Given data

Number of moles of titanium (IV) oxide in 1 jelly bean (n): 1.28 × 10⁻⁵ moles

Step 2: Calculate the mass (in grams) corresponding to 1.28 × 10⁻⁵ moles of TiO₂

To convert moles to mass, we need a conversion factor. In this case, it is the molar mass of TiO₂: 79.87 g/mol.

1.28 × 10⁻⁵ mol TiO₂ × 79.87 g TiO₂/1 mol TiO₂ = 1.02 × 10⁻³ g TiO₂

The steady-state concentration of the pollutant in the lake is approximately 0.000528 mg/L.

To find the steady-state concentration of the pollutant in the lake, we need to consider the inflow and outflow rates of the pollutant and the decay reaction rate.

The total inflow rate of the pollutant is given by:

Qin = Qs * Cs + Qw * Cw

The total outflow rate is equal to the flow rate out of the lake, which is the same as the inflow rate:

Qout = Qin

The change in the pollutant concentration in the lake over time can be expressed by the following equation:

dC/dt = (Qin - Qout)/V - K * C

At steady state, the change in concentration is zero (dC/dt = 0). Rearranging the equation, we have:

(Qin - Qout)/V - K * C = 0

Substituting the given values:

(10.0 m^3/s * 20.0 mg/L + 3.0 m^3/s * 229 mg/L)/12×10^6 m^3 - 0.14 per day * C = 0

Simplifying the equation:

(200.0 mg/s + 687.0 mg/s)/12×10^6 m^3 - 0.14 per day * C = 0

887.0 mg/s / 12×10^6 m^3 - 0.14 per day * C = 0

0.0000739 mg/L/s - 0.14 per day * C = 0

0.0000739 mg/L/s = 0.14 per day * C

Solving for C, the steady-state concentration:

C = (0.0000739 mg/L/s) / (0.14 per day)

C ≈ 0.000528 mg/L

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A Hertzsprung-Russell (H-R) diagram shows the relationship between a star's luminosity (brightness) and its surface temperature.

(1) Luminosity: The entire amount of energy released by a star over the course of one unit of time is known as its luminosity. It is frequently compared to a star's brightness but takes into consideration how far away the star is from Earth. Solar luminosities are commonly used to quantify brightness, with our Sun's energy output being one solar luminosity. Brighter stars are often found closer to the top of an H-R diagram while fainter stars are typically found closer to the bottom.

(2) Surface temperature: The heat or energy emitted from a star's outer layers is measured by its surface temperature. It is frequently categorized by the spectral type of the star, which is established by the absorption lines in the star's spectrum. Classes O, B, A, F, G, K, and M (from hottest to coolest) make up the categorization system for spectral types, which span from hot to cold. The horizontal axis of an H-R diagram is often used to depict the surface temperature or spectral type, with the hottest stars on the left and the coldest stars on the right.

By examining the distribution and properties of stars on the H-R diagram, astronomers can gain insights into stellar evolution, classify stars, determine their life stages, and make predictions about their future evolution. Therefore, a Hertzsprung-Russell (H-R) diagram shows the relationship between a star's luminosity (brightness) and its surface temperature.

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Answer:

a: iron is a substance

Explanation:

because its hard and brittle and classified as a metal

A BCC unit cell with a lattice constant, a, of 2.4 Å. The volume atomic concentration of the unit cell is 0.0722 atoms/ų.

To determine the volume atomic concentration of a Body-Centered Cubic (BCC) unit cell, we need to consider the number of atoms present in the unit cell and the volume occupied by the unit cell.

In a BCC unit cell, there is one atom located at the center of the cube and eight atoms at the corners, but each corner atom is shared among eight adjacent unit cells. Therefore, the total number of atoms present in the unit cell is 1.

The volume of a BCC unit cell can be calculated using the formula:

Volume = a³

where "a" is the lattice constant.

Given that the lattice constant, a, is 2.4 Å, we can calculate the volume of the unit cell as follows:

Volume = (2.4 Å)³

Converting the units to cubic angstroms:

Volume = 13.824 ų

Now, to determine the volume atomic concentration, we need to divide the number of atoms (1) by the volume of the unit cell:

Volume Atomic Concentration = Number of Atoms / Volume

Volume Atomic Concentration = 1 / 13.824 ų

The volume atomic concentration of the BCC unit cell is approximately 0.0722 atoms/ų.

Therefore, the volume atomic concentration of the unit cell in a BCC crystal with a lattice constant of 2.4 Šis approximately 0.0722 atoms/ų.

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The complete question is:

A BCC unit cell with a lattice constant, a, of 2.4 Å. Determine the volume atomic concentration of the unit cell.

The speaker's family behaves in a Chinese manner that they always do. Since their behaviour is who they are, they don't feel guilty about it. The speaker's family seems more outgoing, and the minister's family appears more reserved.

The speaker's guilt about her family's Chinese traditions affects how she depicts the meal and her relatives' actions. The Speaker's family behaved as they would in real life. However, Amy Tan (speaker) felt ashamed of her family's noisy chopstick eating. They behaved as if they were the only ones in the room. The minister family is a more reserved one. It's strange how they make an effort to blend in with Chinese family traditions.

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The end of a very long 5-mm-diameter rod is held at 124°C. The surface of the rod is exposed to ambient air at 30°C, with a convective heat transfer coefficient of 100 W/m2K. Determine the temperature at a radial distance of 2.5 mm from the rod's center. The thermal conductivity of the rod is 15 W/mK.b) What is the temperature gradient in the rod at this location?c).

What is the heat flux at this location?The temperature at a radial distance of 2.5 mm from the rod's center is 79.58°C.The solution for this problem can be found by following the steps below:Solution:a) The temperature of the rod, T, can be calculated using the formula for one-dimensional conduction:q/A = -k (d T/d r)whereq is the heat flux,A is the cross-sectional area of the rod,r is the radial distance from the center of the rod,k is the thermal conductivity of the rod,and T is the temperature of the rod.

Taking the boundary condition into account,T(r=0) = 124°CandT(r=2.5 mm) = 30°C, the solution to the differential equation is:T = T0 + (T1 - T0) (r/R)2whereT0 = 30°CT1 = 124°CR = 2.5 mm/2 = 1.25 mmso,T = 30 + (124 - 30) (r/1.25)2 = 30 + 78 (r/1.25)2at r = 2.5 mm,T = 79.58°Cb) The temperature gradient, d T/d r, is given by the derivative of the above equation:d T/d r = 124 (r/1.25)2 / 1.25where d T/d r = 98.72°C/mat r = 2.5 mmc) The heat flux, q/A, is given by the Fourier's law of heat conduction:q/A = -k (d T/d r)whereq/A = -15 (98.72/1000) = -1.48 W/m2at r = 2.5 mmTherefore, the temperature at a radial distance of 2.5 mm from the rod's center is 79.58°C, the temperature gradient in the rod at this location is 98.72°C/m, and the heat flux at this location is -1.48 W/m2.

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Answer:

The reason is carbon's ability to form stable bonds with many elements, including itself. 

Explanation:

This property allows carbon to form a huge variety of very large and complex molecules. In fact, there are nearly 10 million carbon-based compounds in living things!

Answer:

Explanation:

The reason is that carbon is able to form stable bonds with many elements, including itself, which allows carbon to form a huge variety of very large and complex molecules.

1. The answer to the statement above" Is this system at equilibrium " is yes as the reaction rate for both forward and reverse reaction are all equal.

It follows that from the reactions above, the concentration of A above is 1.5M and that of B is also 1.5, the resultant concentration is still 1.5M. This implies that the concentration of the A and B are constant throughout in the equilibrium reaction.

The major reason why the concentrations of B and that of A are the same is simply because both the forward and backward reactions are the same in all areas.

The major significance of equilibrium reactions is that it helps us to understand the direction of chemical reaction

So therefore, we can now confirm from the explanation above that a system is said to be at equilibrium when there is no change in the amounts or concentration of the reactants and products.

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Answer:

sidjnfjdueirb sidisjsiwijejdkdodif

Liquids and gases were differ when pressure is applied to them in a container because gases easily compress when pressure is applied, but liquids don't. Option A is correct.

When a gas is placed in a container and pressure is applied to it, the gas molecules move closer together, which causes the gas to compress. This is because gas molecules are widely spaced and have weak intermolecular forces, allowing them to easily move and flow. In contrast, liquids are composed of molecules that are already closely packed and have stronger intermolecular forces. As a result, applying pressure to a liquid does not significantly compress it.

However, liquids can still expand slightly when pressure is applied. This is due to the fact that liquids are not completely incompressible, and some compression does occur. But compared to gases, the degree of compression is much less.

Hence, A. is the correct option

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--The given question is incomplete, the complete question is

"How do liquids and gases differ when pressure is applied to them in a container? A. Gasses easily compress when pressure is applied, but liquids don't. B. Gases easily expand when pressure is applied, but liquids don't. C. Liquids easily expand when pressure is applied, but gases don't. D. Liquids easily compress when pressure is applied, but gases don't."--

The pH of a solution that is 0.20 M HOCl and 0.90 M KOCl can be determined using the Henderson-Hasselbalch equation.

The pKa of HOCl is 3.1, which means that the pH of the solution should be around 3.1.

In order for the buffer to have pH=pKa, HCl needs to be added. The quantity of HCl required to reach the desired pH can be determined using the Henderson-Hasselbalch equation.

In this case, the quantity of HCl needed to be added to 1.0L of the original buffer to reach pH=pKa would be 0.05 moles of HCl. Adding HCl to the buffer will shift the equilibrium to the left, resulting in increased concentration of HOCl and decreased concentration of KOCl.

This will decrease the pH of the buffer and bring it closer to the desired pH=pKa.

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Answer:

490

98 for 1 mole, Hence for 5 moles 5 X 98 =490.

Explanation:

Brainliest please?

If you combine the results of this question with the results from question 3B, what do you know about the total pH change caused by adding the last 0. 1 mL of HNO3 it is now an acid.

A chemical that gives off hydrogen ions in water and forms salts by combining with certain metals. Acids have a sour taste and turn certain dyes red. Some acids made by the body, such as gastric acid, can help organs work the way they should. An example of an acid is hydrochloric acid.Nitric acid is a nitrogen oxoacid of formula HNO3 in which the nitrogen atom is bonded to a hydroxy group and by equivalent bonds to the remaining two oxygen atoms.

Nitric acid is made by reaction of nitrogen dioxide ( NO 2) with water. The net reaction is: 3 NO 2 + H 2O → 2 HNO 3 + NO. Normally, the nitric oxide produced by the reaction is reoxidized by the oxygen in air to produce additional nitrogen dioxide.

pH may be defined as a measure of free acidity. More precisely, pH is defined as the negative log of the hydrogen ion concentration. The range of pH extends from zero to 14. A pH value of 7 is neutral, because pure water has a pH value of exactly 7.

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Answer:

\(\large \boxed{\text{21.6 L}}\)

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

         C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:      24.5

(a) Moles of C₆H₁₂O₆

\(\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}\)

(b) Moles of CO₂

\(\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}\)

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\(\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}\)

Answer:

D.) The nuclear model describes the atom as a dense, positive nucleus surrounded by a cloud of negative electrons.

Answer

The nuclear model describes the atom as a dense, positive nucleus surrounded by a cloud of negative electrons.

The volume of 3M from making stock solution is 7.5 liters.

To make a 3M solution from a 6M stock solution of AgNO3, we need to add water to the stock solution. Let us assume x liters of 3M solution from 5 liters of the 6M stock solution.

The amount of AgNO3 in the 5 liters of the stock solution is:

5 L * 6 mol/L = 30 mol AgNO3

To make a 3M solution, we need:

3 mol/L * x L = 3x mol AgNO3

We can use the formula:

C1V1 = C2V2

where,

C1 = concentration of the stock solution

C2 = concentration of the final solution

V1 = volume of stock solution

V2 = volume of the final solution

In this case, we know that:

C1 = 6 M

C2 = 3 M

V1 = x

V2 = 5 L + x

In this case, we know that:

C1 = 6 M

C2 = 3 M

V1 = x

V2 = 5 L + x

let us substitute the above values in the formula

6 M * V1 = 3 M * (5 L + x)

Solving for x, we get:

x = 2.5 L

Therefore, to make a 3M solution from a 6M stock solution, a final volume of 5 + 2.5 = 7.5 liters to get a 3M solution.

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Answer:

Steel often melts at around 1370 degrees C (2500°F)."

Explanation:

your welcome

Answer:

Steel often melts at around 1370 degrees C (2500°F)."

Explanation:

Answer:

see the attached file!!!!!

Answer:

Carbon compounds are defined as chemical substances

Explanation:

Carbon compounds are defined as chemical substances containing carbon. More compounds of carbon exist than any other chemical element except for hydrogen.

The intermolecular forces present in CH\(_3\)NH\(_2\) include dipole-dipole, dispersion, and hydrogen bonding.

Intermolecular forces are the attractive forces between molecules. The types of intermolecular forces present in a compound depend on its molecular structure and polarity.

Analyzing the options:

I. Dipole-dipole forces: CH\(_3\)NH\(_2\) is a polar molecule due to the electronegativity difference between nitrogen and hydrogen. Dipole-dipole forces arise from the attraction between the positive end of one molecule and the negative end of another molecule. CH\(_3\)NH\(_2\) exhibits dipole-dipole forces.

II. Ion-dipole forces: CH\(_3\)NH\(_2\) does not contain ions; therefore, ion-dipole forces are not present.

III. Dispersion forces: Dispersion forces, also known as London dispersion forces, occur in all molecules and arise from temporary fluctuations in electron distribution. CH\(_3\)NH\(_2\) molecules can experience dispersion forces.

IV. Hydrogen bonding: Hydrogen bonding occurs when a hydrogen atom is bonded to a highly electronegative atom (such as nitrogen, oxygen, or fluorine) and interacts with another electronegative atom. CH\(_3\)NH\(_2\) contains a nitrogen atom bonded to hydrogen atoms, allowing for hydrogen bonding.

Based on the analysis, the intermolecular forces present in CH\(_3\)NH\(_2\) are dipole-dipole, dispersion, and hydrogen bonding. Therefore, the correct option is E: I, III, and IV.

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Answer:

No

Explanation:

There is only one Nitrogen on the left side while there are 2 on the right. Also there is zero oxygen on the left side and two on the right. The amounts of hydrgen arent equal to each side either.

Potassium hydrogen phthalate is primary standard.

When standardizing a strong base titrant, the potassium hydrogen phthalate, or KHP, is frequently utilized as the principal standard.

The most typical application of phenolphthalein indicator occurs in titrations where KHP is employed as such an acidimetric standard to also standardize base solution.

An exceedingly pure, stable, devoid of water or hydration, high molecular weight reagent is just a fundamental standard. Sodium carbonate, potassium hydrogen iodate, potassium dichromate, oxalic acid, and various others are some primary standards.

Therefore, Potassium hydrogen phthalate is primary standard.

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Answer:The answer is group 1 and alkali metals

Explanation:The elements in group one are alkali metals and have 1 valance electron

(a) The decay rate of radioisotopes decreases with time according to the exponential decay formula:

decay rate = initial decay rate × (1/2)^(time/half-life)

given Assign a value.

35.0 decay/sec = initial decay rate × (1/2)^(7000 years / 5730 years)

Solving for the initial decay rate, the wood sample when the tree was felled had about 67.8 decay/sec.

(b) The total nuclear binding energy can be calculated using the formula:

Binding Energy = Nuclear Mass - (Proton Mass + Neutron Mass)

We need the nuclear mass. Find the proton mass and neutron mass. The mass of the atomic nucleus can be obtained from the atomic mass listed in the periodic table. For 14C, it is 14.003241 amu. The masses of protons and neutrons can be found by looking them up in the table of atomic masses. A proton has a mass of 1.007276 amu and a neutron has a mass of 1.008665 amu.

Using these values, the total binding energy of 14C can be calculated. Converting to meV, we find that the total binding energy of 14C is about -7.8 meV.

(c) Energy is released when two 7Li nuclei collide to form a 14C nucleus. This can be calculated using the formula:

q value = (final nuclear mass - initial nuclear mass) × c^2

where c is the speed of light.

Substituting the values ​​given in the question, we get: We can see that the emitted energy is about 17.3 mev.

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1.5 moles of \(O_{2}\) gas

Reaction of Al with \(O_{2}\) to form \(Al_{2} O_{3}\) is given below...

2 Al + 1.5 \(O_{2}\) → \(Al_{2} O_{3}\)

Here, 2 moles of Aluminum reacts with 1.5 moles of \(O_{2}\)

atomic weight of aluminum = 27 g / mole

so, 1 mole of aluminum contains 27 g

then how many moles are there if it has 54 g in it.

moles of aluminum in 54 g = 54/27 = 2 moles

As per the equation mentioned above,

2 moles of aluminum reacts with 1.5 moles of \(O_{2}\).

∴1.5 moles of oxygen gas (02) are needed to completely react with 54.0 g of aluminum.

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1.21 mole are there in  7.3 x ×10²³ molecules. It is established that the mole consists of precisely 6.022×10²³ elementary components.

The International System of Units uses the mole (symbol: mol) as the unit of material quantity (SI). How many basic entities of a certain substance are present in an item or sample is determined by the quantity of that material.

It is established that the mole consists of precisely 6.022×10²³ elementary components. An elementary thing may consist of an atom, an molecule, and ion, a ion pair, or perhaps a subatomic particle like an electron dependent on what the material is.

mole = number of particles / 6.022×10²³

       = 7.3 x ×10²³/ 6.022×10²³

      =1.21 mole

Therefore, 1.21 mole are there in  7.3 x ×10²³ molecules.

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The initial temperature in degrees Celsius, of the gas, given that the gas has volume of 0.219 L when the temperature is 30 °C is 182.19 °C

From the question given above, the following data were obtained:

Now, we can obtain the initial temperature of the gas by using the Charles' law equation as shown below:

V₁ /  = V₂ / T₂

0.329 / T₁ = 0.219 / 303

Cross multiply

T₁ × 0.219 = 0.329 × 303

Divide both side by 130

T₁ = (0.329 × 303) / 0.219

= 455.19 K

Subtract 273 to obtain answer in °C

= 455.19 - 273 K

= 182.19 °C

Thus, from the above calculation, we can conclude that the initial temperature is 182.19 °C  

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