Which radioactive sample would contain the greatest remaining mass of the radioactive isotope over 10 years: 2.0 grams of Au-198; 2.0 grams of K-42; 4.0 grams of P-32; or 4.0 grams of Co-60? Why?

It takes a sample of CO₂ 23.3 seconds escape from a container. it will take the same volume of NO₂(g) to escape from the same container is 23.4 sec.

The effusion rate gas is inversely proportional to the square root of molecular mass.

Rate ∝ √1/M

for the same volume of gases,

t1 / t2 = √ M1 / M2

molar mass of  CO₂ = 44 g/mol

molar mas of  NO₂ = 46 g/mol

23.3 / t2 = √ 44 / 46

23.3 / t2 = 6.6 / 6.7

23.3 / t2 = 0.9

t2 = 23.4 sec

Thus, It takes a sample of CO₂ 23.3 seconds escape from a container. it will take the same volume of NO₂(g) to escape from the same container is 23.4 sec.

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another name for a maggot is a worm

The pressure that must be applied to increase the dentisy by 0.10% is 453.48 atm.

It's important to note that isothermal compressibility simply means the fractional differential change that can be seen in the volume as a result if the change in pressure.

It should be noted that the isothermal compressibility formula will be:

= 1/v (dv/dp)

where v = volume

p = pressure

dv = 0.10% = 0.001

This will be:

2.21 × 10^(-6) = 0.001/dp

dp = 0.001 / 2.21 × 10^(-6)

dp = 452.48 atm

New pressure = 1 atm + 452.48 atm

= 453.48 atm

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Complete question

The isothermal compressibility of lead at 293 K. Calculate the pressure that must be applied to increase the dentisy by 0.10%.

+571.6 KJ and 1803.03 J/K are  the entropy (S) of system and the change of S of surrounding.

2H2O(l) --->2H2(g) + O2(g)

first step is to get dHrxn = dHf(Products) - dHf(reactants)

take dHf of reactants and products from standard thermodynamic table

dHf(H2O) = -285.8 KJ/mol

dHf(H2) =0 kj/mol

dHf(O2)= 0 kj/mol

dHrxn = 2dHf(H2) + dHf(O2) - 2dHf(H2O)

  = 2*0 + 0 - 2 [-285.8] kj =  +571.6 KJ

this is for 2 mole of H2O

for 1.88 mols H2O, dHrxn =  +571.6 KJ/2mol×1.88 mol = 537.304 kj

T =298 K

dS surrounding = - dHrxn/T = -537304 j/298 K = -1803.03 J/K

Entropy is the measurement of the amount of thermal energy per unit of temperature in a system that cannot be used for productive labor. Entropy is a measure of a system's molecular disorder or unpredictability since work is produced by organized molecular motion.

In an isolated system—one that does not exchange heat or work with its surroundings—spontaneous change for an irreversible process always moves in the direction of rising entropy, according to the Clausius form of the second law. For instance, the ice block and the stove are two components of a standalone system for which the overall entropy rises as the ice melts.

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A gas evolved during a chemical reaction and has a volume of 651 mL at 38 °C and 869 torr.  the volume of this gas in mL at 110 °C and 3.87 atmospheres was 238 mL.

given that :

volume V1 = 651 mL

temperature T1 = 38 °C = 38 + 273 = 311 K

pressure P1  = 869 torr = 1.14 atm

pressure P2 = 3.87 atm

temperature T2 = 110 °C = 110 + 273 = 383 K

volume V2 ?

using ideal gas equation , we get :

P1 V1 / T1 = P2 V2 / T2

(1.14 × 651 ) / 311 = (3.87 × V2 ) / 383

2.38 = 0.010 V2

V2 = 238 mL

Thus, A gas evolved during a chemical reaction and has a volume of 651 mL at 38 °C and 869 torr.  the volume of this gas in mL at 110 °C and 3.87 atmospheres was 238 mL.

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The most reactive alien metal are:  bazinga (Bz) , minajine (Mdj) , madmenium (Mlr) and Dancingstar (Ds).

When newly prepared, polished, or shattered, a metal exhibits a shiny look and quite good electrical and thermal conductivity. Typically, metals are malleable and ductile. The metallic link formed by the atoms or molecules of the metal gives rise to these characteristics.

Therefore, Ernst (E), highho (Hi), terriblum (T), and sississ are the metalloids (Ss). Sississ has the largest atomic mass of any metalloid. Ernst has the lowest atomic mass of any metalloid. Group 14 includes Highho and Terriblum.

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The ratio of the carbon to the rest of the atoms in the compound would be 2.49375 to 3.9901 of hydrogen, and 1.00 of oxygen.

The ratio of an atom to the rest of the atoms in a compound can be determined by knowing the empirical or molecular formula of the compound.

Now, the compound in question contains C, H, and O.

Let's divide the mol f each atom by the mol of the smallest. the smallest in this case is 2.00 moles.

C = 4.9875/2.00 = 2.49375

H = 7.9802/2.00 = 3.9901

O = 2.00/2.00 = 1.00

Thus, the ratio for carbon is 2.49375 to 3.9901 of hydrogen, and 1.00 of oxygen.

The empirical formula of the compound is [tex]C_2H_4 O[/tex].

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The ratio of C if 4.9875 mol C, 7.9802 mol H, and 2.00 mol O are present in the sample is 2.5.

Empirical formula is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical ratio to derive the empirical formula can be calculated by dividing the number of moles of each element by the smallest value.

According to this question, there are 4.9875 mol C, 7.9802 mol H, and 2.00 mol O in the sample. The smallest mole is 2 moles.

Therefore, the empirical ratio of carbon in the compound is 2.5.

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The speed that reflects the average kinetic energy of the molecules is known as the root-mean-square speed. using N2 at 25 °C.

The square root of temperature and the square root of the molar mass are both exactly proportional to the rms velocity. Thus, doubling the rms velocity of the molecules when the temperature of a specific gas is quadrupled.

The root mean square velocity of gaseous hydrogen molecules equals that of oxygen molecules at 47°C at what temperature.

A gas will effuse more quickly the lighter it is; conversely, the heavier a gas is ,the slower it flows, the more. Helium (He) will have the highest rate of effusion since it has the lowest molecular weight (atomic weight, in this example).

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The mass of antimony present in the sample is 1.65 g

The percentage mass of antimony in the sample is 20.7%.

The number of moles of antimony present in the sample is determined from the mole ratio of the reaction.

The mole ratio of the reaction is obtained from the balanced equation of the reaction as follows:

Balanced equation of the reaction:

BrO₃⁻ (aq) + 3 Sb³⁺ (aq) + 6 H⁺ ⟶ Br⁻ (aq) + 3 Sb⁵⁺ + 3 H₂O

moles of KBrO₃ in the 34.8 mL of a 0.130 M aqueous solution of KBrO₃ is determined as follows:

Moles of KBrO₃ = 0.13 * 34.8/1000

Moles of KBrO₃ = 0.004524 moles

1 mole of  KBrO₃ reacts with 3 moles of Sb³⁺

Moles of Sb³⁺ reacted = 0.004524 * 3

Moles of Sb³⁺ reacted = 0.13572 moles

The mass of antimony present = moles * molar mass

Molar mass of antimony = 121.76 g/mol

Mass of antimony = 121.76 * 0.13572

Mass of antimony = 1.65 g

Percentage mass of antimony = (1.65/7.97) * 100%

Percentage mass of antimony = 20.7%

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In the formation of the sodium chloride, there is the loss of one electron from the sodium atom to the chlorine atom.

An ionic bond is the kind of bond that is formed between a metal and a non metal. The metal is known to donate electrons to the non metal. In the actual sense, the compound that is formed is an ion pair because the metal is positively charged while the non metal is negatively charged.

In the case of the sodium chloride, we know that the both elements, sodium and chlorine are highly reactive. Hence the sodium atom would quickly loose electrons to the chlorine atom and then the sodium chloride compound is formed.

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Answer:

Your break would end at 4:18

Explanation:

The volume of the flask for the preparation of 0.231M solution can be 200ml flask. The equation C1V1 = C2V2 is used for the dilution of the samples.

What is dilution?

Dilution is a process in which a solution is made less concentrated by adding some other constituent to it. To dilute a solution more solvent should be added than the solute. The number of ions of solute remains the same but the volume is increased so the concentration decreases.

For the dilution using,

C1V1 = C2V2

5.2M x 8ml = 0.231M x V2

V2 = 5.2M x 8ml / 0.231M

V2 = 180.08ml.

Therefore, 180.08ml of new solution is made through dilution for which me need 200ml flask or higher volume flasks.

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Morphine is used medicinally as a pain reliever. If an aqueous solution of morphine has [H3O⁺] = 1.1 x 10-10 M, the pH of this solution is 9.6.

given that :

[H3O⁺] = 1.1 x 10-10 M

the pH of the solution is given as :

pH = - log ( [H3O⁺] )

pH = - log (  1.1 x 10-10 M )

pH = - (-9.6 )

pH = 9.6

the pH of the given solution is 9.6.

Thus, Morphine is used medicinally as a pain reliever. If an aqueous solution of morphine has [H3O+] = 1.1 x 10-10 M, the pH of this solution is 9.6.

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The reaction is exothermic in nature and the value of q in kJ is -1.12x10^3 kJ

The reaction is 2H₂S(g) + 3O₂(g)→→→→→→2H₂O(g) + 2SO₂(g)

The amount of heat evolved in the reaction = 1.12 X 103 KJ

2 moles of hydrogen sulphide react.

In an is In an endothermic reaction, energy is consumed. an exothermic reaction, energy evolved Thus, the given reaction is exothermic.

Exothermic reactions are those reactions in which the energy is evolved or released by the reaction.

And endothermic reactions are those reactions in which heat or energy is absorbed in the reaction.

Since the reaction is evolving energy.

Hence the reaction is an exothermic reaction.

The value of q is defined as the amount of heat transferred to the system.

Since the heat is going away from the system.

Hence the value of q will be -ve.

And the magnitude of q will be exactly the same as the amount of heat evolved by the reaction.

Hence q = -1.12 X 103 KJ  (-ve sign shows that heat is being evolved i.e exothermic process)

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Since the A digital signal transmits energy in the form of waves, the conclusion that would be best based on this information is option C:  Digital transmission of information is a fast and reliable method.

Signals, both digital and analog, are transferred via electromagnetic waves. The sound you hear or the graphics you see on a screen are the result of changes in frequency and amplitude. The continuous waves that make up analog signals can be of any frequency and amplitude.

Note that Data transfer, also known as digital transfer or digital communication, is the movement of data between two locations. Digital messages from a data source, such as a computer or keyboard, may be the source of the transmitted data. A phone call or a video signal are examples of analog signals that could also be present.

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The formula to find the mass in grams is =  number of moles × Molecular mass

A mole is defined as 6.02214076 × 1023 of the chemical unit, be it atoms, molecules, ions, or others as well. The mole is a convenient unit to use because of the great number of atoms, molecules, or others that are present in any substance.

To find number of moles we use the formula that is weight divided by molecular mass

From this we can find out the mass as well;

Mass = Number of moles × Molecular mass

Mass = 3.6 × 98.070 ⇒ 353.08 grams

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Answer:

1.468 M

Explanation:

The pressure of 8.06 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half and the final volume of gas is 12.09 L

The volume of gas in a flexible container is 8.06 L

The pressure is P1=X atm

Temperature is T1=YK

Let us calculate the volume V2 if Pressure P2=X/3 atm

Temperature T2=Y/2 K

Using the ideal gas equation, we get

P1V1/T1=P2V2/T2

V2=V1P1T2/T1P2

V2=8.06LX atm×Y/2K/YK.X/3 atm

V2=8.06x3/2

V2=12.09 L

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The weight percent of KCN and NaCN in the original sample are;

The weight percent of KCN and NaCN in the original sample is calculated as follows:

The mass of the original mixture of KCN and NaCN = 0.7256 g

The dissociation equation of the ionic compounds in an aqueous solution is given below:

KCN ⇄ K⁺ + CN⁻

NaCN ⇄ Na⁺ + CN⁻

Mass of precipitate of AgCN = 2.5874 g

Moles of AgCN in a 2.5874 g sample = mass / molar mass

molar mass of AgCN = 133.886 g/mol

Moles of AgCN in the sample = 2.5874 / 133.886

Moles of AgCN in the sample = 0.01932 moles

Moles of CN⁻ in the sample = 0.01932 moles

Mass of CN⁻ present in the sample = 0.01932 * 26

Mass of CN⁻ present in the sample = 0.50232 g

Moles of Na⁺ present = 0.01932 / 2 moles

Moles of Na⁺ present = 0.009663 moles

Mass of Na⁺ present = 0.00966 * 23

Mass of Na⁺ present = 0.2222 g

Moles of K⁺ present = 0.01932 / 2 moles

Moles of K⁺ present = 0.00966 moles

Mass of K⁺ present = 0.009663 * 39

Mass of K⁺ present = 0.3768 g

Mass of sample =  0.50232 g + 0.2222 g + 0.3768 g

Mass of sample = 1.10132 g

weight percent of KCN = (0.3768 + 0.50232/2)/1.10132 * 100%

weight percent of KCN = 57%

weight percent of NaCN = (100 - 57)%

weight percent of NaCN = 43%

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There are 5 oxygen atoms in one nitrogen pentoxide. The number of oxygen atoms in one mole of nitrogen pentoxide is 3.01 × 10²⁴.

Any  substance containing 6.022 × 10²³ atoms is called one mole of the substance. This number is called Avogadro number. Hence, one mole of every element contains Avogadro number of its atoms.

One mole oxygen contains 6.022 × 10²³ oxygen atoms. Similarly one mole of a compound contains 6.022 × 10²³  number of its constituent molecules.

Now, the number of oxygen atoms in one mole of N₂O₅ will be 6.022 × 10²³ × 5 = 3.01 × 10²⁴. Therefore, option b is correct.

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Mole measure the number of elementary entities of a given substance that are present in a given sample. 6.02 x 10²³ oxygen atoms are

present in one mole of dinitrogen pentoxide N[tex]_2[/tex]O[tex]_5[/tex]. The correct option is option 4.

The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity of amount of substance.

One mole of any substance contains  6.02 x 10²³ number of  atoms. 6.02 x 10²³ is called Avogadro number. Mole is inversely proportional to Avogadro number.

Therefore,  6.02 x 10²³ oxygen atoms are present in one mole of dinitrogen pentoxide. The correct option is option 4.

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A weak base equilibrium as shown in the equation below.  the effect on Kb if  BH⁺ were added to the solution; the equilibrium constants  is remains unchanged.

The equation is given as :

B (aq) + H2O (l) ⇌ BH⁺ (aq) + OH⁻ (aq)

K = [ BH⁺ ] [ OH⁻] / [B]

the equilibrium constant  for the given reaction is remain unchanged. the equilibrium constant will only depends on the temperature of the reaction.

Thus, A weak base equilibrium as shown in the equation below.  the effect on Kb if  BH⁺ were added to the solution; the equilibrium constants  is remains unchanged.

B (aq) + H2O (l) ⇌ BH⁺ (aq) + OH⁻ (aq)

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The number of grams of oxygen that can be prepared from 8.50g of pottasium chlorate is 3.33grams.

Stoichiometry is the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions (chemical equations).

According to this question, the decomposition of KClO₃ is generally used to prepare small amounts of O₂ in the laboratory.

2KClO₃ → 2KCl + 3O₂

Based on the above equation, 3 moles of oxygen is produced by the decomposition of 2 moles of pottasium chlorate.

8.50grams of KClO₃ is calculated in moles as follows: 8.50g ÷ 122.55 g/mol = 0.069mol of KClO₃.

This means that 0.069 × 1.5 = 0.104 moles of oxygen is produced.

mass of oxygen produced = 0.104 × 32 = 3.33grams.

Therefore, 3.33grams is the mass of oxygen formed.

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The final concentration of the diluted solution of NaF is equal to 1.24 M.

The concentration and the volume of the stock solution or dilute solution can be determined from the following equation:

M₁V₁ = M₂ V₂

where M₁ and V₁ are the concentration and volume of the stock or concentrated solution respectively and M₂ and V₂ are the concentration and volume of the diluted solution.

Given, a stock solution of NaF of concentration, M₁ = 2.4 M

The concentration of the diluted solution of NaF is M₂.

The volume of the concentrated solution of NaF, V₁ = 1.7 L

The volume of diluted solution of NaF, V₂  = 3.3 L

Substitute the value of the concentration and volume of NaF in equation (1):

(2.4)× (1.7) =  M₂ × (3.3)

M₂ = 1.24 M

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We cannot predict whether a spatial object is a star or a planet based on just its mass. However, Jupiter is already proven to be a planet which revolves around the sun. Thus we can't agree to the statement.

Stars are spatial things with light energy formed by gases and dust particles. They are stationary at a position in the universe. They can form groups with other stars called constellations in various shapes which are connected by the force of attraction.

Planets are revolving around a star such as in solar system, where, 8 planets including earth and Jupiter. Jupiter is the biggest planet in the solar system and it is not star.

Therefore, Jupiter is not a star and the statement is false.

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The amount of the concentrated solution and water needed would be 154.5 mL and 345.5 mL respectively.

The problem here can be solved using the simple dilution equation that is expressed as:

[tex]m_1v_1[/tex] = [tex]m_2v_2[/tex]

Where [tex]m_1[/tex] is the molarity of the undiluted solution, [tex]v_1[/tex] is the volume of the underlined solution needed, [tex]m_2[/tex] is the molarity of the solution to be prepared, and [tex]v_2[/tex] is the volume of the solution to be prepared.

In this case:

[tex]m_1[/tex] = 0.7280 M

[tex]v_1[/tex] = ?

[tex]m_2[/tex] = 0.2250

[tex]v_2[/tex] = 500.0 mL

Rearranging the equation to make [tex]v_1[/tex] the subject:

[tex]v_1[/tex] = [tex]m_2v_2/m_1[/tex]

Now, let's substitute the values:

[tex]v_1[/tex] = 0.2250 x 500.0/0.7280

    = 154.5 mL

This means the volume of the concentrated solution needed is 154.5 mL out of 257.4 mL available. Since our final volume is 500.0 mL, the 154.5 mL stock solution will be made up to 500 mL with water.

Thus, the amount of water required would be:

500.0 - 154.5

  = 345.5 mL

In other words, the volume of the concentrated solution needed is 154.5 while the volume of water needed is 345.5 mL.

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The complete question: If you have 257.4 mL of a 0.7280 M potassium carbonate solution and you need 500.0 mL of a 0.2250 M potassium carbonate solution, how much concentrated solution and how much distilled water do you need?

Mole measures the number of elementary entities of a given substance that are present in a given sample. The mass of hydrogen is 20.2gram.

The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity of amount of substance.

Given number of atoms= 6.02 x 10²⁴atoms

we know one mole of any element contains 6.022×10²³ atoms which is also called Avogadro number

mole =given number of atoms ÷ 6.022×10²³(Avogadro number)

Substituting the values

mole=6.02 x 10²⁴÷ 6.022×10²³

mole = 10 mole

Mole = mass ÷Molar mass

Molar mass = 2.02 g/mol

Substituting the given values

10= mass ÷ 2.02 g/mol

Mass = 20.2gram

Therefore, mass of hydrogen is 20.2gram.

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As a pathologist, Joachim is responsible for performing autopsies which means that he will most likely be involved in dissecting human bodies to determine how someone died and is denoted as option C.

This is referred to as a surgical procedure which involves the thorough examination of a dead body by dissection to determine the cause and manner of death so as to know next line of action.

As a pathologist, the bodies have to be cut open and carefully analyzed so as to ensure that the cause of death is determined and also helps to solve crime cases during investigations.

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The value of hydroxide ion concentration, [OH⁻] in the table, given that [H₃O⁺] is 2.0×10⁻⁶ M, is 5.0×10⁻⁹ M

From the defininion of pH, we understood that the relationship between hydronium ion, H₃O⁺ and hydroxide ion, OH⁻ in a solution is given as:

[H₃O⁺] × [OH⁻] = 10¯¹⁴

With the above formula, we can determine the value of the hydroxide ion, OH⁻. Details below:

[H₃O⁺] × [OH⁻] = 10¯¹⁴

2.0×10⁻⁶ × [OH⁻] = 10¯¹⁴

Divide both side by 2.0×10⁻⁶

[OH⁻] = 10¯¹⁴ / 2.0×10⁻⁶

[OH⁻] = 5.0×10⁻⁹ M

Thus, we can conclude that the hydroxide ion concentration, [OH⁻] is 5.0×10⁻⁹ M

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The new volume of a gas 40ml, if 50 mL at 81.0 kPa has its pressure increased to 101.3 kPa.

What is Boyle's Law?

Boyle's law is a gas law that states that the pressure exerted by a gas (with a given mass and constant temperature) is inversely proportional to its volume. In other words, as long as the temperature and amount of gas remain constant, the pressure and volume of a gas are inversely proportional to each other. Boyle's law was proposed in 1662 by the Anglo-Irish chemist Robert Boyle.

Any change in the volume occupied by a gas (at constant quantity and temperature) results in a change in the pressure exerted by it, according to Boyle's law.

In other words, the product of a gas's beginning pressure and volume is equal to the product of its end pressure and volume (at constant temperature and number of moles).

P1V1 = P2V2

Where,

P1 denotes the gas's initial pressure.

V1 denotes the gas's initial volume.

P2 denotes the gas's final pressure.

V2 denotes the gas's final volume.

The pressure-volume relationship provided by Boyle's law can be used to get this phrase. PV = k for a fixed amount of gas kept at a constant temperature. Therefore,

P1V1 = k (starting pressure * starting volume)

P2V2 = k (last pressure * last volume)

∴ P1V1 = P2V2

According to Boyle's Law

p1v1=p2v2

81x50= 101.3xv2

v2 = 140ml

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