which is considered a population?

Answer:

The half-life of a radioactive substance is the time it takes for half of the substance to decay. In this case, the half-life of Cs-131 is 30 years. After 120 years, which is equivalent to four half-lives (120 years / 30 years/half-life = 4 half-lives), the original mass of Cs-131 would have been halved four times.

Let’s say the original mass of the Cs-131 sample is M. After one half-life, the remaining mass would be M/2. After two half-lives, the remaining mass would be (M/2)/2 = M/4. After three half-lives, the remaining mass would be (M/4)/2 = M/8. And after four half-lives, the remaining mass would be (M/8)/2 = M/16.

Since we know that after four half-lives (120 years), 6.0 g of Cs-131 remain, we can set up an equation to solve for the original mass M: M/16 = 6.0 g. Solving for M, we find that M = 16 * 6.0 g = 96 g.

Therefore, the original mass of the Cs-131 sample was 96 grams.

Explanation:

How close the magnets are together and which direction the poles are pointed to

hibernation exclamation mark

When there are more moles of gaseous products than gaseous reactants in the reaction at equilibrium, it is demonstrated that K\(c\) is smaller than K\(p\).

The relationship between the equilibrium constant (K) for partial pressure (K\(p\)) and the equilibrium constant for concentration (K\(c\)) is as follows,

K\(c\) = K\(p\) (RT)⁻Δn

K  \(p\)= K\(c\) (RT)⁻Δₙ

and here Δₙ = (product - reactant), Δₙ > 0

Therefore, K  \(p\) > K\(c\)

Therefore, it proves that K\(c\) is less than K\(p\) if the reaction has more moles of gaseous products than gaseous reactants at equilibrium.

What is the Equilibrium Constant?

The equilibrium constant of a chemical reaction is found at chemical equilibrium, the state in which, given enough time, a dynamic chemical system approaches and its composition shows no further tendency to change. is equal to the value of the reaction quotient of For a given set of reaction conditions, the equilibrium constants are independent of the initial analytical concentrations of reactant and product species in the mixture. Therefore, given a known equilibrium constant, the composition of a system in equilibrium can be determined from its initial composition. However, the value of the equilibrium constant can be affected by parameters that affect the reaction, such as temperature, solvent, and ionic strength.

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In general, if the change in temperature is proportional to the amount of copper(II) sulfate used, then increasing the amount of copper(II) sulfate used will lead to a corresponding increase in temperature change, and decreasing the amount of copper(II) sulfate used will result in a decrease in temperature change.

Temperature change in substances can be caused by a variety of factors, including the addition or removal of heat energy, changes in pressure, chemical reactions, and phase changes (such as melting, boiling, or freezing).

If this relationship does not hold, it could indicate that other factors are also at play, such as a limiting reagent or a change in reaction conditions. Therefore, comparing the results from parts A and B can help determine whether the change in temperature appears to be proportional to the amount of copper(II) sulfate used or not.

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The complete question is:

Copper(II) Sulfate and Zinc

The reaction between zinc (Zn) and copper sulfate (CuSO₄) is an oxidation-reduction reaction. Zinc reacts with copper sulfate to form zinc sulfate. So, zinc undergoes oxidation. At the same time, copper sulfate reduces to pure metallic copper and so undergoes reduction. Here is the equation for the reaction:

Zn + CuSO₄ → Cu + ZnSO₄.

In this task, you’ll carry out this oxidation-reduction reaction to ±nd the enthalpy of the reaction.

If you need a refresher on using a graduated cylinder and an electronic balance, watch the videos about measuring volume and measuring mass. Before you begin this task, review the lab safety guidelines.

Estimated time to complete: 1 hour

If you’ve purchased an Edmentum lab kit, remove the items that appear in the equipment list. The chemicals are located inside a box within the kit. You’ll also need distilled or tap water and a pen or a fine-tip marker for labeling test tubes.

You’ll need these materials:

Compare the results from parts A and B. Does the change in temperature appear to be proportional to the amount of copper(II) sulfate used? (edmentum lab)

When oil from a tin is poured through a funnel placed tightly into the neck of a bottle, the oil stays in the funnel because the air pressure inside the bottle prevents the oil from entering the bottle.

The force that air, whether compressed or unconfined, applies to whatever surface it comes into touch with is known as air pressure.

The air in the bottles exerts a force on the liquid that is to be poured inside it.

Since the air has no way of escaping, the oil cannot be poured into the bottle.

However, if a little space is allowed by the funnel for the air to escape, the liquid can then be poured into the bottle.

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Answer:

Electronic configurations of three magnesium atoms and two nitrogen atoms.

\({ \sf{.}}\)

Answer:

dude

Explanation:

dude use a calculator, \ USE A CALCULATOR

Answer:

A 33.7L

Explanation:

Hello,

In this case, we can deal with argon as an ideal gas that can be studied via the ideal gas equation:

\(PV=nRT\)

Whereas we have the pressure, volume, moles, ideal gas constant and temperature respectively, so, for us to find the volume we simply solve for it as shown below:

\(V=\frac{nRT}{P} =\frac{3.5mol*0.082\frac{atm*L}{mol*K} *293K}{2.5atm} \\\\V=33.7L\)

Therefore, the answer is A 33.7L.

Best regards.

Answer:

The answer is

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

From the question

Density of aluminum = 2.00 g/mL

volume = 29.8 mL

The mass is

mass = 2 × 29.8

We have the final answer as

Hope this helps you

The electrode classification that identifies a filler metal that can be used in the overhead position is known as an AWS E6010 classification electrode.

AWS E6010 classification electrodeAWS E6010 classification electrode is an electrode classification that identifies a filler metal that can be used in the overhead position. The E6010 electrode classification is ideal for welding thinner sheets of metal, usually, less than 1/8" thick. The welding process is often used in pipes, fuel tanks, and other non-structural welds. The electrodes have a thick and cellulose-based coating, making them ideal for a variety of welding applications.

The E6010 classification electrode is commonly used for root passes and pipe welding. The classification of the electrode indicates how much current is required for proper welding. It also specifies the type of metal that can be welded with the electrode. The classification is also used to identify the filler metal that is used in the electrode, making it easier to determine which electrode is appropriate for a particular welding task.

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Answer:

ΔH = 8.87 x 10₅

Explanation:

q = mcΔT = 20g x 1.00 cal/g°C x 10.6°C = 212 cals x 4.184 j/cals = 8.87 x 10⁵ joules

Answer:

hydrocarbon

Explanation:

these are called hydrocarbons

Answer:

The chemical reaction involving the reduction of ZnO by CO is not feasible thermodynamically: This is because the standard free energy of formation ( ) of CO2 from CO is higher than that of the formation of ZnO from Zn. Thus, CO cannot be used to reduce ZnO to Zn.

Answer:

thermal shock

Explanation:

the temperatures inside the glass jar should have continued to increase over time. Internal stresses due to uneven heating. This is also known as “thermal shock”.

In general, the thicker the glass, the more prone it will be to breaking due to the immediate differences in temperature across the thickness of glass.

Borosilicate glass is more tolerant of this, as it has a higher elasticity than standard silicon glass.

You may also note that laboratory test tubes and flasks are made with thinner walls, and of borosilicate glass, when designated for heating.

The empirical formula of citric acid is C₃H₄O₃.

To determine the empirical formula of citric acid based on its elemental composition, we need to convert the percentages into moles and find the simplest ratio of the elements present.

Given the percentages:

Carbon (C): 37.51%

Hydrogen (H): 4.20%

Oxygen (O): 58.29%

Assume we have 100 grams of citric acid, which allows us to directly convert the percentages into grams.

Carbon (C): 37.51 grams

Hydrogen (H): 4.20 grams

Oxygen (O): 58.29 grams

Next, we calculate the number of moles for each element using their respective atomic masses:

Carbon (C): 37.51 g / 12.01 g/mol = 3.12 mol

Hydrogen (H): 4.20 g / 1.01 g/mol = 4.16 mol

Oxygen (O): 58.29 g / 16.00 g/mol = 3.64 mol

Now, we divide each element's moles by the smallest number of moles (3.12) to obtain the simplest, whole-number ratio:

Carbon (C): 3.12 mol / 3.12 mol = 1

Hydrogen (H): 4.16 mol / 3.12 mol = 1.33

Oxygen (O): 3.64 mol / 3.12 mol = 1.17

Rounding to the nearest whole number, we find the empirical formula of citric acid is:

C₁H₁.₃₃O₁.₁₇

However, since we need to express the formula with whole numbers, we multiply all the subscripts by 3 to get:

C₃H₄O₃

Therefore, Citric acid's empirical formula is C₃H₄O₃.

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The concentration of starch in the original sample is 1 g/L.

The concentration of the original sample is determined as follows:

After diluting the sample 4-fold, the concentration of the starch = 1/4 of the original concentration.

The absorbance of the diluted sample is related to the absorbance of the original sample by the following equation:

A₁ = A₂x dilution factor

where;

Substituting the given values, we get:

A₁ = 0.5 x 4 = 2

Now we can use the Beer-Lambert Law to find the concentration of the original sample:

A = εbc

where

Since we do not know the value of ε or b, we will assume that they are constant and cancel out when we compare two measurements. Therefore:

A₁ / A₂ = c₁ / c₂

Substituting the given values, we get:

2 / 0.5 = c₁ / (c₂ x 4)

Solving for c₁:

c₁ = (2 / 0.5) x (1 / 4)

c₁ = 1 g/L

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The final temperature of the gas, after absorbing 1300 J of heat and doing 2200 J of work, is approximately 375.45 K.

To find the final temperature (T_final) of the gas, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added (Q) minus the work done (W) by the system:

ΔU = Q - W

Since the gas is ideal and monatomic, the change in internal energy is related to the temperature change (ΔT) through the equation:

ΔU = nC_vΔT

where n is the number of moles and C_v is the molar heat capacity at constant volume.

Rearranging the equations and substituting the given values:

nC_vΔT = Q - W

(5.00 mol)(3/2R)ΔT = 1300 J - 2200 J

(5.00 mol)(3/2)(8.3145 J/(mol⋅K))ΔT = -900 J

Simplifying:

(37.9725 J/K)ΔT = -900 J

ΔT = -900 J / (37.9725 J/K)

ΔT ≈ -23.70 K

Since the initial temperature is 126 °C, we convert it to Kelvin:

T_initial = 126 °C + 273.15 = 399.15 K

Now we can find the final temperature:

T_final = T_initial + ΔT

T_final = 399.15 K - 23.70 K

T_final ≈ 375.45 K

Therefore, the final temperature of the gas is approximately 375.45 K.

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Answer:

In equation form, Newton's second law of motion is a=Fnetm a = F net m . This is often written in the more familiar form: Fnet = ma. The weight w of an object is defined as the force of gravity acting on an object of mass m. ... Friction is a force that opposes the motion past each other of objects that are touching.

F = ma, or force equals mass times acceleration, is Newton's second law of motion.

According to the first law, an object's motion will not change unless a force acts on it.

According to the second law, the force acting on an object is equal to its mass multiplied by its acceleration.

When two objects interact, they apply forces of equal magnitude and opposite direction to each other, according to the third law.

His second law states that a force is defined as the change in momentum divided by the change in time. Newton's second law of motion is F = ma, or force equals mass times acceleration.

F → A B = − F → B A. The conservation of momentum is associated with Newton's third law of motion.

Every action must have an equal and opposite reaction, according to the law.

Thus, this is the equation for Newton's second law of motion.

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Na2S2O3 (aq) + 2HCl (aq) ==> SO2(aq) + S(s) + H2O(l) + 2NaCl (aq)    

The balancing equation shows that one of the products of the reaction between aqueous HCl and sodium thiosulfate, which results in a colorless solution, is SOLID sulfur.

The reaction mixture becomes hazy as a result of this. Both of the other byproducts, SO2 and NaCl, are soluble in water.

S in Na2S2O3 has an oxidation number of +2.

S in SO2 has an oxidation number of +4.

S(s) has a zero S oxidation number.

As a result, what is being described as a "disproportionation" reaction is actually a redox reaction.Therefore, a decrease in oxidation state is reduction. Thus, oxygen is being diminished in this situation. As a result, a chemical is simultaneously being reduced and oxidized in a process. Therefore, that is once more referred to as a disproportionation response.

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Answer:

Special type of surface that repels liquid

Explanation:

Answer:

A. Do not change their energy level.

Explanation:

The process in which a solid changes directly to a gas is called sublimation. It occurs when the particles of a solid absorb enough energy to completely overcome the force of attraction between them. Dry ice (solid carbon dioxide, CO2) is an example of a solid that undergoes sublimation.

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CH2Cl2(l) experiences both dipole-dipole interactions and London dispersion forces. These intermolecular forces play a crucial role in determining the physical properties of dichloromethane, such as its boiling point and solubility in polar solvents.

The molecule CH2Cl2(l) is dichloromethane, a polar molecule. It has several types of intermolecular forces present:

1. Dipole-dipole interactions: Due to the presence of polar bonds and an asymmetrical molecular shape, CH2Cl2 exhibits dipole-dipole interactions. These forces arise from the attraction between the positive end of one molecule and the negative end of another. The chlorine atoms are more electronegative than the hydrogen and carbon atoms, creating partial negative charges around the chlorine atoms and partial positive charges around the hydrogen and carbon atoms. These partial charges attract each other, resulting in dipole-dipole interactions.

2. London dispersion forces: Even though CH2Cl2 is a polar molecule, it also experiences London dispersion forces. These forces arise from temporary fluctuations in the electron distribution, creating instantaneous dipoles. These temporary dipoles induce dipoles in neighboring molecules, leading to an attractive force between them. The larger the molecule or the higher the number of electrons, the stronger the London dispersion forces. In the case of CH2Cl2, the chlorine atoms have a greater number of electrons, contributing to the strength of the dispersion forces.

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Answer:oxygen Explanation:The medical condition described here is anaemia. It is a blood cell disorder whereby the red blood cell doesn't function properly and hence doesn't carry enough oxygen to the tissues. This is usually caused when ones body is deficient of iron.The symptoms that may occur to such patients are weakness, fatigue, headache and pale skin.Based on the explanation, the answer is oxygen

Explanation:

Answer:

8,920,000

Explanation:

Answer:8,200,000

Explanation:

The sample of methyl tert-butyl ether has 0.0946 moles of oxygen.

When calculating the amount of oxygen in a sample of methyl tert-butyl ether (MTBE), it is important to take into account the chemical formula \(CH_{3}OC(CH_{3})_{3}.\)

Each molecule of MTBE has one oxygen atom (O), as can be seen from the formula. The sample has 0.0946 moles of carbon (C), according to the information provided.

Since MTBE has a carbon to oxygen content of 1:1, there will be 0.0946 moles of oxygen in total.

Therefore, the sample of methyl tert-butyl ether has 0.0946 moles of oxygen.

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