why do chemical equations have to be equal

Answer:

Bottle is sitting on the shelf for 15 days.

Explanation:

Given data:

Co-60 half life = 5 days

Total amount = 30.0 g

Amount left = 3.75 g

Time taken = ?

Solution:

AT time zero = 30 g

AT first half life= 30g /2 = 15 g

At 2nd half life = 15 g/ 2 = 7.5 g

At 3rd half life = 7.5 g/2 = 3.75 g

Now we will calculate the sitting time of bottle.

Half life = Time taken / number of half lives

3× 5 days = time taken

Time taken = 15 days

Bottle is sitting on the shelf for 15 days.

Ice in a 25.00-gram sample melts at 0.0 °C and subsequently heats to 20.0 °C. The energy taken in is 10.5 KJ. The heating curve in sections 2 and 3 illustrates this issue.

we know that,

melts  25.00g   335J    = 8375 J

                         -1g

warm t=mC\(_{p}\)ΔT=25.00g×4.184 J/g°C×20.0 °C=2092 J

Add together = 10467J=10.47 KJ≅10.5 KJ

Energy is the ability to conduct work in physics. Additionally, there is heat and work, which is energy moving from one body to another. Energy is always identified in accordance with its nature once it has been transferred. Therefore, the heat transported may manifest as thermal energy while work performed may result in mechanical energy.

Motion is a property of all energies. Anybody in motion, for instance, possesses kinetic energy. Even while at rest, a tensioned object like a spring or bow has the capacity to move; this is because of the way it is built.

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Explanation:

more predators, less deers. also starvation, and diseases.

The mass of calcium phosphate, Ca₃(PO₄)₂ that can be produced from the reaction of 3.75 g of calcium chloride with 4.25 g of potassium phosphate is 3.11 g

Let us begin by obtaining the limiting reactant. This obtained as follow

3CaCl₂ + 2K₃PO₄ → Ca₃(PO₄)₂ + 6KCl

From the balanced equation above,

333 g of CaCl₂ reacted with 424 g of K₃PO₄

Therefore,

3.75 g of CaCl₂ will react with = (3.75 × 424) / 333 = 4.77 g of K₃PO₄

From the above calculation, we can see that a higher amount (i.e 4.77 g) of K₃PO₄ than what was given (i.e 4.25 g) is needed to react with 48 g of CaCl₂

Thus, the limiting reactant is K₃PO₄

Finally, we shall determine mass of calcium phosphate, Ca₃(PO₄)₂ produced. Details below:

3CaCl₂ + 2K₃PO₄ → Ca₃(PO₄)₂ + 6KCl

From the balanced equation above,

424 g of K₃PO₄ reacted to produce 310 g of Ca₃(PO₄)₂

Therefore,

4.25 g of K₃PO₄ will react to produce = (4.25 × 310) / 424 = 3.11 g of Ca₃(PO₄)₂

Thus, the mass of calcium phosphate, Ca₃(PO₄)₂ produced is 3.11 g

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The results of the separation using two-dimensional gel electrophoresis reveal the distribution and abundance of proteins in a sample.

Two-dimensional gel electrophoresis is a powerful technique used to separate complex mixtures of proteins based on their isoelectric point (pI) and molecular weight. The first dimension of this technique involves isoelectric focusing (IEF), where proteins are separated based on their charge. A pH gradient is established across the gel, and when an electric field is applied, proteins migrate towards the pH region where their net charge is zero, resulting in their separation according to their pI.

In the second dimension, the proteins from the first dimension gel are placed on top of a polyacrylamide gel, which is then subjected to sodium dodecyl sulfate-polyacrylamide gel electrophoresis (SDS-PAGE). In SDS-PAGE, proteins are separated based on their molecular weight. The proteins from the first dimension gel are now distributed along a single axis according to their pI and separated further by size during electrophoresis.

The resulting gel displays a complex pattern of spots, each representing a specific protein in the sample. By comparing the protein patterns obtained from different samples or conditions, researchers can identify changes in protein expression, post-translational modifications, or protein interactions. These results can provide insights into cellular processes, disease mechanisms, and biomarker discovery.

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a) The BOD5 of the wastewater can be calculated as follows:

Initial DO - Final DO = BOD5
9.0 mg/l - 4.0 mg/l = 5.0 mg/l (BOD5)
b) The ultimate carbonaceous BOD can be estimated by assuming that all the BOD has been utilized. Therefore, it is equal to the BOD5 value.
Ultimate carbonaceous BOD = BOD5 = 5.0 mg/l
c) The amount of BOD remaining after 5 days can be calculated as follows:
Initial DO - DO after 5 days = BOD remaining
9.0 mg/l - 2.0 mg/l = 7.0 mg/l (BOD remaining after 5 days)
d) To estimate the reaction rate constant k, we can use the first-order rate equation:
BODt = BOD5 * e^(-kt)
where BODt is the BOD remaining at time t, and e is the base of the natural logarithm.
Using the data at day 30:
2.0 mg/l = 5.0 mg/l * e^(-k*30)
k = 0.0461 (1/day)
Therefore, the estimated reaction rate constant k is 0.0461 (1/day).
A BOD test was conducted using a mixture of 30 mL wastewater and 270 mL dilution water, with a nitrification inhibitor added. The initial DO was 9.0 mg/L.
a) The BOD5 of the wastewater is calculated by subtracting the DO after 5 days (4 mg/L) from the initial DO (9.0 mg/L), resulting in a BOD5 of 5 mg/L.
b) The ultimate carbonaceous BOD can be determined by subtracting the DO after 30 days (2 mg/L) from the initial DO (9.0 mg/L), giving a value of 7 mg/L.
c) The amount of BOD remaining after 5 days can be determined by subtracting the BOD5 from the ultimate carbonaceous BOD (7 mg/L - 5 mg/L), which equals 2 mg/L.
d) To estimate the reaction rate constant k (1/day), more data points are needed. Based on the information provided, a reliable estimation of k cannot be made.

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Answer:

sorry if this is wrong bt j would say rising sea levels

The unhybridized pz orbitals on each carbon overlap to a π bond (pi).The sigma bond framework of the ethylene molecule is produced by the overlap of hybrid orbitals or by the interaction of a hybrid orbital and a 1s hydrogen orbital.

Each carbon still has its unhybridized pz orbital, though. Sigma bond are typically the only types of single bonding between atoms. One sigma bond and two pi bonds make up triple bonds. One sigma () bond makes up a single bond, one and one pi () bond makes up a double bond, and one and two bonds make up a triple bond.

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Answer:

5.399772727e22 atoms/particles or 5.39 x 10^22 atoms/particles

Explanation:

so to convert you’re going to convert grams to moles and moles to atoms/particles:

grams to moles:

23g CF4 x 1 mol

———— = 0.26136 mol

88g (molar mass of CF4)

moles to atoms/particles:

[2.066 • 10^23 is avogadro's number and is always used to convert to atoms/particles]

0.26136 mol x 2.066 • 10^23

———————

1 mol

this gives you 5.399772727e22 atoms/particles

so 5.399772727e22 atoms/particles or 5.39 x 10^22 atoms/particles for scientific notation

hope this helps :)

The arrows that represent the phase transitions in which heat energy is gained is : (A). 1, 2 and 3

Phase transitions can be defined as the changing of matter from one form to another due to the addition or gaining of heat or due to the removal or loss of heat.

Phase transitions in matter are majorly between three phases which are: Solid, Liquid and Gas.

In conclusion, The arrows that represent the phase transitions in which heat energy is gained is : (A). 1, 2 and 3

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Answer:

3 m/s

Explanation:

I found the unit rate which is 9/3 = 3

The concentration of lead in a solution that contains 193.0 mg pb per 2.35 kg of solution is equals to the 82.13 ppm.

We have, a solution with presence of lead, pb. This solution contains 193.0 mg pb per 2.35 kg of solution. We have to determine the concentration of lead in solution in ppm. The mass fraction of a substance in a mixture/solution is the ratio of the mass of that substance to the total mass of the mixture. Formula of mass fraction is

Mass Fraction (x) = Mass of Salute/Mass of Salution (0)

Now, Mass of lead, pb = 193.0 mg

Mass of solution = 2.35 kg

So, mass fraction, y = 193 × 1/10³ g/2.35 × 10³g

= 82.13/10⁶

Now in ppm , y× 10⁶ ppm

= 82.13 ×10⁶/10⁶ ppm

= 82.13 ppm

Hence, required concentration is 82.13 ppm.

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The law of conservation states that the mass of the element remains the same during the course of the reaction.

The law of reaction conformed by the following:-

According to the question, the amount of carbon is 15g in the reactant and the amount will remain same at the end of reaction, which is product that is 15g.

Hence, the amount remain same in the reaction that is 15g.

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The frequency of a red laser that has a wavelength of 676 nm would be 4.43 x \(10^{14\) hertz.

The frequency and wavelength of a wave are related by the following equation:

λf = c

Where λ is the wavelength of the wave in meters, f is the frequency in Hertz, and c is the speed of light in a vacuum.

in this case, λ = 676 nm = 6.76 x \(10^{-7\) m

c = 299,792,458 m/s

Making f the subject of the formula:

f = c/λ

  = 299,792,458/6.76 x \(10^{-7\)

  = 4.43 x \(10^{14\) hertz

In other words, the frequency of a red laser that has a wavelength of 676 nm would be 4.43 x \(10^{14\) hertz.

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Answer:

True

Explanation:

Permanent artificial, temporary artificial and natural. They are classified by the way in which they have achieved magnetism and by how long they remain magnetic.

Answer:51.1%

Explanation:

Mass percent : It is defined as the mass of the given component present in the total mass of the compound. Formula used : First we have to calculate the mass of  and . Mass of  = 18 g/mole Mass of  = 7 × 18 g/mole = 126 g/mole Mass of  = 246.47 g/mole Now put all the given values in the above formula, we get the mass percent of  in . Therefore, the mass percent of  in  is, 51.1%

Answer: Mechanical Energy

Explanation:

    To produce electricity, a generator needs an input of mechanical energy. Mechanical energy is a combination of potential energy and kinetic energy. Generators use this to create electricity to power light bulbs, technological devices, and more!

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On the basis of their chemical and physical characteristics, the elements in the periodic table can be divided into groups or families.

Chemists can predict how elements will interact based on their positions in the periodic table by understanding elemental families. For example, alkali metals such as sodium and potassium are extremely reactive because they have a single valence electron.

Alkali metals, alkaline earth metals, halogens, noble gases, transition metals, and other substances are included in these families.

Because all members of a family have the same number of valence electrons—the outermost electrons involved in chemical bonding—the elements in that family share comparable chemical characteristics.

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When a student needs more of a hazardous material, the recommended procedure is to follow proper safety protocols and guidelines, including consulting with the instructor or supervisor.

Handling hazardous materials requires careful consideration of safety measures to minimize the risks involved. In this situation, the student should first consult with their instructor or supervisor to communicate their need for more of the hazardous material. The instructor or supervisor can provide guidance on the appropriate procedures to follow, including assessing the necessity for the material and ensuring that the student has the necessary training and knowledge to handle it safely.

It is crucial to understand the risks associated with the hazardous material and follow all safety guidelines and protocols. This may include wearing appropriate personal protective equipment (PPE), working in a well-ventilated area, and using proper storage and disposal methods.

Furthermore, it is important to adhere to any legal requirements or regulations regarding the procurement and handling of hazardous materials. This may involve obtaining the necessary permits or licenses, ensuring compliance with safety standards, and maintaining proper documentation.

By following these recommended procedures, students can ensure their safety and the safety of others while obtaining the required amount of hazardous materials.

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Answer:

The Coriolis effect is an apparent deflection in the paths of moving objects caused by Earth's faster speed near the equator than near the poles. In the solar system, the greatest expression of this effect occurs in planetary atmospheres. On Earth, for example, the air around us moves freely above the ground.

Explanation:

Answer:

Two ions that produced similar colors in the flame test were Ca+2 and Sr+2. 3.

Explanation:

The colors are produced when an electron jumps to a higher level and then jump back down.

Ba2+ and Cu2+ and Sr2+ and Li+ were the pair with the similar color. Sr and Li displayed red colors, while Ba and Cu had mild greenish yellowish hues.

Why do some ions in the flame test generate colors that are similar?

The precise sizes of the potential energy jumps differ from metal to metal. As a result, the flame color of each metal will differ due to its unique spectral line pattern. The movement of the electrons in the metal ions contained in the compounds results in the hues of the flame.

The energy released by each electron when it returns to its initial condition determines the hue of the light that is produced.

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Answer:

Explanation:

A covalent bond consists of the mutual sharing of one or more pairs of electrons between two atoms. These electrons are simultaneously attracted by the two atomic nuclei. A covalent bond forms when the difference between the electronegativities of two atoms is too small for an electron transfer to occur to form ions.

Answer:

175 meters

Explanation:

Applying,

v = d/s............... Equation 1

Where v = velocity of the arrow, d = distance traveled by the arrow, t = time

make d the subject of the equation

d = vt............. Equation 2

From the question,

v = 35 m/s, t = 5 seconds.

Substitute these values into equation 2

d = (35×5)

d = 175 meters.

Answer: 0.27 M

Explanation:

molarity = (0.350)/(1.30) = 0.27 M

The acid that contains three hydrogen ions are called triprotic acid.

White solid pure anhydrous phosphoric acid melts into a viscous liquid at 42.35 degrees Celsius. Phosphoric acid, which contains three ion hydrogen atoms, acts as a triprotic acid in an aqueous solution. The hydrogen ions disappear one at a time.

H₃PO₄ (aq)  ⇄  H⁺ (aq) + H₂PO₄⁻ (aq)          Kₐ₁ = 7.5 × 10⁻³

H₂PO₄⁻ (aq)  ⇄  H⁺ (aq) + H₂PO₄²⁻ (aq)       Kₐ₂ = 6.2 × 10⁻³

HPO₄²⁻ (aq)  ⇄  H⁺ (aq) + PO₄³⁻ (aq)            Kₐ₃ = 7.5 × 10⁻³

The first dissociation constant of phosphoric acid indicates that it is not an especially potent acid. It is a weaker acid than hydrochloric acid and sulfuric acid, but stronger than acetic acid. The ease with which each subsequent dissociation stage happens decreases. As a result, H₂PO₄⁻  is a relatively weak acid but HPO₄²⁻ is an extremely weak acid.

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Tetrahedral arrangement for electron pair geometry is more stable for methane when compared between tetrahedral and square planar.

In comparison, the square planar arrangement is not as stable as the tetrahedral arrangement. The reason behind this is as each C-H bond (molecular orbital) can be defined as a region of high electron (negative charge) density. It is just like repelling where each bond will repel the others and will move as far away from the other bonds as much as they could.

In a methane molecule, the carbon atom forms covalent bonds with four hydrogen atoms that are not similar to one another and there are no lone pairs on the carbon atom. Hence, the four hydrogen atoms are at the four corners of the tetrahedron, and carbon is at the center.

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Approximately 2.22 x \(10^2^5\) molecules of nitrogen and 6.29 x \(10^2^4\)molecules of oxygen are present in the International Space Station (ISS). If the oxygen and CO2 scrubber system fails, four astronauts on the ISS have approximately 73 hours before the air becomes unbreathable and they risk losing consciousness.

First, we convert the volume of the ISS to liters: 932 \(m^3\) = 932,000 liters.

Next, we calculate the number of moles of each gas using the ideal gas law equation. The pressure is given as 1 atm, and the temperature is 293 K.

For nitrogen:

PV = nRT

(1 atm)(932,000 L) = n(0.0821 L·atm/(mol·K))(293 K)

n ≈ 2.93 x \(10^4\) moles

For oxygen:

PV = nRT

(1 atm)(932,000 L) = n(0.0821 L·atm/(mol·K))(293 K)

n ≈ 8.29 x \(10^3\) moles

Finally, we convert the number of moles to the number of molecules by multiplying by Avogadro's number (6.022 x \(10^2^3\) molecules/mol).

For nitrogen:

2.93 x \(10^4\) moles × 6.022 x \(10^2^3\) molecules/mol ≈ 2.22 x \(10^2^5\) molecules

For oxygen:

8.29 x \(10^3\) moles × 6.022 x \(10^2^3\) molecules/mol ≈ 6.29 x \(10^2^4\) molecules

Therefore, approximately 2.22 x \(10^2^5\) molecules of nitrogen and 6.29 x \(10^2^4\) molecules of oxygen are present in the ISS.

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