Practice 3: What is the mass of the liquid?
What is the volume of the liquid?
What is the density of the liquid?

Answer:

Oxygen

Explanation:

The 1st 2 are compounds

The last one "air" can be used to say oxygen, however, "air" is not on the periodic table of elements.

Oxygen is on the periodic table of elements

The electron orbit, is the path of an electron around the nucleus of an atom.

Explanation:

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Property could be used to separate a mixture of salt and sand is solubility in water

Salt and sand mixture can be separated by filtration followed by an evaporation process

To separate the salt and sugar based on the solubility and the substances is soluble that means it dissolved in solvent and salt is ionic compound that means soluble in water

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1-Bromobutane reacts with water to form 1-butanol according to the following equation overall rate law for this proposed mechanism is rate = k [C₄H₉Br]

The reaction intermediate in the chemical species that is formed in one elementary step and consumed in subsequent step and the slowest step in the reaction mechanism is rate determining step and the rate determining step limit the overall rate and therefore determines the rate law for the overall reaction

Here given reaction

C₄H₉Br(aq) + 2H₂O(l) → C₄H₉OH(aq) + H₃O⁺(aq) + Br⁻(aq)

And the following is a proposed mechanism for the reaction

C₄H₉Br(aq) → C₄H₉⁺(aq) + Br⁻(aq)     slow

C₄H₉⁺(aq) + H₂O(l) → C₄H₉OH₂⁺(aq)   fast

C₄H₉OH₂⁺(aq) + H₂O(l) → C₄H₉OH(aq) + H₃O⁺(aq)   fast

So in that 1st step is the rate determining step means slow step

So rate expressions take the general form: rate = k . [A]a . [B]b

So given equation rate expressions is k [C₄H₉Br]

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Volume of oxygen would be collected at STP is 293.3 ml

V1 = 293 mL

V2 = ?

P1 = 102 kPa

P2 = 100 kPa

T1 = 10°C = ( 273 + 10 ) = 283 k

T2 = 273 k

The Combined Gas Law combines Charles Law, Boyle s Law and Gay Lussac s Law. The Combined Gas Law states that a gas pressure x volume x temperature = constant

The combined gas law is an amalgamation of the three previously known laws which are- Boyle's law PV = K, Charles law V/T = K, and Gay-Lussac's law P/T = K. Therefore, the formula of combined gas law is PV/T = K, Where P = pressure, T = temperature, V = volume, K is constant.

P1V1/T2 = P2V2/T2

102 x 293 / 283 = 100 x V2 / 273

105.6 = 0.36 x V2

105.6/0.36 = 293.3

V2 = 293.3 ml

Hence, volume of oxygen would be collected at STP is 293.3 ml

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Gallium has two naturally occurring isotopes: 69ga with a mass of 68.9256 amu and a natural abundance of 60.11% and 71ga. Using the mass of gallium from the periodic table,  the mass of Ga-71 isotope is 70.9246 amu.

There are two naturally occurring isotopes of gallium: Ga-69 isotope having mass 68.9256 and percentage abundance is  60.11%, let say 'X' be the mass of  isotope - Ga-71.

Its percentage abundance can be calculated as:

%Ga-71=100-60.11=39.89%

The atomic mass of an element can be calculated by adding the atomic masses of its isotopes multiplied by their percentage abundance.

In this case we have

Atomic mass = m(Ga-69)×%(Ga-69) + X ×%(Ga-71)

The atomic mass of Ga is 69.723 amu.

Now putting the values, we will get

69.723 amu = (68.9256amu)(60.11/100) + X (39.89/100)

Hence,

69.723 amu = 41.431amu + + X (39.89/100)

Now rearranging the above equation we get,

X = 69.723 amu -  41.431amu/0.3989 = 70.9246amu

Thus, the mass of Ga-71 isotope is 70.9246 amu.

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The complete question is:

Gallium has two naturally occurring isotopes: 69ga with a mass of 68.9256 amu and a natural abundance of 60.11% and 71ga. use the atomic mass of gallium from the periodic table to find the mass of gallium-71. express the mass in atomic mass units to two decimal.

Answer:

true

Explanation:

The physical and chemical composition of matter determines which wavelength (or color) is reflected.

The total mass of the mixture is 80.5 g. The analysis of the calculation shows that there should be three significant figures in the end result.

This is an addition operation, and when dealing with addition and subtraction operations it is necessary to consider the last significant figure for each of the numbers. The result of the addition/subtraction should have the same last significant figure as the number with the leftmost (or highest) last significant figure. When applied to this calculation, the principle looks like this:

30.05 - the last significant figure is 5, and it is in the hundredths

50.0 - the last significant figure is 0, in the tenths

0.4006 - the last significant figure is 6, in the ten thousandths.

Of the three, the leftmost/highest significant figure is 0 in the tenths, so the end result should also have its last significant figure in the tenths.

30.05 g + 50.0 g + 0.4006 g = 80.4506 g, but we round up to the tenths, and because the numbers after 5 are greater than 0, we have to round up to reach the end result of 80.5 g

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Answer:

Boyle's law

Explanation:

The hydronium ion concentration, [H₃O⁺ ], of the solution is  3.83 * 10⁻⁶.

The hydronium ion concentration of a solution is the concentration in mol/dm³ of the hydronium ions of a solution.

The hydronium ion is formed when a water molecule combines with a hydrogen ion as shown below:

H₂O + H⁺ ---> H₃O⁺

The hydronium ion concentration of a solution is also known as the hydrogen ion concentration of the solution or pH.

The product of the hydronium ion concentration of a solution and the hydroxide ion concentration of a solution gives the ion product of water.

[H₃O⁺ ] * [OH⁻] = Kw

where Kw = 1.0 * 10⁻¹⁴ M²

The hydronium ion concentration of the given solution can be calculated as follows:

[H₃O⁺ ] * [OH⁻] = Kw

[OH⁻] = 2.61 * 10⁻⁹ M

Kw = 1.0 * 10⁻¹⁴ M²

[H₃O⁺ ] = Kw / [OH⁻]

[H₃O⁺ ] = 1.0 * 10⁻¹⁴ M² / 2.61 * 10⁻⁹ M

[H₃O⁺ ] = 3.83 * 10⁻⁶

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Lewis structures are diagrams that illustrate the bonding between atoms in a molecule as well as any lone pairs of electrons that may exist. They are also known as Lewis dot formulae, Lewis dot structures, electron dot structures, or Lewis electron dot structures (LEDs). Lewis structures are diagrams that illustrate the bonding between atoms in a molecule as well as any lone pairs of electrons that may exist. They are also known as Lewis dot formulae, Lewis dot structures, electron dot structures, or Lewis electron dot structures (LEDs). A Lewis structure can represent any covalently attached molecule as well as coordination compounds. who proposed it in his 1916 article The Atom and the Molecule.

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The temperature of 1.585 K is required for 5 moles of Helium at a pressure of 760 torrs and volume of 650 mL.

The Ideal Gas Equation is the equation for hypothetical gases expressed mathematically by combining empirical and physical constants and is also known as the general gas equation.

The ideal gas equation can be represented for gas as:

PV = nRT

Where P is the pressure, V is the volume, n is the amount of ideal gas  (in moles), R is the gas constant and T is the temperature (in Kelvin) of the ideal gas.

Given, the pressure of helium gas, P = 760 torr = 1 atm

The volume of helium gas = 650 ml = 0.65 L

The number of moles of helium gas, n = 5

The value of the gas constant, R = 0.082 L atm /K mol

Substitute the values in the ideal gas equation:

(1atm) × (0.65 L) = (5 mol) ×(0.0832 L atm /K mol) × T

T = 1.585 K

Therefore, the temperature of the helium gas at a given presurre and volume is equal to 1.585 K.

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yes I agree thank you for sharing

When the temperature of a gas decrease the pressure also decrease. That is option B.

Temperature is defined as the quantity that measures the degree of how hot or cold an object or medium is.

When the temperature of a medium increases there would be increase in the collision between the molecules that is found in the medium.

The increase in intermolecular collision leads to increase in pressure of the medium. Subsequently, when there is decrease in temperature, the pressure of the medium also falls.

Therefore, it can be said that the temperature of a gas decreases which causes the pressure to also decrease.

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Answer:

I think "they are held together by chemical bonds"

Two different samples of matter are measured in each of these glass containers. Both of the containers show the same measure for both samples of matter. These samples have volume in common.

In chemistry, volume is defined as the three-dimensional space taken up by matter. Volume is measured in cubic meters.

If you were to put two different samples in a graduated glass container and both of them have the particular digit that shows their volume then it means their volume is the same and they take up the same amount of space irrespective of their chemical composition.

For example, if you pour the sample into a glass container and the number you get is 250 m³.

Then this 250 m³ refers to the volume of the sample.

In general, volume is measured as volume =  mass/density and its SI unit is m³.

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The momentum of the bowling ball is projected upward with a velocity of 2.0 m/s is 10 kg m/s.

The amount of force will that bowling ball have as it falls back to earth is 10 N.

We know that,

Momentum is the product of speed of an object and its mass. This means that an object at rest has zero momentum since the velocity is also zero.

It is given as -

Momentum = Mass of an object x speed/ velocity

Momentum = 5 kg x 2.0 m/s

                   = 10 kg m/s

To calculate amount of force bowling ball have when it falls back to Earth is -

Force exerted by an object is equal to product of its mass and velocity.

The second law of motion states that change in momentum of the object is given by mass multiplied by the velocity, which is equal to force.

F = mv

where,

F = force

m = mass

v = velocity

and,

p = mv

where,

p = momentum

m = mass

v = velocity

Hence, the amount of force that bowling ball have as it falls back to Earth is 10 N.

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This circumstance frequently arises when an enzyme or a solid surface is used to catalyze a process (heterogeneous catalysis).

The unit for the rate of reaction depends on the odder of the reaction .for the zero order reaction a→b,a plot of 1/⊂a⊃vs.time will produce a straight line. the rate of a zero order reaction decreases as the reaction proceeds

This circumstance frequently arises when an enzyme or a solid surface is used to catalyze a process (heterogeneous catalysis).

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Answer:

Explanation:

A

Explanation:

Plenary lecture paper

Combustion dynamics of large-scale wildfires

Author links open overlay panelLiuNaianXieXiaodong

https://doi.org/10.1016/j.proci.2020.11.006

Get rights and content

Under a Creative Commons licenseOpen access

Abstract

A fact often overlooked is that large-scale wildfires, although occurring infrequently, are responsible for the overwhelming majority of fire-related suppression costs, economic losses, and natural resources damages. Fortunately, the increasingly severe problems of large-scale wildfires worldwide have been receiving ever-growing academic attention. The high-intensity burning behaviors in wildfires stem from the significant interaction of combustion with heat transfer and atmospheric flow under complicated fuel, meteorology, and topography conditions. Therefore, mitigating measures against large-scale wildfire disasters have grown into a challenging research focus for combustion scientists. Research over the past century has resulted in incrementally enhanced insights into the mechanisms of combustion dynamics underlying the various erratic behaviors in large-scale wildfires, with theories and models of fire accelerations developed and validated. These advances are expected to improve the efficacy of large-scale wildfire predictions significantly. Nevertheless, the physical interpretation of the acceleration of large-scale wildfires is far from adequate and complete. This paper intends not to make a comprehensive review of the entire wildfire research field, but to depict an overall pattern of the essential factors that lead an initial small-scale spreading flame to a large-scale wildfire beyond control. It is outlined that the complicated transformation of fuel preheating mechanisms determines the growth of surface fire spread, while varied large-size flame fronts and unique spread modes induced in specific fire environments play an essential role in fire spread acceleration. Additionally, multiple fires burning and merging often act as crucial steps for accelerating surface fire spread, generating large-size flames, and triggering unique spread modes. These major potential factors strike the energy balance of a low-intensity wildfire and push it to a high-intensity state. Several issues regarding intensely burning behaviors in large-scale wildfires are selected for in-depth discussions, for which an overview of the progress and challenges in research is presented. It is concluded that the fundamental exploration targeted at developing application tools capable of dealing with large-scale wildfires remains at its early stages. Opportunities for innovation are abundant, yet systematic and long-term research programs are required.

In this situation, wash the separating funnel with soap and water once more before using it for the subsequent experiment and rinse the funnel with acetone into a waste container.

Separating funnels are employed in the extraction process and must be free of contaminants that could reduce the yield of the extraction. The funnel could be placed in a desiccation after being cleaned and rinsed to allow the water on the body to dry before being used for the following extraction procedure. Washing laboratory equipment before use irrespective of its state (whether it is clean or not) contributes to standard laboratory practices.

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Answer:

During metaphase, the cell's chromosomes align themselves in the middle of the cell through a type of cellular "tug of war." The chromosomes, which have been replicated and remain joined at a central point called the centromere, are called sister chromatids.

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Answer: 

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The parts of the chemical equation to the correct description are done below.

Column A              Column B

1. Cl₂ + Ail₃ ----------- c. Reactants

2. I2 + AlCl3. ---------- a. Products

3. 3.(g) or (aq) ---------  d. State of Matter

4. 4.--> -------------------- b. Shows which way the reaction progresses.

Chemical equations tell us the factors and/or compounds that can be reacting and the product(s) of the response. The coefficients on the materials in the response tell us the mole ratio or molecular ratio of the elements/compounds inside the reaction. A chemical equation is the symbolic representation of a chemical response in the form of symbols and chemical formulation.

Chemical equations are an efficient way to explain chemical reactions. This module explains the shorthand notation used to express how atoms are rearranged to make new compounds at some stage in a chemical reaction. It indicates how balanced chemical equations convey the proportions of each reactant and product concerned.

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One of these neurotransmitters, dopamine, is released in the reward area of the brain and results in pleasurable feelings and an uplifted mood. More nicotine is required to feel good the more you smoke.

The nightshade plant family naturally produces nicotine, an alkaloid that is widely used recreationally as a stimulant and anxiolytic. It is a medication meant to decrease withdrawal symptoms in smokers who desire to quit.

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The stages of a cell's lifecycle are prophase, prometaphase, metaphase, anaphase, and telophase.

The cell cycle is a sequence of events that arise in a cell as it develops and splits. Cells expend most of their period in the interphase, where they expand, replicate chromosomes, and prepare for cell division.

The cell then exits the interphase and experiences mitosis to conclude division. The consequential cells, known as daughter cells, enter the interphase and initiate a new cell cycle.

The cell cycle is the name a cell gives to the procedure by which it replicates and produces two new cells.

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Answer:

C. Cl changing to Cl-

Explanation:

Cl gains 1 electron, resulting in an increase in radius.

The other 3 give up electrons.

The balanced chemical equation of the reaction between aluminum metal with aqueous sulfuric acid to form aqueous aluminum sulfate and hydrogen gas is 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

A chemical equation is a symbolic representation of a chemical reaction where reactants are represented on the left, and products on the right.

According to this question, aluminum metal reacts with aqueous sulfuric acid to form aqueous aluminum sulfate and hydrogen gas.

This reaction will be balanced only if the atoms of each element on both sides of the equation are the same. The chemical reaction is as follows:

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

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If the gases are ideal and the end pressure is the same as the initial pressure, the partial pressures are PA=nART/(VA+VB) and PB=nBRT/(VA+VB), respectively.

If the gases are ideal and the end pressure is the same as the initial pressure, the partial pressures are PA=nART/(VA+VB) and PB=nBRT/(VA+VB), respectively.

There are two ways to calculate partial pressures: 1) To determine the individual pressure of each gas in a mixture, use PV = nRT. 2) Determine the proportion of the total pressure that may be assigned to each individual gas using the mole fraction of each gas.

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The given examples depict law of reciprocal proportions.

Jeremias Ritcher put forward the law of reciprocal proportions in 1792. The ratio of the masses in which they do so is either the same or a simple multiple of the mass ratio in which they combine, according to the statement, "If two different elements combine independently with the same weight of a third element."

a. Metal oxide contains

M = 0.46 g.

O = 0.77−0.46 = 0.31 g

Therefore, weight of oxygen that combines with 1g of

Metal = 0.31/0.46 = 0.674 g

b. 0.82g of Metal liberates H₂ = 0.00009 × 760 = 0.068g

Thus, the weight of H₂ that will be liberated by 1g of

Metal = 0.068/0.82 = 0.083 g

According to (a) and (b) the ratio of the weight of O:H that combine with or is displaced by the same weight of metal is 0.674:0.083 or 8:1.

Again according (c ), 1.25g of water contains

O= 1.12g and

H 1.25 − 1.11 = 0.14g

Therefore, The ratio of O:H::1.11:0.14 or 8:1

Hence, that ratio being the same, the law of reciprocal proportions is illustrated.

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