5 students are working together to build 5 models. All models will represent the molecular structure of sodium bicarbonate, NaHCO3, commonly known as baking soda. The students are using toothpicks to represent bonds and jelly beans to represent atoms.


Red jelly beans represent sodium atoms (Na) White jelly beans represent hydrogen atoms (H) Black jelly beans represent carbon atoms (C) Blue jelly beans represent oxygen atoms (O)


Directions: Identify the number of each color of jelly bean needed to make All five models of NaHCO3. For your answer: Please write down the color of the jelly bean, the element symbol and the number of jelly beans. ex. Y (He) = 3

The empirical formula is the simplest formula for a compound which is defined as the ratio of subscripts of smallest possible whole number of the elements present in the formula. It is also known as the simplest formula.

write a balanced equation for the reaction between tantalum pentachloride and water?

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You would multiple the percentages times the atomic mass and then add each of them together.

Ex. (23.995)(.7899)+(24.986)(.10)+(25.992)(.1101). = 24.3139 amu

199.86 0r 200g/mol molar mass for the compound cabr2

The orbitals or shells that enclose an atom's nucleus are home to electrons. In the nucleus, protons and neutrons are present. In the atom's nucleus, they coalesce. An atom's nucleus is surrounded by electrons in distinct orbits. The molar mass is calculated by dividing the mass (g) of a certain chemical element or chemical compound by the amount of the substance (mol). The standard atomic masses (in g/mol) of the component atoms can be added to determine the molar mass of a compound.

Thus, the molar mass of CaBr2 can be calculated as shown below.

Molar mass=(1×40.078gmol+(2×79.904gmol)

=40.078gmol+159.808gmol

=199.886gmol

Hence, the molar mass of CaBr2 is 199.886gmol.

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The mass of the water can be obtained as 31 g.

I would want us to have the first law of thermodynamics at the back of our minds. I said this because the energy that have been lost by the brass is the energy that the water is going to have to gain and this is the first law of thermodynamics.

Thus we have;

H = mcdT

m = mass of the water

c = Specific heat capacity of water

dT = temperature change

Then we have;

2690 = m * 4.18 * (42 - 21)

m = 2690/4.18 * (42 - 21)

m = 31 g

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The initial voltage of the electrochemical cell would be less than the voltage under standard conditions.

An electrochemical cell is a device that converts chemical energy into electrical energy, or vice versa, through a redox reaction. The voltage of an electrochemical cell can be determined by measuring the potential difference between the two half-cells that make up the cell. The voltage of an electrochemical cell is determined by the half-cell potentials and the concentration of the ions in the solutions. The half-cell potential of a metal is the potential of a metal electrode in a solution of its ions at standard conditions, which is defined as a concentration of 1 M and a temperature of 25°C. In the case of zinc, the half-cell potential is -0.76 V, and for copper, the half-cell potential is +0.34 V. In the given cell, one half-cell is made of a strip of pure zinc in 500 milliliters of 0.10 M ZnCl2 solution, and the other half-cell is made of a strip of pure copper in 500 milliliters of 0.010 M Cu(NO3)2 solution. The concentration of the ions in the solutions is not at the standard concentration of 1 M, which means that the voltage of the cell will be different from the standard voltage. A deviation from the standard concentration results in a deviation from the standard potential, with a lower concentration leading to a lower potential. Therefore, the initial voltage of the electrochemical cell would be less than the voltage under standard conditions because the concentrations of the ions in the solutions are not at the standard concentration of 1 M.

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Answer:

Hydrogen (H20), Carbon Dioxide (CO2), ATP

2.8×10−2 M will the cyclopropane concentration be after 235. 0 min.

The cycloalkane cyclopropane has the chemical formula (CH2)3, and it is made up of three methylene groups (CH2) that are joined together to create a ring. The structure experiences significant ring strain as a result of the ring's modest size.

[tex]k=3.36*{10^{-5}}{s^{-1}}\\\\K=\frac {2.303 }{t} log \frac{a}{a-x}\\\\3.36*{10^{-5}}=\frac {2.303 }{235min} log \frac{0.445}{a-x}\\\\\\3.36*{10^{-5}}=\frac {2.303 }{235*60} log \frac{0.445}{a-x}\\\\\\20.571*{10^{-2}}=log \frac{0.445}{a-x}\\\\\\1.58=\frac{0.445}{a-x}\\\\\\(a-x)=0.281\\\\(a-x)=2.8*{10^{-2}}M[/tex]

This meant that the quick and painless induction of anaesthesia using cyclopropane and oxygen was possible Ralph Waters, an American anaesthetist, developed cyclopropane and put it to use in medicine. To save this then-expensive substance, he devised a closed system with carbon dioxide absorption. With a blood/gas partition value of 0.55 and a minimum alveolar concentration of 17.5%, cyclopropane is a reasonably strong, non-irritating, and sweet-smelling substance. .8×10−2 M will the cyclopropane concentration be after 235. 0 min.

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The balanced equation for the reaction of lithium and bromine to produce lithium bromide is:

Li + Br2 → LiBr

The total mass of the reactants and products must be equal in order for the rule of conservation of mass to apply. As a result, we can determine the mass of the other reactant and product if we know the mass of one reactant and one product.

In this instance, we are aware that the initial concentrations of the reactants are 25.0 g of lithium and 25.0 g of bromine. Lithium has a molar mass of 6.939 g/mol, while bromine has a molar mass of 159.808 g/mol. So that we may convert each substance's mass to a unit of moles:

Li: 25.0 g / 6.939 g/mol = 3.6 moles

Br2: 25.0 g / 159.808 g/mol = 0.156 moles

We can then convert the moles back to grams using the molar mass of each substance:

Li: 3.6 moles x 6.939 g/mol = 24.96 g

LiBr: 3.6 moles x (6.939 g/mol + 159.808 g/mol) = 627.38 g

Br2: 0.156 moles x 159.808 g/mol = 24.96 g

So, when the reaction is complete, we have 24.96 g of lithium, 24.96 g of bromine, and 627.38 g of lithium bromide.

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Answer: The elements in Groups 2,15,16 and 17 of the Periodic Table are called alkali metals. They form ionic compounds when they react with non-metals. Their ions have a single positive charge. For example, sodium forms sodium ions, Na+

Explanation:

hope you understood

The energy diagram for a solution formed between molecules A and B indicates that the solution is energetically favorable.

This means that when A and B are mixed together, the total energy of the system is lower than if they were kept separate.

This is due to the molecules forming strong intermolecular forces, such as hydrogen bonds, that make the solution more stable. Additionally, this type of solution can be described as being “miscible”, meaning that the two molecules are able to mix together to form a homogenous solution.

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Noble gases does not have any oxidation number. And elements in pure form does not have any oxidation numbers.

Elements in their pure form, such as gold (Au), do not have oxidation numbers because they are not chemically bonded to any other elements. Some other examples of elements that do not have oxidation numbers include carbon (C) in a diamond, and hydrogen (H) in molecular hydrogen (H2).

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The empirical formula of nicotine is C0.96H1N1

The empirical formula of a compound is the simplest whole-number ratio of the atoms in the compound.

To find the empirical formula of nicotine, you need to determine the number of moles of each element present in the sample and then divide by the lowest number of moles.

First, you need to convert the mass of nicotine, CO2, and H2O to moles. To do this, you can use the molar mass of each element:

C = 12.01 g/mol

H = 1.01 g/mol

N = 14.01 g/mol

O = 16.00 g/mol

5.250 mg of nicotine is equal to:

[tex]\frac{ 5.250 mg}{(12.01+1.01+14.01)g/mol}[/tex])   = 0.0027 moles

14.242 mg of CO2 is equal to:

[tex]\frac{14.242mg}{(12.01+16.00*2)g/mol}[/tex]     = 0.0024 moles

4.083 mg of H2O is equal to:

 [tex]\frac{4.083mg}{(1.01*2+16.00)g/mol}[/tex]       = 0.0013 moles

Now you need to find the number of moles of each element present in the sample:

C = 0.0027 moles × 1

H = 0.0013 moles × 2

N = 0.0027 moles × 1

And finally, you can divide these values by the lowest number of moles to get the simplest whole-number ratio:

C =    [tex]\frac{ 0.0027 moles}{0.0027 moles}[/tex]      = 1

H =    [tex]\frac{0.0026 moles}{ 0.0027 moles}[/tex]      = 0.96

N =    [tex]\frac{ 0.0027 moles}{0.0027 moles}[/tex]      = 1

The empirical formula of nicotine is C0.96H1N1

It's important to note that the empirical formula represents the simplest ratio of atoms in a molecule but doesn't necessarily represent the actual molecular formula which could have more atoms or a different arrangement of atoms.

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Answer:

I think it is b chromosomes are made of DNA. I'm not for sure tho sorry

Molar mass of Cu2CO3 is 187.05 g/mol

Molar mass M is the product of the mass of a chemical compound and the molecular weight of the substance. MB = m/nB, where m is the total mass of a sample of a pure material and nB is the molecular weight of substance B, is the formula for it. Pure substance is covered under the definition. Any material has a molecular weight of 6.023 x 1023. (Avagadro number). It can be used to quantify the results of a chemical reaction. Mol is used to indicate the unit.

= 63.5*2+12.01+16*3

= 187.05

so the molar mass of Cu2Co3 is 187.05

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A chemist must dilute the 67.3 ml of the 341 mM of aqueous copper(ii) sulfate solution until the concentration falls to 256 mM is 89.6 mL.

The initial concentration, C1 = 341 mM

The final volume, V1 = 67.3 mL

The final concentration, C2 = 256 mM

The final volume, V2 = ?

By adding the distilled water to the solution until it will reaches a certain final volume :

The expression is given as :

C1 V1 = C2 V2

V2 = C1 V1 / C2

V2 = ( 341 × 67.3 ) / 256

V2 = 89.6 mL

Thus, the final volume in milliliters is 89.6 mL.

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The number of hemoglobin molecules can be calculated by multiplying the number of moles of hemoglobin by Avogadro's number (N_A), which is the number of entities (such as atoms or molecules) in one mole of a substance.

So, the number of hemoglobin molecules in the woman's blood is:

7.5 × 10^-3 mol * 6.02 x 10^23 molecules/mol = 45.15 x 10^20 molecules

To convert the number of moles of hemoglobin to grams, we can multiply the number of moles by the molar mass of hemoglobin:

7.5 × 10^-3 mol * 64,456 g/mol = 483.17 g

So, the woman has approximately 483.17 grams of hemoglobin in her blood.

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Carbon dioxide and water combine to form glucose and oxygen (C6H12O6).

Reaction C6H12O6+6O26CO2+6H2O C 6 H 12 O 6 + 6 O 2 → 6 C .

The equation for aerobic respiration is O 2 + 6 H 2 O, which is an oxidation or combustion reaction.

Carbon atom balance

6CO 2 +H 2 O C   6 H 12 O 6 +O 2

Equilibrate the hydrogen atom.

6CO 2 +6H 2 OC  6 H 12 O 6 +O2

The oxygen atom must be balanced.

6CO 2 + 6H 2 O + C  6 H 12 O 6 + 6O 2

The chemical formula for equilibrium is thus:

6CO 2 + 6H 2 O + C 6 H 12 O 6 + 6O 2.

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The percent yield of product if a student had not heated the test tube for as long as the directions instructed is the result of the product will not match even the percent yield of the product being tested can fail.

A test tube is a tool used to carry out biochemical tests and grow microbes. The media that can be put into the test tube are solid media and liquid media and for the test tube lid you can use cotton, aluminum foil, metal caps, or plastic caps.

How to use a test tube can be heated into a beaker that has been filled with water and heated with a spirit burner for the heating process. Heating the test tube is important for laboratory tests such as purification of solvents by distillation and chemical reactions. One function of heating the solution is to increase the solubility of the solute in it. If a student does not heat the test tube according to the instructions given, the results of the product being tested will not be maximized. Even trials regarding the product fail with a high percentage.

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All the statements given above are correct.

Atoms within resonance structures are the same. The species' structure is not accurately reflected by a single resonance structure.

A representation of a molecule or an ion called a resonance structure depicts several electron combinations that are not conceivable in a single, static structure.

Resonance hybrid refers to the actual structure of the molecule or ion, which is an average of the potential resonance structures.

Resonance structures are not different in terms of atoms, but rather in the arrangement of electrons.

Resonance structures vary, and no single resonance structure can fully capture the structure of an entire species.

The species that make up resonance structures have isomers.

The species could have any of these structures at any time since resonance forms quickly interconvert.

It is not real to have individual resonance shapes.

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Inorganic chemicals in sunscreen can reflect or scatter the light away from the skin, and organic (carbon-based) ones can absorb UV rays

Answer:

It helps prevent the skin from absorbing UV waves.

Explanation:

When the atom, or system, emits or absorbs energy, a transition between two energy levels is said to take place. A system's ground state is the lowest energy level, and excited states have higher energy levels.

Only specific discrete values of energy, known as energy levels, can be absorbed by a bound or spatially constrained quantum mechanical system or particle. Unlike classical particles, which can have any amount of energy as well as excited and ground states, Electrons may be found at specific intervals from an atom's nucleus known as energy levels, also referred to as electron shells. In essence, this means that nuclear power facilities operate at full capacity for more than 92% of the year. This is approximately roughly two times more than natural gas and coal units, and approximately three times as reliable as wind and solar facilities.

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24g of mass of the ammonia evaporates as it absorbs 8. 2 kilo joules of energy.

q = m .ΔHf

[tex]8.2KJ=m(399)\frac{J}{g}\\\frac {8.2*10^{3} J}{339\frac{J}{g}}=m\\0.024*10^{3}g=m\\24g=m[/tex]

With the formula NH3, ammonia is a nitrogen and hydrogen inorganic chemical. Ammonia, the simplest pnictogen hydride and a stable binary hydride, is a colourless gas with a strong, pungent odour.When molecules on a liquid's surface escape into the vapour phase, evaporation takes place. When clusters of liquid molecules erupt into vapour bubbles, boiling happens within the liquid's bulk rather than at its surface. Below, at, and above their boiling points, all liquids will evaporate.It contributes considerably to the nutritional demands of terrestrial creatures by serving as a precursor to 45% of the world's food and fertilisers. Biologically, it is a frequent nitrogenous waste, especially among aquatic animals.

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The age of a sample of charcoal in which 90% of the carbon-14 that was originally present has decayed is about 18,935 years.

A = A₀ (1/2)ᵗ/ᵏ (1) Half-life formula.

0.1A₀ = A₀ (1/2)ᵗ/⁵⁷⁰⁰ (2)Substitute A=0.1A₀ and k=5700

0.1 = (1/2)ᵗ/⁵⁷⁰⁰ (3)Divide both sides by A₀.

log₁/₂0.1 = t/5700 (4)Convert to log from using x=bʸ - logbx = y

5700log₁/₂0.1 = t (5)Multiply both sides by 5700.

18,935 = t

Use the half-life formula A = A₀(1/2)ᵗ/ᵏ where A is the initial amount, t is the number of years, and k is the half-life of the substance. It is given that 90% has decayed which means 10% remains so let A = 0.1A₀. Carbon-14 has a half-life of 5700 years so let k=5700. Substitute these values into the formula and solved for t to get 18,935 years as its value.

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The activity after 26 days can be calculated as 1.1 x 10^4 cpm.

Radioactive is a type of energy that is emitted from certain elements when they undergo radioactive decay. Radioactive elements, such as uranium and radium, contain unstable atoms that are in a constant state of decay. When these atoms break apart, they release energy in the form of radiation.

The decay constant (k) of the sample can be calculated as follows:
k = ln(2.00 x 10^6/9.0 x 10^5) / 4.0
k = 0.073
Therefore, the activity after 26 days can be calculated as follows:
A = 2.00 x 10^6 e^(-0.073 x 26)
A = 1.1 x 10^4 cpm

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Answer:

In 125 grams of feCO3  287.30 g of FeCO3 oxygen is produce

Explanation:

The Balance Chemical Equation is as follow,

                           FeCl₂ + Na₂CO₃    →    FeCO₃ + 2 NaCl

Calculate Mass of FeCl₂ as,

                           Molarity  =  Moles ÷ Volume

Solving for Moles,

                           Moles  =  Molarity × Volume

Putting Values,

                           Moles  =  2 mol.L⁻¹ × 1.24 L

                           Moles  =  2.48 mol

Also,

                            Moles  =  Mass ÷ M.Mass

Solving for Mass,

                           Mass  =  Moles × M.Mass

Putting Values,

                            Mass  =  2.48 mol × 126.75 g.mol⁻¹

                            Mass =  314.34 g of FeCl2

Calculate Mass of FeCO₃ formed as,

According to equation,

          126.75 g (1 mole) FeClZ produces  =  115.85 g (1 mole) FeCO₃

So,

               314.34 g of FeCl₂ will produce  =  X g of FeCO₃

Solving for X,

                    X =  (314.34 g × 115.85 g) ÷ 126.75 g

                     X =  287.30 g of FeCO₃

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Answer:

the first statement is correct...

it only supports the main idea of the law of conservation of mass...

hope I can help you...

The experimental evidence that led Rutherford to develop his atomic model is; from a gold foil, alpha particles were shown to disperse backward.

Alpha particles were seen to disperse in the opposite direction after being emitted from a gold foil in the now-famous experiment. In his explanation, which he published in May 1911, Rutherford proposed that the dispersion was brought about by a rigid and compact core at the center of the atom known as the nucleus.

The reasoning that went into developing the atomic model was that the only way that alpha particles could be deflected at enormous angles as if they were impacting on a dense opaque region, and this dense region was located at the center of each atom.

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