methyl isocyanate reacts with strong acids, such as sulfuric acid, to form a cation. for protonation on , which resonance contributor is most important?

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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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 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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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:

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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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 mass of oxygen in the room would be 46.2 kg.

The mass of a component as a percentage of the mass of the substance itself is known as the percent composition. Mathematically, this is expressed as:

Mass of component/mass of material multiplied by 100% gives the percent composition.

The percentage of oxygen in the air in this instance is represented as 21%, while the total mass of air in the classroom is provided as 220 kg. This formula can be used to determine how much oxygen is present in the classroom:

% oxygen composition = mass of oxygen/mass of air

Making the oxygen mass the focus of the formula:

Mass of air x percent oxygen content equals mass of oxygen

Let's replace the values now:

Weight of oxygen = 21% times 220

= 0.21 x 220

= 46.2 kg

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Answer:

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

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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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 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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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

Answer:

29 ouns

Explanation:

it has 950 billion so it will be scaled in grams not kilograms

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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The white solid formed when an alkaline earth metal reacts with iodine is an ionic compound.

The white solid formed when an alkaline earth metal reacts with iodine is an ionic compound. This compound is composed of polyatomic anions and cations in an ionic lattice, which gives it its solid form. The cations come from the alkaline earth metals, which are group two elements on the periodic table, and the anions come from the iodine, which is a halogen. The compound is typically composed of a ratio of 1:1 or 2:1, depending on the metals in reaction. The compound is usually insoluble in water and can be used in various applications. It is often used for its chemical inertness, which makes it ideal for processes that involve corrosive or reactive materials.

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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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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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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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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 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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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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Copper sulphate has oxygen elements in it while copper sulphide doesn't have.

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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 pressure of the gas if the volume were decreased to 6L is 6 atm.

Boyle's law of ideal gases states that the pressure of an ideal gas is inversely proportional to its volume at constant temperature.

The Boyle's law equation is given as follows;

P₁V₁ = P₂V₂

Where;

P₁ = initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume

3 × 12 = 6 × P

36 = 6P

P = 6atm

Therefore, the pressure of the hydrogen gas is 6 atm.

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C. 2 ATP.  At the cellular level, energy is used and stored as adenosine triphosphate (ATP). Adenine, ribose, and three serially bound phosphate groups make up.

the structure of ATP, which is a nucleoside triphosphate. An organic substance called adenosine triphosphate provides energy for a variety of biological functions in living cells, including muscular contraction, nerve impulse transmission, condensate dissolving, and chemical synthesis. The main molecule for storing and transmitting energy in cells is adenosine 5'-triphosphate (ATP). At the cellular level, energy is used and stored as adenosine triphosphate (ATP). Adenine, ribose, and three serially bound phosphate groups make up. It is frequently referred to as the cell's energy currency and is like the money that is kept in a bank.provides energy for a variety of biological functions in living cells, including muscular contraction, nerve impulse transmission,

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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 wire is 91826.77 N/m stiff, One layer of a cross section contains 1.0158e13 atoms, Number of bonds per unit length = 9.96e9, rigidity = 9.035 e-9 N/m

a) The force acting on the bar is given by (F) = mg = 119 * 9.8 = 1166.2 N.

The wire's extension (L) is 1.27 cm, or 0.0127.

Therefore, stiffness of wire = F/L = 1166.2 / 0.0127.

The wire is 91826.77 N/m stiff.

b) The ratio of the number of atoms in one cross-sectional layer to 8e-4 (2.51e-10)2

One layer of a cross section contains 1.0158e13 atoms.

c) The length's bond count is 2.5 / 2.51e-10.

Number of bonds per unit length = 9.96e9

d) Bond stiffness equals force acting on the bond / bond strain

bond-applied force = 1.152e-20 N

strain is equal to (0.0127/2.5)*2.51e-10 = 1.275e-12 m.

Stiffness is calculated as 1.152e-20 / 1.275e-12.

rigidity = 9.035 e-9 N/m

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