Use standard enthalpies of formation to calculate ΔH∘rxn for the following reaction. C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(g)

Answer:

2. At pH near the pI, nearly all the molecules carry no net charge

Explanation:

The general equilibrium for a molecule with multiple ionizable groups could be:

H₃A⁺ ⇄ H₂A + H⁺ ⇄HA⁻ + H⁺⇄A²⁻ + H⁺

Where each equilibrium has its own pKa

The equilibriums in which H₂A are involved (The neutral molecule), are used to determine the isoelectrical point.

The isoelectrical point is defined as the value of pH in which the molecule has no net charge.

Thus, true option is:

The units for atomic mass are atomic mass units (amu).  the atomic mass of copper for this location is 63.55 amu.

The chemical symbol Cu stands for copper. Copper is a soft, malleable, ductile metal that is a good conductor of heat and electricity. Copper is one of the most widely used metals in electrical and electronic equipment due to its superior conductivity and non-corrosive properties. This metal is widely used in wiring, roofing, plumbing, and electronic applications. Its atomic mass is 63.55 amu.The atomic mass of copper for this location can be determined using the following formula:

Atomic mass = (mass of isotope 1 x relative abundance of isotope 1) + (mass of isotope 2 x relative abundance of isotope 2)The atomic mass of copper for this location

= (62.93 x 0.6915) + (64.93 x 0.3085) = 63.55 amu

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The chemical formula for x and y is Cu₂O and CuO. The mass cooper which can react with 0.5g of oxygen to yield x and y is 2.745 g.

A chemical formula is a phrase that lists the constituent parts of a compound together with their relative quantities. No subscript is used if there is just one atom of a certain kind. A subscript is added to the symbol of an atom if it contains two or more of a certain type of atom.

1. 1.535 g of X → 1.365 g of Copper

1.535 – 1.365 = 0.170g of Oxygen

Atomic weight of Cu = 63.5,

Atomic weight of Oxygen = 16

For Cu 1.365 g / 63.5 = 0.02 mol

For Oxygen 0.170 g / 16 = 0.01 mol

X = Cu₂O

1.450 g of Y → 1.160 g of Cu

1.450 – 1.160 = 0.290 g of Oxygen

For Cu = 1.160 g / 63.5 = 0.018 mol

For Oxygen = 0.290 g / 16 = 0.018 mol

Y = CuO

2. The total mass of Oxygen = 0,170 g + 0,290 g

= 0.460 g

Total mass of Cu = 1.160 g + 1. 365 g

= 2.525 g

0.460 g of Oxygen → 2.525 g of Cu

0.500 g of Oxygen → (2.525 x 0.5) / 0.460

= 2.745 g of Cu

Thus, The law of multiple proportions was formulated by John Dalton in 1804.

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Answer:..

Explanation:

A bouncing ball gradually stops bouncing because its energy is converted to heat energy.

According to the law of conservation of energy, energy is neither created nor destroyed but can be transformed from one form to another.

When a ball is bouncing up and down, we notice that the ball will slow down gradually. This is because, the energy in the bouncing ball is transferred to the small air molecules inside the ball as heat. Hence, the ball looses energy consistently until it comes to a stop.

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Answer:

electrophile is pe do file

Explanation:

elecricity = needing a professor= creepy profesor= pe edo

Answer:

The answer is A.

Answer:

The correct option is a: Hg-183.

Explanation:

The alpha decay is given by:

\(^{A}_{Z}X \rightarrow ^{A-4}_{Z-2}Y + ^{4}_{2}He\)

Hence, if Polonium-191 decays by emitting an alpha particle we have:

\(^{191}_{84}Po \rightarrow ^{187}_{82}Pb + ^{4}_{2}He\)        

Now, if Pb-187 also decays by emitting an alpha particle we have:

\(^{187}_{82}Pb \rightarrow ^{183}_{80}Hg + ^{4}_{2}He\)        

Therefore, the correct option is a: Hg-183.

I hope it helps you!

Answer:

Two pi bonds, one sigma bond.

Explanation:

A triple bond has three bonds:

2 pi bonds, and one sigma bond.

Answer:

Two pi bonds & one sigma bond.

A triple bond is made of:

Hope it helps!

The enthalpy change of the reaction C(diamond) → C(graphite) is -2.9 kJ.

The given information is ΔHrxn for the reaction C(diamond) → C(graphite) can be calculated with the given equations:Equations:  C(diamond) + O2(g) → CO2(g) ΔH = −395.4 kJ  2 CO2(g) → 2 CO(g) + O2(g) ΔH = 566.0 kJ  C(graphite) + O2(g) → CO2(g) ΔH = −393.5 kJ  2 CO(g) → C(graphite) + CO2(g) ΔH = −172.5 kJThe required reaction can be obtained by adding the equations (1) and (4), as follows:C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)Addition of the two equations (1) and (4) results in a reaction whose products are C(graphite) and CO2.

To get the final equation that involves only the required reactants and products, the equation (2) should be added, which consumes CO2 and produces O2, as shown below:C(diamond) + O2(g) → CO2(g)  ΔH = −395.4 kJ [eq. (1)]  2 CO(g) → C(graphite) + CO2(g)  ΔH = −172.5 kJ [eq. (4)]  2 CO2(g) → 2 CO(g) + O2(g)  ΔH = 566.0 kJ [eq. (2)]  C(diamond) + O2(g) + 2CO(g) → C(graphite) + 3CO2(g)  ΔHrxn=ΣΔHf(products)−ΣΔHf(reactants)  ΔHrxn=[(3 mol CO2)(-393.5 kJ/mol) + (1 mol C(graphite))(0 kJ/mol)] − [(1 mol C(diamond))(0 kJ/mol) + (1 mol O2)(0 kJ/mol) + (2 mol CO(g))(−172.5 kJ/mol)] − [(2 mol CO2)(566.0 kJ/mol)]  ΔHrxn=−2.9 kJ.

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Answer:

NaCl may be i guess so....

Answer:

K₂SO₄ + Pb(C₂H₃O₂)₂ →PbSO₄ + 2KC₂H₃O₂

A chemical equation is a symbolic representation of a chemical reaction. The chemical equation for the reaction between potassium sulfate (\(K_2SO_4\)) and lead(II) acetate (\(Pb(CH_3COO)_2\)) can be written as follows:

\(K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK\)

A basic chemical equation consists of two main parts: the reactant side (left side) and the product side (right side), separated by an arrow indicating the direction of the reaction. Reactants are substances that undergo a chemical change, while products are substances formed as a result of the reaction.

In this reaction, potassium sulfate reacts with lead(II) acetate to form lead(II) sulfate and potassium acetate. It is important to note that the equation is balanced with stoichiometric coefficients, ensuring that the number of atoms of each element is the same on both sides of the equation.

Therefore, the chemical equation for the reaction between potassium sulfate (\(K_2SO_4\)) and lead(II) acetate (\(Pb(CH_3COO)_2\)) can be written as follows:

\(K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK\)

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Answer:

1st one is...

Explanation:

The 1st one is change in size or shape. 2nd one is formation of precipitate. 3rd one is physical change. 4th one is formation of gas.

Answer:

Br-CH2-CH(CH3)2-C(Cl)H-CH(CH3)2-CHO

Explanation:

The molecule has a total of 14 carbon atoms, 13 hydrogen atoms, and 1 bromine atom. The carbon atoms are arranged in a chain with a methyl group attached to the second carbon atom, a chlorine atom attached to the third carbon atom, and two methyl groups attached to the fourth carbon atom. The fifth carbon atom has a carbonyl group attached to it.

The molecule is an aldehyde, which means that it has a carbonyl group (C=O) at the end of the chain. The carbonyl group is polar, and the oxygen atom has a partial negative charge. The hydrogen atom has a partial positive charge. This polarity makes the aldehyde group susceptible to nucleophilic attack.

The bromine and chlorine atoms are both electrophilic, which means that they have a partial positive charge. This makes them susceptible to nucleophilic attack.

The methyl groups are non-polar and do not have any significant reactivity.

The molecule is a chiral molecule, which means that it has a mirror image that is not superimposable on itself. This is because the carbon atom with the carbonyl group is attached to four different groups.

The molecule is a liquid at room temperature and has a strong odor. It is used in a variety of products, including perfumes, flavorings, and plastics.

Explanation:

Force=mass *acceleration

F=300*5

f

F=1500N

Answer:

The coefficient of Z₂ is 1.

Explanation:

From the question given above:

X + ZY —> XY + Z₂

Next, we shall balance the equation to obtain the coefficient of Z₂. This can be obtained as follow:

X + ZY —> XY + Z₂

There is 1 atom of Z on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of ZY as shown below:

X + 2ZY —> XY + Z₂

There are 2 atoms of Y on the left side and 1 atom on the right side. It can be balance by putting 2 in front of XY as shown below:

X + 2ZY —> 2XY + Z₂

Now, we have 1 atom of X on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of X as shown below:

2X + 2ZY —> 2XY + Z₂

Now the equation is balanced.

Thus, the coefficient of Z₂ is 1.

Answer:

Explanation:cool thre answer is 009000000000.1 because you have to add them so ye

Answer:

By heating the mixture to maximum boiling point and then the solution is distilled at a constant temperature without having a change in composition.

Explanation:

An azeotropic mixture is also called a constant boiling mixture and it is a mixture of two or more liquids whose proportions cannot be altered by simple distillation due to the fact that when an azeotropic mixture is boiled, the vapor has the same proportions of constituents as the unboiled mixture.

Now, maximum boiling azeotropic mixture are the solutions with negative deviations that have an intermediate composition  for which the vapor pressure of the solution is minimum and as a result, the boiling point is maximum. At that point, the solution will distill at a constant temperature without having a change in composition.

Answer:

C) Continental drift :)

Explanation:

Answer:

Last one

hope it helps mvhsbshsjjak

The partial pressure of nitrogen is approximately 733 mm Hg. Please note that the answer is rounded to the nearest whole number, which is 733 mm Hg. Option D)

To determine the partial pressure of nitrogen, we need to consider the Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is the sum of the partial pressures of each gas present.

Given that the total pressure in the laboratory is 775 mm Hg and the partial pressure of water at 35.0°C is 42.2 mm Hg, we can calculate the partial pressure of nitrogen.

To find the partial pressure of nitrogen, we subtract the partial pressure of water from the total pressure:

Partial pressure of nitrogen = Total pressure - Partial pressure of water

Partial pressure of nitrogen = 775 mm Hg - 42.2 mm Hg

Partial pressure of nitrogen = 732.8 mm Hg

Therefore, the partial pressure of nitrogen is approximately 733 mm Hg.

Please note that the answer is rounded to the nearest whole number, which is 733 mm Hg. Option D) is correct.

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Diatomic

consist of two atoms bonded together. In contrast, monatomic elements consist of single atoms (e.g., Ar, He). Many compounds are diatomic, such as HCl, NaCl, and KBr. Diatomic compounds consist of two different elements. There are seven pure elements that form diatomic molecules.

Key Takeaways: Diatomic Elements

Diatomic elements are pure elements that form molecules consisting of two atoms bonded together.

There are seven diatomic elements: hydrogen, nitrogen, oxygen, fluorine, chlorine, iodine, bromine.

These elements can exist in pure form in other arrangements. For example, oxygen can exist as the triatomic molecule, ozone.

This is a list of the seven diatomic elements. The seven diatomic elements are:

Hydrogen (H2)

Nitrogen (N2)

Oxygen (O2)

Fluorine (F2)

Chlorine (Cl2)

Iodine (I2)

Bromine (Br2)

All of these elements are nonmetals, since the halogens are a special type of nonmetallic element. Bromine is a liquid at room temperature, while the other elements all gases under ordinary conditions. As the temperature is lowered or pressure is increased, the other elements become diatomic liquids.

Astatine (atomic number 85, symbol At) and tennessine (atomic number 117, symbol Ts) are also in the halogen group and may form diatomic molecules. However, some scientists predict tennessine may behave more like a noble gas.

While only these seven elements routinely form diatomic molecules, other elements can form them. However, diatomic molecules formed by other elements are not very stable, so their bonds are easily broken.

compounds

are formed by the chemical combination of two or more element. example water is a compound because two elements water and hydrogen come to form it.

By using the ideal gas law and Avogadro's number, we find that there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

To determine the number of molecules of CO2 in 2.5 L at STP (Standard Temperature and Pressure), we can use the ideal gas law and Avogadro's number.

Avogadro's number (N_A) is a fundamental constant representing the number of particles (atoms, molecules, ions) in one mole of substance. Its value is approximately 6.022 × 10^23 particles/mol.

STP conditions are defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (1 atm).

First, we need to convert the volume from liters to moles of CO2. To do this, we use the ideal gas law equation:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Since we have STP conditions, we can substitute the values:

(1 atm) × (2.5 L) = n × (0.0821 L·atm/(mol·K)) × (273.15 K).

Simplifying the equation:

2.5 = n × 22.4149.

Solving for n (the number of moles):

n = 2.5 / 22.4149 ≈ 0.1116 moles.

Next, we can calculate the number of molecules using Avogadro's number:

Number of molecules = n × N_A.

Number of molecules = 0.1116 moles × (6.022 × 10^23 particles/mol).

Number of molecules ≈ 6.72 × 10^22 molecules.

Therefore, there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

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Answer:

Germanium

Explanation:

One mole of carbon dioxide would require 84.01 grams of NaHCO₃.

It is discovered that the ratio of moles of CO₂ generated to moles of NaHCO₃ reacted is 1:2.

We can observe from this equation that 1 mole of NaHCO₃ results in 1 mole of CO₂. As a result, NaHCO₃ and CO₂ have a molar ratio of 1:1.

Na2CO₃(s) + H₂O(g) + CO₂ = 2 NaHCO₃(s)(g)

CO₂ has a molar mass of about 44.01 g/mol. As a result, we must determine how much NaHCO₃ weighs in relation to one mole of CO₂. Using the molar mass of NaHCO₃, the following can be calculated:

Molar mass of NaHCO₃ is 84.01 g/mol.

The mass of NaHCO₃ needed to create one mole of CO₂ is as follows:

(84.01 g NaHCO₃/1 mole NaHCO₃) = 84.01 g CO₂/mol for 1 mole of NaHCO₃.

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Answer:

1.674 x 10^-23 grams

Explanation:

Hydrogen-1 is called Protium

wikipedia

atomic mass of Protium is 1.00794 amu

sciencedirectcom

atomic mass of 10 Protiums is 10.0794 amu

10.0794 amu in grams is

1.6737236x10^-23 grams