Nitrosyl bromide decomposes according to the following equation. 2NOBr (g) 2NO (g) Br2 (g) A sample of NOBr (0. 64 mol) was placed in a 1. 00-L flask containing no NO or Br2. At equilibrium the flask contained 0. 46 mol of NOBr. How many moles of NO and Br2, respectively, are in the flask at equilibrium

Explanation:

T = 409.5 K, P = 1.50 atm: V = 22.4 L The ideal gas law is: PV = nRT where. P = pressure. V = volume n = number of moles.

The amount of energy that is required each year for the purpose of agriculture is only about 10%.

Energy is a very important part of civilization. In the current time, energy is necessary for all kinds of productions and agriculture is not an exception.

As such, the amount of energy that is required each year for the purpose of agriculture is only about 10%.

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There are approximately 0.000486 moles of hydrogen present in the sample.

To find the number of moles of hydrogen present in the sample, you need to use the ideal gas law equation, which is PV = nRT where P is the pressure of the gas (in atm), V is the volume of the gas (in L), n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas (in K). First, you need to convert the pressure from torr to atm. You can do this by dividing the pressure in torr by 760, since 1 atm is equal to 760 torr. In this case, the pressure in atm would be: P = 764.5 torr / 760 = 1.0059 atm Next, you need to convert the volume from ml to L. You can do this by dividing the volume in ml by 1000, since 1 L is equal to 1000 ml. In this case, the volume in L would be: V = 30.42 ml / 1000 = 0.03042 L Finally, you need to convert the temperature from degrees Celsius to Kelvin. You can do this by adding 273.15 to the temperature in Celsius. In this case, the temperature in K would be:T = 25.58 degrees Celsius + 273.15 = 298.73 K Now that you have all the necessary values, you can plug them into the ideal gas law equation to solve for the number of moles of hydrogen: n = (PV) / (RT) = (1.0059 atm * 0.03042 L) / (8.31 J/mol*K * 298.73 K) = 4.86 x 10^-4 mol Therefore, there are approximately 0.000486 moles of hydrogen present in the sample

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Answer:

16L of 70% solution

32L of 25% solution

32L of 15% solution

Explanation:

x = volume of 70% acid solution

y = volume of 15% acid solution

2x = volume of 25% acid solution

2x + x + y = 80

3x + y = 80 (×8)

24x + 8y = 640 → 24x = 640 - 8y

0.7x + 0.25(2x) + 0.15y = 0.3(80)

⁷/₁₀.x + ¹/₂.x + ³/₂₀.y = 24

¹²/₁₀.x + ³/₂₀.y = 24 (×20)

24x + 3y = 480

(640 - 8y) + 3y = 480

640 - 5y = 480

5y = 160

y = 32

3x + 32 = 80

3x = 48

x = 16

Answer: 1. Nurse practitioner.

2. Data scientist.

3. Information security analyst.

4. Physician assistant (PA)

5. Software developer.

6. Actuary.

7. Financial manager.

Explanation:

Answer:

Testable questions can be answered through investigations that involve experiments, observations, or surveys. Testable questions are answered by collecting and analyzing evidence that is measurable. Testable questions relate to scientific ideas rather than personal preference or moral values.

Explanation:

hope this helps

The molar mass of Cu(OH)2 is 97.56 g/mol. Using the given information, the mass of Cu(OH)2 precipitated can be calculated as 0.524 × 0.0560 × 97.56 g ≈ 2.97 g.

To calculate the grams of Cu(OH)2 precipitated, we need to determine the limiting reactant. From the balanced equation, we can see that the stoichiometric ratio between CuI2 and Cu(OH)2 is 1:1.

First, we calculate the number of moles of CuI2:

moles of CuI2 = concentration × volume

moles of CuI2 = 0.524 M × 0.0560 L

Since the stoichiometric ratio is 1:1, the moles of Cu(OH)2 will be the same as the moles of CuI2.

Next, we calculate the mass of Cu(OH)2:

mass of Cu(OH)2 = moles of Cu(OH)2 × molar mass of Cu(OH)2

By substituting the moles of Cu(OH)2 with the moles of CuI2 obtained earlier and the molar mass of Cu(OH)2, we can calculate the mass of Cu(OH)2 precipitated.

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Answer:

the creation of oxygen.

Explanation:

According to the given information CaCl₂(s) are the particles arranged in a crystal structure.

The correct option is A.

A crystal system is composed of a number of axes. Or, to put it another way, the structure is an arranged grouping of atoms, ions, or molecules. The crystal structure is made up of bonded atoms, atom clusters, or molecules. Due to the constituent particles' innate propensity to arrange themselves in symmetrical patterns, this structure forms.

This inorganic salt is crystallized calcium cations and chlorine anions. Ions are grouped in a systematic way.

Ethane (C₂H₆) is a gas; the symbol for a gas in chemistry or its physical state is g. Gases lack a solid structure.

Methanol (CH₃OH) is a liquid; the symbol for liquids in chemistry is l, and it lacks a structural structure.

Since water dissolves calcium iodide, the resulting aqueous solution (aq) lacks a crystal structure.

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Answer:Some examples of the force of gravity include;

-The force that holds the gases in the sun.

-he force that causes a ball you throw in the air to come down again.

-The force that causes a glass you drop to fall to the floor.

-The force that keeps the Earth and all of the planets in line in the proper position in their orbits around the sun.

-The force that causes an apple to fall downward from an apple tree.

-he force that keeps you walking on Earth instead of floating away into space.

Answer; 2.531 moles

Explanation:

The trend of atomic radius refers to the predictable pattern of the size of atoms as one moves across or down the periodic table of elements.

Moving across a period (from left to right), the atomic radius of an element generally decreases. This is because the number of protons and electrons in the atom increases, leading to a stronger attractive force between the positively charged nucleus and negatively charged electrons. The increased pull causes the electron cloud to be pulled in closer to the nucleus, making the atomic radius smaller.

Moving down a group (from top to bottom), the atomic radius generally increases. This is because each successive element has one additional energy level than the element above it, and the electrons in these energy levels are farther from the nucleus. The increase in distance between the outermost electrons and the nucleus results in an increase in the atomic radius.

Therefore, the general trend of atomic radius is that it decreases from left to right across a period and increases from top to bottom down a group.

The following would have a larger atomic radius in the pair;

1) Ge

2) Fe

3) Kr

4) Sr

5) Na

6) I

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Carbon-14 has a half-life of 5,730 years. A fossil with 4 g of C initially started with 32 g of C is 17200 years old.

Half life of a substance tells about how much time taken by a sample to reduced to 50%.

The 32g of C is decayed and it remains 4g of it. Since carbon decays follows first order reaction. To find out the age of fossil, we need to use the formula which is given as

ln(N_t/N_0)=-λt(i)

Where, N_t = 4g

N_0 = 32 g

We need to find λ using half life of carbon which is given as

λ=(0.693/t(1/2))

where t(1/2)=5730 years

Plug all values in the formula

λ=(0.693 year⁻¹/5730)

λ=0.0012094 year⁻¹

Plug all the available values in the equation (i)

ln(4 g/32 g)=-0.00012094 year⁻¹×t

ln(0.125)=-0.00012094 year⁻¹×t

-2.07944=-0.00012094 year⁻¹×t

t=(-2.07944 year/-0.00012094)

t = 17193.9805 year

t = 17200 year

Therefore, a fossil with 4 g of C initially started with 32 g of C is 17200 years old.

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The rate of change of the concentration of Na₂C₂O4 is -0.0551 M/s.

To determine the rate of change of the concentration of sodium oxalate (Na₂C₂O4), we can use the rate equation:

Rate = (Δ[Na₂C₂O4]) / (Δt)

where Δ[Na₂C₂O4] represents the change in concentration of Na₂C₂O4 and Δt represents the change in time.

In this case, the initial concentration of Na₂C₂O4 is 1.34 M, and after 19.3 seconds, the concentration is 0.276 M.

Substituting the values into the rate equation, we have:

Rate = (0.276 M - 1.34 M) / (19.3 s - 0 s)

Rate = (-1.064 M) / (19.3 s)

Rate = -0.0551 M/s

Therefore, the rate of change of the concentration of Na₂C₂O4 is -0.0551 M/s.

The negative sign indicates that the concentration of Na₂C₂O4 is decreasing over time, as the reactant is being consumed in the reaction.

It's important to note that the rate of a reaction is influenced by various factors, such as the reaction mechanism, temperature, and presence of catalysts. The rate can be determined experimentally by measuring the change in concentration of a reactant or product over a specific time interval.

The given information allows us to calculate the rate of change for the specific reaction involving Na₂C₂O4. However, without additional information about the reaction, it is not possible to determine the exact nature or stoichiometry of the reaction, as well as any other reactants or products involved.

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It is no doubt Cyclo-2-en-1-one. It can be concluded from the value of C=O stretching value. For the other two compound, the C=O stretching will come more than 1720 cm-1 region. The peak at 3096 cm-1 is for C-H stretching (alkene). The peak at 3010 cm-1 is for alkane C-H stretching.

Compound is a chemical term for a substance composed of similar molecules containing atoms from various chemical elements.

A molecule made up of just one element's atoms is not referred to as a compound.

Alkanes, including hexane and butane, are substances made of carbon and hydrogen atoms bound together by only one bond.

During chemical reactions, it gets changed into new compounds. According to the chemical bonds they include, there are four main types of compounds. These are:

1) Atoms in molecular compounds linked by covalent bonds.

2) Atoms in ionic compounds connected by ionic bonds.

3) Atoms bound together by metallic bonds in intermetallic compounds.

4) Atoms are kept together by coordinate bonds in complexes for co-ordination.

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The molar mass of a 0.314-gram sample of gas having a volume of 1.6L at 287K and 0.92 atm is 0.503g/mol.

The temperature and pressure of a gas that behaves like an ideal gas is directly proportional to one another.

First, calculating molar mass :

The molar mass of a substance can be calculated by dividing the mass of the substance by its number of moles.

However, the number of moles can be calculated using the formula:

PV = nRT

From where;

P = pressure

V = volume

R = gas law constant

T = temperature

n = number of moles

0.92 × 1.6 = n × 0.0821 × 287

1.47 = 23.56n

n = 1.47/23.56

n = 0.624 moles

Molar mass = 0.314 g ÷ 0.624 mol

Molar mass = 0.503 g/mol

Therefore, the molar mass of a 0.314-gram sample of gas having a volume of 1.6L at 287K and 0.92 atm is 0.503g/mol.

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As we go down the group, the electropositivity of the cations increases. This is because the atomic size increases

Ionization energy is defined as the minimum energy required to remove an electron from a gaseous atom or ion in its ground state. The first ionization energy (IE1) is the energy required to remove the outermost or valence electron from an atom or ion.

The ionization energy generally increases as we move across a period or from left to right in the periodic table due to the increasing effective nuclear charge experienced by the valence electrons. This is because the number of protons in the nucleus increases, while the number of electrons in the same energy level remains the same. This results in a greater attraction between the nucleus and the valence electrons, making it more difficult to remove an electron.

However, as we move down a group or column in the periodic table, the ionization energy generally decreases. This is because the atomic size increases due to the addition of new energy levels, and the valence electrons become farther away from the nucleus. As a result, the outermost electron is shielded from the attraction of the positively charged nucleus by the inner electrons, reducing the effective nuclear charge experienced by the valence electrons.

Cesium (Cs) is the largest atom in Group 1 (alkali metals) and has the lowest effective nuclear charge. As a result, it has the lowest first ionization energy among all the elements. The valence electron in Cs is located farther away from the nucleus compared to other elements in the group, making it easier to remove the electron.

Regarding the electropositivity of cations, electropositivity refers to the tendency of an element to lose electrons and form positive ions (cations). As we move down a group, the electropositivity of the cations generally increases. This is because the atomic size increases, and the valence electron becomes farther away from the nucleus, making it easier to remove the electron and form a positive ion. Additionally, the effective nuclear charge experienced by the valence electrons decreases as we move down the group, which further facilitates the formation of cations.

Therefore, the cations of Cs and other elements in the same group will have a higher electropositivity compared to those of elements in the upper groups. Alkali metals like Cs readily form cations with a +1 charge by losing their single valence electron, which has a noble gas configuration, making them highly reactive and important elements in various industrial and biological processes.

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Cesium (Cs) has the lowest first ionization energy among the elements in its group (Group 1, the alkali metals), this is because Cs has the largest atomic size and the highest number of electron shells, resulting in its valence electron being farthest from the nucleus.

Due to this increased distance and electron shielding, the electrostatic attraction between the valence electron and the nucleus is significantly weaker. Consequently, less energy is needed to remove the electron. The ionization energy of an element refers to the energy required to remove one electron from its outermost shell, forming a cation

As you move down Group 1, the electropositivity of cations increases. Electropositivity is the tendency of an element to lose electrons and form a positive ion. With each additional electron shell, the attraction between the nucleus and the valence electron becomes weaker, facilitating electron loss. So therefore, alkali metals' electropositivity increases as you go down the group, making Cs the most electropositive element in this group.

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Answer:

Experiment one.

You want to find out the absence of a certain mineral so you will take away the mg of nutrients.:

The question gives us the reaction to produce sodium sulfide (Na2S) from sodium sulfate (Na2SO4) and carbon (C) and provides the amount of reactants used (15.0 g of Na2SO4 and 7.50 g of C), asking the limiting reactant, the excess reactant and the amount of Na2S produced.

I) The first step for this type of question is checking if the given equation is balanced. For this case, we don't need to adjust the coefficients as the equation is already balanced.

II) Next, we need to calculate the molar mass of the compounds we'll be using. To calculate the molar mass, I'll be using the following atomic masses and considering the number of each atom in the molecules:

Na: 22.99 u

S: 32.07 u

O: 15.99 u

C: 12.01 u

Now, we can calculate the molar mass of Na2SO4, C and Na2S:

Na2SO4: molar mass = (2 * 22.99) + (1 * 32.07) + (4 * 15.99) = 142.01 g/mol

C: molar mass = (1 * 12.01) = 12.01 g/mol

Na2S: molar mass = (2 * 22.99) + (1 * 32.07) = 78.05 g/mol

iii) The third step is to convert the masses given for Na2SO4 and C into the correspondent number of moles using the molar mass of these compounds:

IV) On the forth step, we must define the limiting reactant for this reaction considering the amounts used of each one and the stoichiometric coefficients:

1 mol Na2SO4 reacts with 1 mol C

0.106 mol Na2SO4 reacts with...?

Solving this calculation, we have that we would need 0.106 mol of C to react with 15.0 g of Na2SO4. Since we there are 0.624 mol of C available to react, we can conclude that carbon is the reactant in excess (there is an excess of 0.518 mol) and sodium sulfate is the limiting reactant.

V) At last, we can calculate the amount of Na2S produced from the limiting reactant amount used (0.106 mol of Na2SO4) and the stoichiometric coefficients:

1 mol Na2SO4 ----------1 mol Na2S

0.106 mol Na2SO4 ---- y

Solving for y, we have that 0.106 mol of Na2S will be produced.

vWe can convert this amount into mass of Na2S using its molar mass:

In summary:

Na2SO4 is the limiting reactant and there is an excess of 0.518 mol or 6.22 g of C;

8.27 g of Na2S will be produced from 15.0 g of Na2SO4.

After 1999, the willow tree area started to decrease again. A possible reason for this decrease could be a change in environmental factors, such as climate change, leading to less favorable conditions for the growth and reproduction of willow trees.

Droughts, floods, or temperature fluctuations may have impacted the trees' ability to thrive in their habitat. Additionally, human activity, such as deforestation, urbanization, or agricultural expansion, could have led to the loss of suitable land for willow trees to grow. Pests and diseases could have also played a role in the decline of the willow tree population.

Invasive species, which may outcompete native species for resources or introduce new pathogens, could have contributed to the reduction in willow tree numbers. Finally, changes in land management practices, such as shifting to other types of vegetation or the removal of willow trees for commercial purposes, could have contributed to the decrease in willow tree area after 1999. So therefore environmental factors, such as climate change, leading to less favorable conditions for the growth and reproduction of willow trees, possible reason after 1999, the willow tree area started to decrease again.

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The given unknown sample is not iron as its density is 7.34156 g/cm3.

The density of a material is the way of measuring of how firmly it is packed together. It is expressed as mass per unit volume.

Density is typically classified into two types: absolute density and relative density.

Relative density, also referred to as specific gravity, is the ratio of a material's density to the density of a background material. Water is commonly used as a reference material.

Three of an object's most fundamental properties are mass, volume, and density. Mass describes how bulky something is, volume describes its size, and density is defined as mass divided by volume.

As in the given scenario,

The volume of sample = 2.25(1.5)(2.25) = 7.59375 cm3

We know that,

Density = mass / volume

Density = 55.75 / 7.59375 = 7.34156 g/cm3

Thus, here, one can conclude that the given sample is not iron.

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Answer:

“GMOs have helped farmers reduce their environmental footprint by allowing them to use fewer inputs and enabling a shift to reduced tillage. These practices have led to less time spent on a tractor, less fuel used and fewer emissions.

Hope this helps!!!

Answer:

Oxygen, Hydrogen, Carbon, and Nitrogen

Explanation:

Living cells are made up primarily of four elements that are Oxygen, Hydrogen, Carbon, and Nitrogen

Among the given reactions of alkenes, the reaction of addition of HBr (treatment with HBr) to an alkene is stereo-specific.

What is stereo-specific?

Stereo-specificity is a chemical concept that describes a reaction's capability to differentiate between stereo-isomers. A stereo-specific reaction is one in which only one stereo-isomer is generated from a given stereo-isomeric reactant. The stereo-chemistry of the reactant, the product, or both is generally associated with stereo-specificity. Stereo-specific reactions are also referred to as stereo-selective reactions.

When an alkene reacts with HBr in the presence of a peroxide initiator, the hydrogen (H) and the bromine (Br) atoms add to the carbon atoms of the double bond in a syn addition. This means that the hydrogen and the bromine atoms are added to the same side of the double bond, resulting in the formation of a single product with a specific stereo-chemistry.

Therefore, the reaction of addition of HBr to an alkene is stereo-specific, because the stereo-chemistry of the product is determined by the syn addition of the hydrogen and the bromine atoms to the double bond. The other reactions listed are not stereo-specific.

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Answer:

Hope this helps! :)

Explanation:

The ground state electron configuration is the arrangement of electrons around the nucleus of an atom with lower energy levels. Electrons occupying the orbitals of varying energy levels naturally fall towards the lowest energy state or ground state. For example, the ground state electron configuration of oxygen is 1s2 2s2 2p4

(But to simplify it, A short hand notation to indicate the electron orbitals which are filled in a particular atom.)

The diagonal rule basically says that electrons fill orbitals in order of increasing "quantum number sum" (n + ℓ). When two orbitals share the same "quantum number sum", they will be filled in order of increasing n.

1. The absolute error is 128.76 g

2. The relative error is 0.99

3. The percentage error is 99%

From the question given above, the following data were obtained:

Absolute error = |Accepted - Measured|

Absolute error = |130 - 1.24|

Absolute error = 128.76 g

Relative error = Absolute error / accepted value

Relaive error = 128.76 / 130

Relative error = 0.99

Percenrage error = relative error × 100

Percenrage error = 0.99 × 100

Percenrage error = 99%

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Answer:

\(\rm MgBr_2 \cdot 6\, H_2O\).

Explanation:

Let \(x\) denote the number of \(\rm H_{2}O\) formula units for every \(\rm MgCl_{2}\) formula unit. The formula of the hydrate would be \({\rm MgBr_2} \cdot x\, {\rm H_2O}\).

Look up the relative atomic mass of each element on a modern periodic table:

Formula mass:

\(\begin{aligned}M({\rm MgBr_{2}}) &= 24.305 + 2 \times 79.904 \\ &= 184.113\; \rm g\cdot mol^{-1}\end{aligned}\).

\(\begin{aligned}M({\rm H_{2}O}) &= 2 \times 1.008 + 15.999 \\ &= 18.015\; \rm g\cdot mol^{-1}\end{aligned}\).

The anhydrous compound would be \(\rm MgBr_{2}\). There was \(27.21\; \rm g\) of this compound. Calculate the number of moles of formula units in that much of this compound:

\(\begin{aligned}n({\rm MgBr_{2}}) &= \frac{m({\rm MgBr_{2}})}{M({\rm MgBr_{2}})} \\ &= \frac{27.21\; \rm g}{184.113\; \rm g \cdot mol^{-1}} \\ &\approx 0.1478\; \rm mol\end{aligned}\).

During heating, \(43.19\; \rm g - 27.21\; \rm g = 15.98\; \rm g\) of water \(\rm H_{2}O\) escaped from the hydrate. Calculate the number of formula units in that \(15.98\; \rm g\) of \(\rm H_{2}O\!\):

\(\begin{aligned}n({\rm H_{2}O}) &= \frac{m({\rm H_{2}O})}{M({\rm H_{2}O})} \\ &=\frac{43.19\; \rm g - 27.21\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \\&= \frac{15.98\; \rm g}{18.015\; \rm g \cdot mol^{-1}} \\ &\approx 0.887\; \rm mol\end{aligned}\).

In other words, approximately \(0.1478\; \rm mol\) of \(\rm MgBr_{2}\) formula units (and hence the same number of moles of \({\rm MgBr_2} \cdot x\, {\rm H_2O}\) formula units) contain approximately \(0.887\; \rm mol\) of \(\rm H_{2}O\) formula units. Calculate the ratio between the two:

\(\begin{aligned}x &= \frac{n({\rm H_{2}O})}{n({\rm MgBr_2})} \\ &\approx \frac{0.1478\; \rm mol}{0.887\; \rm mol} \approx 6\end{aligned}\).

Hence, the formula of the hydrate would be \(\rm MgBr_2 \cdot 6\, H_2O\).

Chemistry is the scientific discipline involved with elements and compounds composed of atoms, molecules and ions: their composition, structure, properties, behavior and the changes they undergo during a reaction with other substances.