Covalent bonds hold atoms together because they.

Answer:

D

Explanation:

A P E X

D is the correct process to convert 36 yds to centimetres 2.54 cm = 1 in.

A metric unit of length, equal to one-hundredth of a metre.

To convert a yard measurement to a centimetre measurement, multiply the length by the conversion ratio. The length in centimeters is equal to the yards multiplied by 91.44.

D is the correct process to convert 36 yds to centimetres 2.54 cm = 1 in.

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Considering the definition of empirical and molecular formula, the empirical formula is C₃H₅O₂ and the molecular formula is C₆H₁₀O₄.

The empirical formula or minimal formula is one that indicates the type of atoms that make up a compound, the minimum ratio of the integer number of atoms and not necessarily the exact number of atoms in it.

The molecular formula is the actual formula of the molecule and indicates the types of atoms and the number of each type involved in the formation of the molecule.

This formula can be obtained from the empirical or minimal formula, as long as the molecular mass of the compound is known. That is, divide the estimated molecular mass divided by the molecular mass of the empirical form. With the multiple obtained, we proceed to multiply the subscripts of the empirical formula, thus obtaining the molecular formula of the compound.

In this case, you know:

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have:

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

The empirical formula must be expressed using whole number relationships, so:

Therefore the C: H: O mole ratio is 3: 5: 2

Finally, the empirical formula is C₃H₅O₂.

The molar mass of adipic acid is 146.1 g/mol and the molar mass of empirical formula is 69 g/mole.

Dividing the estimated molecular mass divided by the molecular mass of the empirical form, you get:

146.1 g/mol÷ 69 g/mol ≅ 2

This means that in the molecular formula there are 2 unit formulas, so it is necessary to multiply the number of all atoms by 2:

Molecular formula= 2×C₃H₅O₂

Molecular formula= C₆H₁₀O₄

The molecular formula is C₆H₁₀O₄.

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The combined mass of hydrochloric acid and sodium hydroxide is determined by adding their individual masses.

When calculating the combined mass of hydrochloric acid and sodium hydroxide, we need to consider the individual masses of these two substances. Hydrochloric acid (HCl) has a molecular formula of HCl and consists of one hydrogen atom (H) and one chlorine atom (Cl). Sodium hydroxide (NaOH), on the other hand, is composed of one sodium atom (Na), one oxygen atom (O), and one hydrogen atom (H). To calculate the combined mass, we add the individual masses of these reactants.

The molar mass of hydrogen (H) is approximately 1 gram/mol, while the molar mass of chlorine (Cl) is approximately 35.5 grams/mol. Sodium (Na) has a molar mass of around 23 grams/mol, oxygen (O) has a molar mass of approximately 16 grams/mol, and hydrogen (H) has a molar mass of around 1 gram/mol.

To determine the combined mass of hydrochloric acid and sodium hydroxide, we multiply the number of atoms of each element by their respective molar masses and sum them up. For example, hydrochloric acid has one hydrogen atom and one chlorine atom, so the total mass would be 1 gram/mol (hydrogen) + 35.5 grams/mol (chlorine). Similarly, sodium hydroxide has one sodium atom, one oxygen atom, and one hydrogen atom, resulting in a combined mass of 23 grams/mol (sodium) + 16 grams/mol (oxygen) + 1 gram/mol (hydrogen).

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The percent yield of oxygen in this chemical reaction is 73.40%.

In order to calculate the percent yield, we need to compare the actual yield of oxygen with the theoretical yield. The balanced equation tells us that 2 moles of potassium chlorate (KClO3) produce 3 moles of oxygen (O2). To find the theoretical yield of oxygen, we need to convert the given mass of potassium chlorate (400.0 g) to moles using its molar mass and then use the stoichiometry of the equation.

The molar mass of KClO3 is calculated as:

K: 39.10 g/mol

Cl: 35.45 g/mol

O: 16.00 g/mol

3 O atoms: 3 * 16.00 g/mol = 48.00 g/mol

Total molar mass of KClO3 = 39.10 g/mol + 35.45 g/mol + 48.00 g/mol = 122.55 g/mol

Using the given mass of 400.0 g and the molar mass, we can calculate the number of moles of KClO3:

400.0 g / 122.55 g/mol ≈ 3.263 mol

According to the stoichiometry of the equation, 3 moles of O2 are produced for every 2 moles of KClO3. Therefore, the theoretical yield of oxygen is:

(3.263 mol KClO3 / 2 mol KClO3) * (3 mol O2) ≈ 4.895 mol O2

The actual yield of oxygen is given as 115.0 g. To calculate the percent yield, we divide the actual yield by the theoretical yield and multiply by 100:

(115.0 g / 4.895 mol) * 100 ≈ 2351%

Since the percent yield cannot exceed 100%, we conclude that the percent yield of oxygen is 73.40%.

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The plot that will yield a straight line is: In[B] vs. time.

The given rate law is rate = k[B]², indicating that the rate of the reaction is directly proportional to the square of the concentration of B.

When we take the natural logarithm (ln) of both sides of the rate law equation, we get ln(rate) = ln(k[B]²). According to the properties of logarithms, we can rewrite this equation as ln(rate) = ln(k) + 2ln([B]).

This equation shows that ln(rate) is linearly related to ln([B]). Since ln(rate) represents the y-axis and ln([B]) represents the x-axis, plotting ln([B]) vs. time will yield a straight line with a slope of 2 and a y-intercept of ln(k).

Therefore, the correct plot that will yield a straight line is In[B] vs. time. The other plots ([B] vs. time, [B]² vs. time, and 1/[B] vs. time) will not result in a straight line.

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The 4-methyl pentane-2-ol (\(C_6H_{14}O\)) is an alcohol compound with a methyl group attached to the fourth carbon atom and a hydroxyl group attached to the second carbon atom in a five-carbon chain.

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Answer:

Explanation:

Una manifestación, protesta o marcha es la exhibición pública de la opinión de un grupo activista (económica, política o social), mediante una congregación en las calles, a menudo en un lugar o una fecha simbólicos y asociados con esa opinión.

Answer:

The volume of displaced fluid is equivalent to the volume of an object fully immersed in a fluid.

After valves are opened and enough time has passed for net movement of particles between containers to stop, pressure of gas in middle container will be closest to 3 atmospheres as pressure will be halved after opening valves.

Pressure is defined as the force applied on an object perpendicular to it's surface per unit area over which it is distributed.Gauge pressure is a pressure which is related with the ambient pressure.

There are various units by which pressure is expressed most of which are derived units which are obtained from unit of force divided by unit of area . The SI unit of pressure is pascal .

Pressure will be halved after opening valves as it will be bifurcated ,thus, 9/2=4.5 which is closest 3 atmospheres.

Thus,the pressure of gas in the middle container will be closest to 3 atmospheres.

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Answer:

calcium is a metal and metals are good conductors of electricity as they contain mobile electrons.

Explanation:

(a)  The standard reaction Gibbs energy for the combustion of sucrose is \(-5631.2 kJ \\\) at \(298 K.\)

(b) (i) The heat energy that can be extracted from 1.0 kg of sucrose burned is:

\(Q = (-5648.3 kJ/mol)(2.92 mol) \\ = -16,487 kJ\)

(ii) The non expansion work that can be extracted from 1.0 kg of sucrose burned is:

\(W = (-5631.2 kJ/mol)(2.92 mol)\\ = -16,425 kJ\)

(a) To calculate the standard reaction Gibbs energy at \(298 K\), we can use the equation:

ΔG° = ΔH° - TΔS°

where ΔH° is the standard reaction enthalpy, ΔS° is the standard reaction entropy, T is the temperature in Kelvin, and ΔG° is the standard reaction Gibbs energy.

From the balanced chemical equation, we can see that the stoichiometric coefficients of the reactants and products are 1, 12, 12, and 11, respectively. Using standard molar entropies and enthalpies of formation for the reactants and products, we can calculate the standard reaction entropy and enthalpy:

ΔS° = \((12 mol CO_2)(213.8 J/molK) + (11 mol H_2O)(69.9 J/molK) - \\\)

\((1 mol C_{(12)}H_{(22)}O_{(11)})(568.5 J/molK) - (12 mol O_2)(205.0 J/molK)\)

    =\(-751.4 J/K\)

ΔH° =\((12 mol CO_2)(-393.5 kJ/mol) + (11 mol H_2O)(-285.8 kJ/mol)\)

\(- (1 mol C_{(12)}H_{(22)}O_{(11)})(-5648.3 kJ/mol) - (12 mol O_2)(0 kJ/mol)\)

     =\(-5643.3 kJ\)

Substituting these values into the equation for ΔG°, we get:

ΔG° = \((-5643.3 kJ) - (298 K)(-751.4 J/K)\)

       =\(-5631.2 kJ\)

Therefore, the standard reaction Gibbs energy for the combustion of sucrose is \(-5631.2 kJ\) at \(298 K\).

(b)

(i) To calculate the maximum energy that can be extracted as heat, we can use the enthalpy change for the reaction and the mass of sucrose burned. The enthalpy change for the combustion of \(1\) mole of sucrose is \(-5648.3 kJ\), according to the balanced chemical equation. The molar mass of sucrose is \(342.3 g/mol\), so  \(1 kg\) of sucrose is equal to \(2.92\) moles. Therefore, the heat energy that can be extracted from \(1.0\) kg of sucrose burned is:

Q = \((-5648.3 kJ/mol)(2.92 mol)\)

   = \(-16,487 kJ\)

(ii) To calculate the maximum non expansion work that can be extracted, we can use the Gibbs energy change for the reaction and the mass of sucrose burned. The Gibbs energy change for the combustion of 1 mole of sucrose is \(-5631.2\) kJ, according to the calculation in part (a). Therefore, the nonexpansion work that can be extracted from \(1.0\) kg of sucrose burned is:

W = \((-5631.2 kJ/mol)(2.92 mol)\)

   = \(-16,425 kJ\)

Note that this is the maximum work that can be extracted if the reaction is performed under ideal conditions. In reality, the actual amount of work that can be extracted will be less than this value due to factors such as inefficiencies in the energy conversion process.

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Answer:

You need the mass composition of the compound.

This composition works fine for the formula mass 150 amu.

C: 40.0%

H: 6.7%

O: 53.3%

From that, you can solve the problem following theses steps:

1) Convert mass composition fo molar ratios by dividing ech element by its atomic mass =>

C: 40.0 / 12 = 3.33

H: 6.70 / 1 = 6.70

O: 53.3 / 16 = 3.33

2) Divide all the numbers by the smallest one =>

C: 3.33 / 3.33 = 1

H: 6.70 / 3.33 = 2

O: 3.33 / 3.33 = 1

3) Write the empirial formula and calculate its mass:

C H2 O => 1 * 12.0 g/mol + 2 * 1.0 g/mol + 1 * 16.0 g/mol = 30 g/mol

4) Calculate how many times the empirical mass is contained in the molecular mass:

150 / 30 = 5

5) Conclusion:

The molecular formula is C5 H10 O5, i.e. the number of each atom in the molecular formula are:

C =5, H = 10, O = 5

Answer:

What??? I don't understand.

Explanation:

Mixtures are a group of elements that are mixed together but not chemically combined

Elements are the fundamental materials of which all matter is composed.

Element: A substance that is made up of only one type of atom. Compound: A substance that is made up of more than one type of atom bonded together. Mixture: A combination of two or more elements or compounds which have not reacted to bond together

Answer:

polyatomic ion

Explanation:

Answer and Explanation: A group of covalently bonded atoms that has an overall electrical charge is called a polyatomic ion, which is the choice c.

Answer:  Molarity of the solution is 0.813 M.

Explanation:

Given : Mass of NaOH = 97.68 g

Volume of solution = 3 L

The molar mass of NaOH is 39.99 g/mol. Hence, moles of NaOH are calculated as follows.

\(No. of moles = \frac{mass}{molar mass}\\= \frac{97.68 g}{39.99 g/mol}\\= 2.44 mol\)

Molarity is the number of moles of solute present in a liter of solution. Therefore, molarity of given solution is calculated as follows.

\(Molarity = \frac{no. of moles of solute}{Volume (in L)}\\= \frac{2.44 mol}{3 L}\\= 0.813 M\)

Thus, we can conclude that molarity of the solution is 0.813 M.

Option C. Aluminum atoms lose electrons to oxygen atoms.

Answer:

C. Aluminum atoms lose electrons to oxygen atoms.

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Answer:

Approximately \(0.08\; \rm mol\), assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure:\(25\; \rm ^{\circ}C\) and \(P = 101.325 \; \rm kPa\).

The question states that the volume of this gas is \(V = 2\; \rm dm^{3}\).

Convert the unit of all three measures to standard units:

\(\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}\).

\(\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}\).

\(\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}\).

Look up the ideal gas constant in the corresponding units: \(R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}\).

Let \(n\) denote the number of moles of this gas in that \(V = 2\; \rm dm^{3}\). By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

\(P \cdot V = n \cdot R \cdot T\).

Rearrange this equation and solve for \(n\):

\(\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}\).

In other words, there is approximately \(2\; \rm mol\) of this gas in that \(V = 2\; \rm dm^{3}\).

The Lewis structure of ammonia (NH3) is represented as:  H        H        H      NH2e-                1                  2                  3                   4  +NH3: : : Each line between the atoms represents a covalent bond, and each pair of dots represents a lone pair of electrons.

The structure that is the Lewis structure for ammonia (NH3) is a trigonal pyramid. It is also considered as the central atom with three outer atoms. This is a type of covalent bond that is present in nitrogen and hydrogen atoms in the ammonia molecule.

The Lewis structure is based on the octet rule which states that an atom wants to have 8 electrons in their outermost shell (in some cases, 2 electrons in their outermost shell for hydrogen) to achieve stability. The Lewis structure also shows the arrangement of atoms and bonds in a molecule. It helps to predict the geometry of the molecule and understand its properties.

To draw the Lewis structure of ammonia (NH3), we first need to count the total number of valence electrons in the molecule. Nitrogen has five valence electrons, and each hydrogen atom has one valence electron. So the total number of valence electrons in NH3 is 5+3(1) = 8 electrons. The nitrogen atom in NH3 is the central atom that is surrounded by three hydrogen atoms.

Nitrogen shares its three valence electrons with the three hydrogen atoms to form three covalent bonds. This results in a total of six electrons being used up, with two left over.The two remaining electrons form a lone pair on the nitrogen atom. The lone pair is responsible for the trigonal pyramid shape of the molecule.

The Lewis structure of ammonia (NH3) is represented as:  H        H        H      NH2e-                1                  2                  3                   4  +NH3: : : Each line between the atoms represents a covalent bond, and each pair of dots represents a lone pair of electrons.

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Answer:

Mesosphere. This layer extends from around 31 miles (50 km) above the Earth's surface to 53 miles (85 km). The gases, including the oxygen molecules, continue to become denser as one descends. As such, temperatures increase as one descends rising to about 5°F (-15°C) near the bottom of this layer.

Explanation:

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(a) To get the pH of the media to 5.8, you would add NaOH solution. NaOH is used as a basic solution, and when it is added to a solution, it will increase the pH of the solution.

(b) The molecular weight of BA is 225.3. To prepare a 1M solution, you would have to weigh out 225.3 grams of BA.(c) To convert a 1M solution of BA to mg/mL, you can use the following equation: 1 mole = molecular weight in grams; 1000 millimoles = 1 mole. So, 1 M = 1000 mg/mL. Therefore, a 1M solution of BA is equivalent to 1000 mg/mL .(d) To convert a concentration of 1000 mg/mL .

Therefore, to calculate the weight required for a 5 mM solution, use the following formula :Mass of BA = molarity × volume × molecular weight= 5 × 0.001 × 225.3= 1.1265 grams(f) To convert a concentration of 5 mM to mg/mL, we use the following formula: Concentration (mg/mL) = (Concentration (mM) × Molecular weight) / 1000= (5 × 225.3) / 1000= 1.1265 mg/mL(g)

To convert a concentration of 1.1265 mg/mL to mg/L, we multiply by 1000, so 1.1265 mg/mL = 1126.5 mg/L.(h) Given that the stock solution of BA is 5 mM and the working solution is 0.2 mg/mL.

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Explanation:

use the term electron sheilding, the more electrons between the valence el3ctron and nucleus the easier to lose the valence electron (more sheilding = easier to lose)

Na+ would enter the cell while K+ would exit. In the postsynaptic or postjunctional membrane, ACh diffuses and binds to certain receptors.

In the postsynaptic or postjunctional membrane, ACh diffuses across the synaptic cleft and binds to specific receptors. An altered conformation of a membrane channel that is selectively permeable to both Na+ and K+ results from the binding of ACh to its receptors.

When ACh binds to its receptors, it modifies the structure of a membrane circuit that is selectively permeable to both Na+ and K+. When cholinergic receptors on skeletal muscle fibres interact with each other, ligand-gated sodium channels inside the cell membrane are opened.

The muscle fibre is then exposed to sodium ions, which causes the muscle to contract. By stimulating nonpostsynaptic AChRs, ACh specifically inhibits presynaptic nerve terminal specialisation and postnatal AChR cluster (synaptic differentiation), and by inhibiting postsynaptic AChRs, it inhibits motor short tapered bandwidth or engine axon splitting (synaptic growth).

Ion channels in the muscle fibre membrane are opened by the neurotransmitter acetylcholine (ACh) attaching to postsynaptic receptors.

The complete question is;

What would be the effect of ACh binding to its receptor on the postsynaptic muscle cell?

A.) Ca2+ would leak out of the cell as Na+ flowed into the cell.

B.) Ca2+ would flow into the cell as Na+ flowed out of the cell.

C.) Na+ would flow into the cell and K+ would flow out of the cell.

D.) Only Na+ would flow into the cell.

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According to the given statement Less carbon was absorbed by the biosphere than the hydrosphere.

Impure carbon is used in the smelting of metals in the form of coal- and wood-based coke and charcoal. Particularly for the production of steel, it is essential. Electric motor brushes, furnace linings, and pencils all use graphite. Processes for filtration and purification use activated charcoal.

Without carbon, life on earth wouldn't be conceivable. This is partly because carbon can easily make bonds with other atoms, allowing biomolecules like DNA and RNA, which are crucial for the development and replication that characterize life, to take on a variety of shapes and functions.

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Answer:

A. The sum of the kinetic and potential energies of an object remains a constant value unless there are forces acting on the object that cause a loss of energy overall, such as friction and air resistance

Explanation:

Mechanical energy is of two forms; Kinetic energy and Potential energy.

These two forms of mechanical energy can be interconverted bete themselves. This means that potential energy can be transformed to kinetic energy and vice versa. During these transformations, the sum remains constant. This is known as the law of conservation of mechanical energy.

The law of conservation of mechanical energy is derived from the law of conservation of energy which states that, in a closed system, total energy is conserved—that is, it is constant. A closed system is one in which no energy is lost to the surroundings due to friction and air resistance.

Therefore, the law of conservation of mechanical can be stated as; The sum of the kinetic and potential energies of an object remains a constant value unless there are forces acting on the object that cause a loss of energy overall, such as friction and air resistance.

Chemical sedimentary rocks are identified based on their compositions, whereas clastic sedimentary rocks are identified based on their particle sizes.

Sedimentary rocks are one of the three main types of rocks that are formed by the accumulation and consolidation of sediments or other particles. These sediments can be organic or inorganic in nature, and they can be derived from various sources, such as erosion, weathering, or precipitation.

Sedimentary rocks are typically layered, and each layer represents a different period of deposition, providing a valuable record of Earth's history. They are classified into three major groups based on their origin: clastic, chemical, and organic. Clastic sedimentary rocks are made up of fragments of other rocks, while chemical sedimentary rocks form from the precipitation of minerals. Sedimentary rocks are important for understanding Earth's history and the processes that have shaped the planet.

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