In the 1920s, western powers began to expand their influence in the middle east primarily due to __________. A. The discovery of oil in the regionb. The rising number of terrorist attacks against the westc. European desires to create a jewish homelandd. Rising fundamentalist movements in iran and saudi arabia please select the best answer from the choices provided. Abcd.

The way that using locally made goods can lower CO2 emissions is by reducing the emissions from the transport needed to take foreign made goods to a place.

The use of locally made goods can lower CO2 emissions by reducing the transportation distance needed to transport goods. When goods are manufactured and consumed locally, they do not need to be transported over long distances, which can greatly reduce the carbon emissions associated with transportation.

Transportation is a significant source of greenhouse gas emissions, particularly emissions from the burning of fossil fuels like gasoline and diesel. By reducing the amount of transportation needed, the use of locally made goods can help to lower the overall carbon footprint of an economy.

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Answer:

7,3g

Explanation:

8.336x10^-13 moles x 88 g/mol =7,3 g

Answer:

0.13 grams.

Explanation:

To calculate the mass of iron(III) oxide that contains a trillion oxygen atoms, we first need to determine the molar mass of iron(III) oxide, also known as ferric oxide or Fe2O3. The molar mass of Fe2O3 is approximately 159.69 g/mol.

Next, we need to convert the number of oxygen atoms to moles using Avogadro's number, which is 6.022 x 10^23 atoms/mol. One mole of Fe2O3 contains two moles of oxygen, so the number of moles of Fe2O3 can be calculated as:

(1 trillion oxygen atoms) / (2 * 6.022 x 10^23 atoms/mol) = 8.37 x 10^-12 moles

Finally, we can calculate the mass of iron(III) oxide as:

(8.37 x 10^-12 moles) * (159.69 g/mol) = approximately 0.13 g

So, the mass of iron(III) oxide that contains a trillion oxygen atoms is approximately 0.13 grams.


Allen i little forget about this Sorry

Answer:

Ni(NO3)2 (aq) + Na2S (aq) -> NiS (s) + 2NaNO3 (aq)

Explanation:

Answer:

A) 6.50 moles 0₂

Explanation:

In the reaction, 2 moles of Calcium oxide decomposes to produce 2 moles of Calcium and 1 mole of Oxygen gas.

So, when 6.50 moles of Calcium oxide decomposes, 6.50 x (1 mole O₂ / 2 moles Calcium oxide) = 6.50 x 0.5 = 3.25 moles of Oxygen gas will be produced.

And as 3.25 x 2 =  moles O₂,

Answer: Single Replacement

Explanation:

This reaction is single replacement because the Zinc pushes out the Copper from the CuCl2 to form ZnCl. This happens because Zinc is above Copper in the activity series.

The balanced equation for the reaction between iron (III) hydroxide and water vapor to produce iron (III) oxide is Fe(OH)₃ -> Fe₂O₃ + 3H₂O

a. 26.5 grams of Fe(OH)₃ used.

b. 40 grams of Fe2O3 produced

Determine the number of moles of water produced: 0.75 L of water at STP (standard temperature and pressure) is equal to 0.75 moles.

Find the number of moles of Fe(OH)3: The reaction produces 3 moles of water for every 1 mole of Fe(OH)3, so 0.75 moles / 3 = 0.25 moles of Fe(OH)3.

Find the mass of Fe(OH)3: Fe(OH)3 has a molar mass of 106 g/mol, so 0.25 moles * 106 g/mol = 26.5 grams of Fe(OH)3.

So the answer to part a is 26.5 grams of Fe(OH)3 were used.

Determine the number of moles of Fe2O3: 0.25 moles of Fe(OH)3 produced 0.25 moles of Fe2O3.

Find the mass of Fe2O3: Fe2O3 has a molar mass of 160 g/mol, so 0.25 moles * 160 g/mol = 40 grams of Fe2O3.

So the answer to part b is 40 grams of Fe2O3 were produced.

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Answer:

the final volume of the gas inside the cylinder after 0.500 mole has leaked out is 37.0 L.

Explanation:

The ideal gas law states that 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. Since the pressure and temperature are constant, we can assume that they remain constant during the process of the gas leaking out. So, the initial state (1) and the final state (2) of the gas can be described as follows:

(1) PV1 = n1RT

(2) PV2 = (n1 - 0.500)RT

Now, we can find the final volume of the gas by substituting the known values into the equation and solving for V2:

V2 = (n1RT)/P = (1.90 * 8.31 * T)/P

V2 = (1.90 * 8.31 * T)/P = (1.90 * 8.31 * T)/P = 49.0

V2 = 49.0 L * (1.90 - 0.500) / 1.90

V2 = 37.0 L

The mass of the HgO is 434 g.

Stoichiometry allows us to calculate the amount of each reactant and product involved in a reaction based on its chemical formula. This is done by using the balanced chemical equation for the reaction, which shows the ratio of moles of each reactant and product involved in the reaction.

The reaction equation is; 2HgO(s)→Hg(l)+O2(g).

The number of moles of the oxygen is; 32.1g/ 32 g/mol

= 1 mole

We have to know form the reaction equation that;

2 moles of the HgO produces 1 mole of oxygen

x moles of HgO would produce 1 moles of oxygen

x = 2 moles

Mass of the compound = 2 moles * 217 g/mol

= 434 g

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The total number of atoms in the formula is 2 + 4 + 4 = 10 atoms.

What are atoms?

An atom is the basic unit of matter and the defining structure of elements. The term "atom" comes from the Greek word for indivisible, because it was once thought that atoms were the smallest things in the universe and could not be divided.

Atoms are composed of a nucleus, which contains positively charged protons and uncharged neutrons, surrounded by a cloud of negatively charged electrons. The number of protons in an atom's nucleus determines what element the atom is, as each element has a unique number of protons.

To count the number of atoms in the formula 2Ca(OH)2, we need to determine the number of each type of atom present.

In this formula, there are:

So, the total number of atoms in the formula is 2 + 4 + 4 = 10 atoms.

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The formulas for the compounds formed between platinum ions and bromide ions are Platinum (II) bromide (PtBr2) and Platinum (IV) bromide (PtBr4)

In both of these compounds, the bonding between the platinum ion and the bromide ions is primarily ionic in nature.

In platinum (II) bromide (PtBr2), each platinum ion is surrounded by two bromide ions, and each bromide ion is bonded to one platinum ion. The platinum ion has a +2 charge, and the two bromide ions have a -1 charge each, so the overall charge of the compound is neutral.

In platinum (IV) bromide (PtBr4), each platinum ion is surrounded by four bromide ions, and each bromide ion is bonded to one platinum ion. The platinum ion has a +4 charge, and the four bromide ions have a -1 charge each, so the overall charge of the compound is neutral.

The bond between the platinum ion and the bromide ions is a result of the attraction between their opposite charges.

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Answer:

792 moles.

Explanation:

When 66 moles of HNO3 are consumed, 6 x 66 = 396 moles of NO2 are produced.

Since the equation shows a 1:2 mole ratio between HNO3 and H2O, the number of moles of H2O produced is 2 x 396 = 792 moles.

That's a good summary of how to represent amino acids in Fischer projections and determine their D/L configuration.

Just to clarify, the orientation of the amino group is actually determined by looking at the lowest chiral center of the molecule, which is usually the alpha carbon (the carbon next to the carboxyl group). If the amino group is on the right side of this carbon in the Fischer projection, it is an L-amino acid; if it's on the left side, it's a D-amino acid. This may seem counterintuitive, but it's because the Fischer projection is actually a 2D representation of a 3D molecule, and the labels "D" and "L" were historically assigned based on the orientation of the molecule in space rather than the orientation on paper.

Additionally, the R/S system is a different method of assigning absolute configuration (not D/L) to chiral centers in molecules, including amino acids. In this system, the orientation of the groups around the chiral center are ranked by priority (based on atomic number), and the molecule is oriented so that the lowest priority group is pointing away from the viewer. The remaining three groups are then prioritized in a clockwise or counterclockwise direction, and the molecule is assigned an R or S configuration based on the direction of the priority sequence.

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If the reaction yield is 95.7%, then 2.37 grams of lead (II) oxide will be produced by the decomposition of 2.50 grams of lead (II) carbonate.

Answer:

2.54g

Explanation:

Lead carbonate (PbCO3) can be decomposed to produce lead oxide (PbO) and carbon dioxide (CO2). The equation for this reaction is as follows:

PbCO3(s) -> PbO(s) + CO2(g)

Let's assume that the reaction yield is 95.7%, meaning that 95.7% of the theoretical amount of lead oxide produced in the reaction is actually obtained. To find the actual amount of lead oxide produced, we first need to find the theoretical amount of lead oxide produced by the reaction.

The theoretical amount of lead oxide can be calculated using the stoichiometry of the reaction, where the number of moles of reactants is balanced with the number of moles of products. If we assume that 2.50g of lead carbonate is decomposed, the number of moles of lead carbonate can be calculated as follows:

n = m/M

where n is the number of moles, m is the mass of the substance, and M is the molar mass of the substance. For lead carbonate, the molar mass is:

M = 207.19 g/mol

So, the number of moles of lead carbonate is:

n = 2.50 g / 207.19 g/mol = 0.01202 mol

Since the reaction is balanced, the number of moles of lead oxide produced should be equal to the number of moles of lead carbonate. The mass of lead oxide produced can be calculated using the number of moles and the molar mass of lead oxide:

m = n x M

where M = 223.20 g/mol is the molar mass of lead oxide. So, the mass of lead oxide produced is:

m = 0.01202 mol x 223.20 g/mol = 2.68 g

Since the reaction yield is 95.7%, the actual amount of lead oxide produced is:

actual_mass = 0.957 x 2.68 g = 2.54 g

So, approximately 2.54g of lead oxide will be produced by the decomposition of 2.50g of lead carbonate.

hope it helped .r.a.t.e

The amount of iodine-123 that will still be active 60 hours later is 0.0766 mCi.

The effective half-life of iodine-123 is 12 hours, which means that after each 12-hour period, the activity of the sample will be reduced by half. After 60 hours, or 5 half-lives, the activity will be reduced to:

(1/2)^5 = 1/32

So only 1/32 of the original activity will remain. To find the activity that remains, we can multiply the initial activity by the fraction of the original activity that remains:

Remaining activity = (1/32) x 2.45 mCi = 0.0766 mCi

Therefore, the amount of iodine-123 that will still be active 60 hours later is 0.0766 mCi.

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About 0.153125 mCi of iodine-123 will still be active 60 hours later.

Effective half-life of a radioactive substance is the amount of time it takes for the activity of the substance to decrease to half of its initial value, taking into account both the physical half-life of the substance and any biological processes that may affect its decay rate.

In this case, the effective half-life of iodine-123 is 12 hours. This means that after 12 hours, the activity of the substance will be reduced to half its initial value, and after another 12 hours (i.e., 24 hours after the initial dose), it will be reduced to one-quarter of its initial value, and so on.

To calculate the amount of iodine-123 that will still be active 60 hours later, we can use the following formula:

activity = initial activity x (1/2)^(t / t1/2)

where

Plugging in the values, we get:

activity = 2.45 mCi x (1/2)^(60 hours / 12 hours)

activity = 2.45 mCi x 0.0625

activity = 0.153125 mCi

Therefore, about 0.153125 mCi of iodine-123 will still be active 60 hours later.

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Answer:

5.125 moles of water

Explanation:

Given the reaction:

Cu + 8HNO3 --> 3 Cu(NO3)2 + 2 NO + 4 H2O

When 41 moles of HNO3 are consumed, we can calculate the number of moles of water produced using the stoichiometric coefficients in the balanced chemical equation.

According to the coefficients, 1 mole of Cu reacts with 8 moles of HNO3 to produce 4 moles of H2O. Therefore, when 41 moles of HNO3 are consumed, we can calculate the number of moles of water produced as follows:

41 moles HNO3 / 8 moles HNO3 per mole H2O = 5.125 moles H2O

So, 5.125 moles of water can be made when 41 moles of HNO3 are consumed.

(a) Expanded structure:

CH3–CH2–CH–CH3

Condensed structure:

CH3CH2CHCH3

(b) Carbon a is a primary carbon and Carbon b is a secondary carbon.

Generally, Chemical formulas are symbols used to represent elements and molecules.

They usually consist of the symbols for the elements in the molecule, with subscripts indicating the number of each atom.

For example, the chemical formula for water is H2O, indicating two hydrogen atoms and one oxygen atom.

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An equation that shows the formation of a copper (ii) ion from a neutral copper atom is Cu ⇒ Cu⁺² + 2e⁻.

Oxidation occurs when a reactant loses electrons during a reaction. Reduction occurs when a reactant accumulates electrons during a reaction. When metals react with acid, this is a common occurrence. Oxidation occurs when a reactant loses electrons during a reaction.

When an atom looses an electron to form positive ion, this process is called as an oxidation reaction.

Copper will lose 2 electron to form +2 ion. The equation for the formation of copper (II) ion from neutral copper atom follows:

Cu ⇒ Cu⁺² + 2e⁻

Thus, Cu ⇒ Cu⁺² + 2e⁻ is an equation that shows the formation of a copper (ii) ion.

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According to the recipe, to make 4 moles of lemonade, you use 2 moles of water, one mole of sugar and one mole of lemon juice, expressed in grams:

2 water  + sugar + lemon juice = 4 lemonade

2*(236.59) + 225g + 257.83g  = 4*(719.42)g

   473.18g + 225g + 257.83g = 2877.68g

So for every 2877.68g of lemonade made, they use 473.18g of water, 225g of sugar, and 257.83g of lemon juice.

You know that they made a batch of 2050.25g, so to detect the limiting reactant, first, you have to calculate, in theory, how much of each ingredient you need to make the given amount of lemonade:

Use cross multiplication

Water:

2877.68g lemonade → 473.18g water

2050.25g lemonade → X= (2050.25*473.18)/2877.68= 337.12g water

Following the recipe, to elaborate 2050.25g of lemonade, you need to use 337.12g of water.

Sugar:

2877.68g lemonade → 225g sugar

2050.25g lemonade → X= (2050.25*225)/2877.68= 160.30g sugar.

To elaborate 2050.25f of lemonade you need to use 160.30g of sugar.

Lemon juice:

2877.68g lemonade → 257.83g lemon juice

2050.25g lemonade → X= (2050.25*257.83)/2877.68= 183.69g lemon juice.

To elaborate 2050.25f of lemonade you need to use 183.69g lemon juice.

Available ingredients vs. theoretical yields for 2050.25g of lemonade:

Water 946.36 g → 337.12g

Sugar 196.86 g → 160.30g

Lemon Juice 193.37 g → 183.69g

The lemon juice will be the first ingredient to be used up, there will be a surplus of water and sugar.

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The Avogadro constant, which is 6.022 x 10²³ molecules/mol, must be utilized in order to determine the number of formula units or molecules present in a sample.

The number of entities (such as atoms, molecules, or ions) that make up one mole of a substance is represented by the Avogadro constant, a fundamental physical constant. The letter "NA" stands for the Avogadro constant value, which is approximately 6.022 x 10²³ entities per mole. This indicates that a substance has exactly 6.022 x 10²³ entities in a single mole.

Part A:

Molar mass of H₂O = 18.015 g/mol

Number of moles of H₂O = 60.1 g / 18.015 g/mol

                                  = 3.33 mol

Number of molecules of H₂O = 3.33 mol x 6.022 x 10²³ molecules/mol

                                    = 2.00 x 10²⁴ molecules

N = 2.00 x 10²⁴ molecules H₂O

Part B:

Molar mass of Na₃PO₄ = 163.94 g/mol

Number of moles of Na₃PO₄ = 127 mg / 1000 mg/g / 163.94 g/mol

                                   = 7.73 x 10⁻⁴ mol

Number of formula units of Na₃PO₄ = 7.73 x 10⁻⁴ mol x 6.022 x 10²³ formula units/mol = 4.65 x 10²⁰ formula units

N = 4.65 x 10²⁰ formula units Na₃PO₄

Part C:

Molar mass of C₆H₁₂O₆ = 180.16 g/mol

Number of moles of C₆H₁₂O₆ = 29.6 kg / 1000 g/kg / 180.16 g/mol

                                        = 0.164 mol

Number of molecules of C₆H₁₂O₆ = 0.164 mol x 6.022 x 10²³ molecules/mol = 9.88 x 10²² molecules

N = 9.88 x 10²² molecules C₆H₁₂O₆

Part D:

Molar mass of N₂ = 28.02 g/mol

Number of moles of N₂ = 46.9 g / 28.02 g/mol

                                  = 1.67 mol

Number of molecules of N₂ = 1.67 mol x 6.022 x 10²³ molecules/mol

                         = 1.00 x 10²⁴ molecules

N = 1.00 x 10²⁴ molecules N₂

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The percent yield of Cl2 in the electrolytic decomposition of HCl is 54.18%.

Electrolytic decomposition is a type of chemical reaction that uses electricity to break down a compound into its component elements or ions. This process involves passing an electric current through an electrolyte, which is a substance that conducts electricity when dissolved in a solvent or melted.

To calculate the percent yield of Cl2 in the electrolytic decomposition of HCl, we need to compare the actual yield of Cl2 obtained in the experiment to the theoretical yield of Cl2 that could be obtained if all the HCl reacted completely.

After balancing,

2 HCl (aq) → Cl2 (g) + H2 (g)

The molar mass of HCl is 36.46 g/mol, so 25.8 g of HCl is equal to 25.8 g / 36.46 g/mol = 0.708 mol of HCl.

According to the balanced equation, 2 moles of HCl should produce 1 mole of Cl2, so the theoretical yield of Cl2 can be calculated as:

Theoretical yield of Cl2 = 0.5 × 0.708 mol = 0.354 mol

The molar mass of Cl2 is 70.90 g/mol, so the theoretical yield of Cl2 in grams is:

Theoretical yield of Cl2 = 0.354 mol × 70.90 g/mol = 25.1 g

The actual yield of Cl2 obtained in the experiment is 13.6 g.

The percent yield of Cl2 can be calculated as:

Percent yield = (Actual yield / Theoretical yield) × 100%

Percent yield = (13.6 g / 25.1 g) × 100%

Percent yield = 54.18%

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Answer: 23

Explanation:

because it is been multiplyed by 3 has s 31 nucleons and has atomic numbe equal to 15. The number of electron having m = 0 i

Answer:

The number of electrons with magnetic quantum number m = 0 can be calculated from the atomic number of the element. The atomic number gives the number of protons, and thus the number of electrons in a neutral atom. For the anion X-3, we have 15 - 3 = 12 electrons. The magnetic quantum number m can have integer values from -j to +j, where j is a half-integer representing the total angular momentum quantum number of the electron. In this case, j can be equal to 1/2, 3/2, 5/2, and so on. For the lowest value of j, which is 1/2, the magnetic quantum number m can have two values, +1/2 and -1/2. Thus, there are two electrons with m = 0. The number of electrons with m = 0 is equal to the number of electrons in the lowest energy level, which is the 1s orbital. In this case, two electrons occupy the 1s orbital, and both have m = 0.

Explanation:

Answer:

Fatty acids methyl esters (FAMEs) are a class of biofuels derived from vegetable oils or animal fats and are commonly used as a diesel fuel substitute. FAMEs are suitable for analysis using gas chromatography (GC) because of their relatively high boiling points, which allow for efficient separation and detection of individual components.

Answer:

Fatty acids methyl esters (FAME) are suitable for analysis using gas chromatography (GC) due to their high volatility and thermal stability. These characteristics make FAMEs ideal for GC analysis because they can be vaporized and separated easily into their individual components. This is because the esterification of fatty acids to form FAMEs results in the introduction of a methyl group, which reduces the molecular weight and increases the volatility of the fatty acids. Additionally, the methyl ester functional group has a high smoking point, meaning that it can withstand high temperatures without breaking down, making it suitable for GC analysis where high temperature conditions are used to vaporize and separate the components. The use of GC in the analysis of FAMEs provides accurate and precise information on the composition of fatty acids in various samples, which is useful in applications such as the analysis of edible oils, biodiesel, and other lipid-containing materials.

Explanation:

Answer:

0.041 grams of carbon-14 remaining.

Explanation:

The half-life of carbon-14 is the time it takes for half of the original amount of carbon-14 to decay. In this case, the half-life of carbon-14 is 5,730 years. This means that after 5,730 years, half of the original amount of carbon-14 will remain, and half will have decayed.

In this problem, we are given that a fossilized tree branch originally contained 4.30 grams of C-14, and we are asked how much C-14 would be left after 50,000 years. To find this, we need to calculate how many half-lives have occurred over this time period, and then find the resulting amount of C-14.

We can use the formula N = N0 * (1/2)^(t/t1/2) to calculate the amount of C-14 after a given number of half-lives, where N0 is the initial amount of C-14, t is the total time elapsed, and t1/2 is the half-life of the substance.

Plugging in the given values, we get N = 4.30 g * (1/2)^(50000 years / 5730 years/half-life) = 0.041 g.

So, after 50,000 years, the tree branch will have 0.041 grams of carbon-14 remaining.


ALLEN

One of the main terms used in physics to describe the radioactive decay of a specific sample or element over a predetermined amount of time is half-life, also known as half-life period. After 50,000 years, the tree branch will have 0.041 grams of carbon-14 remaining.

A radioactive material's half-life is typically described as the amount of time it takes for one half of its atoms to decay or change into another substance. Ernest Rutherford made the first discovery of the theory in 1907. It is typically denoted by the letters Ug or t1/2.

Understanding half-lives is crucial because they allow you to determine if a sample of radioactive material is safe to handle. A sample is deemed safe when its radioactivity is below detection thresholds. Ten half-lives later, something occurs.

We can use the formula N = N0 * (1/2)^(t/t1/2)

N = 4.30 g * (1/2)^(50000 years / 5730 years/half-life) = 0.041 g.

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The CCl4 molecule is larger and has more electrons hence it is more polarizable and has a larger boiling point

A higher molar mass does imply a higher boiling point for a substance, assuming that the other conditions (pressure, temperature, etc.) are constant.

This is because the boiling point is a measure of the temperature at which a substance changes from a liquid to a gas at a given pressure. As the molar mass of a substance increases, the intermolecular forces between the molecules tend to be stronger, which makes it more difficult to separate the molecules and vaporize the substance. As a result, the boiling point generally increases with increasing molar mass.

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According to the molecular geometry, with the help of structure of molecules which provide information on site of attachment with other molecules, one can determine their mode of reaction.

Molecular geometry can be defined as a three -dimensional arrangement of atoms which constitute the molecule.It includes parameters like bond length,bond angle and torsional angles.

It influences many properties of molecules like reactivity,polarity color,magnetism .The molecular geometry can be determined by various spectroscopic methods and diffraction methods , some of which are infrared,microwave and Raman spectroscopy.

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Answer: The shape of a molecule helps to determine its properties, which affect how a molecule interacts with other molecules, such as polarity and bonding.

According to the question the molar mass of C2H5OH is 46.07 g/mol.

Molar mass is the mass of a given substance (expressed in grams) divided by the amount of substance (expressed in moles). It is also known as the molecular weight of a substance, and is the sum of the atomic masses of all the atoms in a molecule. Molar mass is used to calculate the mass of a compound in a given volume, and is commonly used in the fields of chemistry, physics, and biology to identify, quantify, and measure substances. It is also used to determine the density of a given substance, and is an important part of formulas used to calculate the concentration of a particular substance in solution.

To calculate the number of molecules, we must divide the given mass of 11.5 g by the molar mass of C2H5OH. This gives us a result of 250.03 molecules of C2H5OH.

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The team provides two groups of students with the same materials, but one group must make a bridge while the other group makes a building. The bridge model is correct.

A gene is the fundamental and structural unit of the hereditary that codes for specific information and proteins required for signaling and cell function.

A mutation or change in the sequence of the region will result in the modification of the proteins coded, resulting in a change in the genotype and phenotype.

If a gene is mutated and the proteins change, the entire sequence and the information for which it codes will change in the building model.

Thus, gene mutation will impact the proteins.

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The molality and density of an aqueous solution of glucose (mm = 180. 2 g/mol) are 5. 21 and 1. 20 g/ml, respectively. Therefore, the molarity of glucose in the solution is 0.0355 M.

Molality is the number of moles of solute per kilogram of solvent, while molarity is the number of moles of solute per liter of solution.

molarity (M) = molality (m) * (density of solvent in g/mL) / (molar mass of solute in g/mol)

molality (m) = 5.21 m

the density of solvent (g/mL) = 1.20 g/mL

molar mass of solute (g/mol) = 180.2 g/mol

So, molarity (M) = 5.21 m * (1.20 g/mL) / (180.2 g/mol)

molarity (M) = 5.21 m * 0.0067 g/mL/g mol

molarity (M) = 0.0355 M

Therefore, the molarity of glucose in the solution is 0.0355 M.

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The average atomic mass  for the new element is given 81.5529 amu. The percentage abundance for the third isotope is 13.6 %. Then its atomic mass is 83.9 amu.

Isotopes are atoms of same element with different atomic mass numbers. Almost all elements have two or more isotopes but not all of them are stable in nature.

The average atomic mass can be calculated from the isotopic mass and and percentage abundance as follows:

atomic mass = ∑ (isotopic mass × %abundance /100 )

Apply the isotopic mass and abundance of all isotopes given as follows:

atomic mass of the element = (76.12/100 ×81.6643 amu) +  (10.27/100 × 81.1239 amu)  +  (13.6 /100 × X amu) =81.5529 amu.

Then,

(13.6 /100 × X amu) = 70.41 amu.

X  = 83.9 amu.

Therefore, the atomic mass of the third isotope is 83.9 amu.

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