How many grams are in 0.0823 moles of Ar?

The property X is an extrinsic property.

The term extensive property refers to the type of property that depends on the amount of the substance present. We must recall that an intensive property does not in any way depend on the number or the amount of the substance present in the sample.

If we have a good look at the table that is being referred to here, we would see that the mass of the substance tends to increase as a certain un named property of the substance X is increasing. This implies that the property X is affected by the mass of the object which is the quantity of matter in the object. This gives us an idea that the property that we are referring to can not be an intrinsic  property since it varies with the mass of the substance that is under study as shown.

We can then conclude that the property X is an extrinsic property.

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Missing parts;

Carrie measured mass and another property, X, of three pure samples of the same compound. She recorded her data in the table. Which of the following best describes property X?

A. X  is an intensive property because it does not vary with the size of the sample.

B. X is an extensive property because it does not vary with the size of the sample.

C. X is an extensive property because it varies directly with the size of the sample.

D. X is an intensive property because it varies directly with the size of the sample.

Answer: If a neutral atom gains electrons, then it will become negatively charged

Explanation: If a neutral atom gains electrons, then it will become negatively charged. If a neutral atom loses electrons, then it become positively charged.

Answer:

Hence, 0.455 mol of aluminum chloride could be produced from 12.0 grams of aluminum.

Explanation:

Given data:

•Mass of aluminum is 12.0 grams.

To find:

•Number of moles of aluminum chloride produced from Al.

The given reaction is as follows:

2A1+ 3Cl₂ → 2AlCl3

First, we have to find the number of moles of aluminum by the relation given below:

n = m M;

Where:

•n is the number of moles.

•m is the mass in grams.

•M is the molar mass in g/mol.

The molar mass of Al is 26.98 g/mol. Substitute the values in above equation to obtain the number of moles of aluminium as follows:

n = (12.0g)/(26.98g /mol)

= 0.445 mol

From the reaction stoichiometry, we can see that 2 moles of Al forms 2 moles of aluminium chloride (A1C13). This means, the number of moles of Al and AlCl3 is equal.

So, 0.445 mol Al will give 0.455 mol AlC13.

When hydrogen bromide (HBr) is added to 2-methylbut-2-ene ((CH3)2CCHCH3), an electrophilic addition reaction takes place, where the π bond of the alkene is broken, and the hydrogen and bromine atoms are added to the resulting carbocation.

The reaction proceeds through a Markovnikov addition, where the hydrogen atom attaches to the carbon atom with the greater number of hydrogen atoms.

In this case, the initial addition of HBr to 2-methylbut-2-ene leads to the formation of a primary carbocation, as the positively charged carbon atom only has one alkyl group attached to it. The primary carbocation is relatively unstable, and it can undergo a rearrangement to form a more stable secondary carbocation.

The major product that is typically obtained is the 2-bromo-2-methylbutane. The hydrogen atom from HBr adds to the carbon with three hydrogen atoms (the more substituted carbon), resulting in the formation of a secondary carbocation.

On the other hand, a minor product is also formed, which is 3-bromo-2-methylbutane. This product arises from the addition of HBr to the primary carbocation, which is less stable. Although the primary carbocation is less favored, it can still be formed and lead to the formation of the minor product.

In summary, the addition of HBr to 2-methylbut-2-ene yields two products: the major product is 2-bromo-2-methylbutane, resulting from the addition of HBr to the more stable secondary carbocation, and the minor product is 3-bromo-2-methylbutane, originating from the less stable primary carbocation.

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The temperature change of the resulting solution, given that the soluton has a specific heat of 1 cal/g·°C is 18.7 °C

We know that heat released is related to change in temperature according to the following formula:

Q = MCΔT

Where

Using the above formula, the temperature change of the solution can be obtained as follow:

Q = MCΔT

-8.8×10² = 47 × 1 × ΔT

-8.8×10² = 47 × ΔT

Divide both side by 47

ΔT = -8.8×10² / 47

ΔT = -18.7 °C

Thus, the temperature change is 18.7 °C

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Answer:

…mixture of two stable isotopes—boron-10 (19.9 percent) and boron-11 (80.1 percent); slight variations in this proportion produce a range of ±0.003 in the atomic weight.

Explanation:

Answer: selenium dichloride

Explanation:

The two processes  which contributed oxygen to earth’s second atmosphere are photosynthesis and photolysis.

It is defined as a process by which plants and other photosynthetic organisms convert the light energy in to chemical energy  through the process of cellular respiration.

Some of the energy which is converted  is stored in molecules of carbohydrates like sugar and starches  which are made up of from carbon dioxide and water . Photosynthetic organisms which can perform photosynthesis  are algae and cyanobacteria. Photosynthesis is largely responsible for producing and maintaining the content of oxygen in earth's atmosphere.

The process begins with proteins absorbing light energy  which are called reaction centers and contain a green pigment which is called chlorophyll . In plants ,these pigments are present inside organelles called chloroplasts while in bacteria they are present in plasma membrane.

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Answer:

(a) GaF3, gallium(III) fluoride

(b) LiH, lithium hydride

(c) AlI3, aluminum(III) iodide

(d) K2S, potassium sulfide

Answer:

Polarity is caused by the difference in electronegativity (EN) of a compound. For example, water, H2O, is polar because oxygen's EN is much higher than hydrogens. Therefore, the electrons spend more time around the oxygen and less time around the hydrogen causing a slightly positive pole at the hydrogens and a slightly negative at the oxygen. Most polar molecules are not symmetrical (like water). Non-polar molecules like CH4 (methane) do not have poles.

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Answer:

1) volume of MgCl2, V = 1.66*10^2 mL = 0.166 L use: number of mol in MgCl2, n = Molarity * Volume = 0.381*0.166 = 6.325*10^-2 mol volume of NaOH, V = 47.33 mL = 4.733*10^-2 L use: number of…

Explanation:

The mass of Mg(OH)₂ that can be produced from the reaction is 0.841 g

The percent yield of Mg(OH)₂ is 79.6%.

The mass of Mg(OH)₂ that can be produced when 186.0 mL solution of 0.107 M MgCl₂ reacts with a 36.50 mL solution of 0.787 M NaOH to produce Mg(OH)₂ and NaCl is calculated below as follows:

Equation of reaction:

MgCl₂ (aq) + 2 NaOH (aq) ⟶ Mg(OH)₂ (s) + 2 NaCl (aq)

Mole ratio of MgCl₂ and NaOH is  1 : 2

Moles of MgCl₂ reacting = 0.107 * 186/1000

Moles of MgCl₂ reacting = 0.02 moles

Moles of NaOH reacting = 0.787 * 36.5/1000

Moles of NaOH reacting = 0.029 moles

NaOH is the limiting reactant

Mass of Mg(OH)₂ that can be produced = 0.029 * 1/2 * 58

Mass of Mg(OH)₂ that can be produced = 0.841 g

Percent yield = 0.67/0.841 * 100%

Percent yield = 79.6%

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The answers include the following:

This is referred to as an allotrope of carbon and it is the most stable form. it is used in the production of substances such as pencil, electrode etc.

Graphite is able to conduct electricity because of the presence of delocalized electrons and is slippery due to its atoms being able to slide over one another.

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Answer:

The interpretation of the particular subject is covered in the subsection below in detail.

Explanation:

Answer: metal turn orange and weaker as it gets oxidised

Explanation:

Answer:

I think that it might be 2.7

Explanation:

A. Cell reaction:\(2 Cd^2+(aq) + 2 Cr(s) - > 2 Cd(s) + 2 Cr^2+(aq) ; E^0cell = -0.51 V\)

B. Cell reaction: \(2 Fe^2+(aq) + Br2(g) - > 2 Fe(s) + 2 Br^-(aq) ; E^0cell = +1.507\)

Let's calculate the standard cell potential, E^0cell, for each pair of half-reactions and write the balanced equations for the overall cell reactions:

A. Half-reactions:

\(Cd^2+(aq) + 2e^- - > Cd(s) (E^0 = -0.403 V)\\Cr^2+(aq) + 2e^- - > Cr(s) (E^0 = -0.913 V)\)

To calculate the standard cell potential, we subtract the reduction potential of the anode (oxidation half-reaction) from the reduction potential of the cathode (reduction half-reaction).

\(E^0cell = E^0cathode - E^0anode\\E^0cell = (-0.913 V) - (-0.403 V) = -0.51 V\)

The balanced equation for the overall cell reaction is obtained by multiplying the half-reactions by coefficients to ensure that the number of electrons transferred is the same:

\(2 Cd^2+(aq) + 2 Cr(s) - > 2 Cd(s) + 2 Cr^2+(aq)\)

B. Half-reactions:

\(Fe^2+(aq) + 2e^- - > Fe(s) (E^0 = -0.447 V)\\Br2(g) + 2e^- - > 2 Br^-(aq) (E^0 = +1.06 V)\\E^0cell = E^0cathode - E^0anode\\E^0cell = (+1.06 V) - (-0.447 V) = +1.507\)V

The balanced equation for the overall cell reaction is:

\(2 Fe^2+(aq) + Br2(g) - > 2 Fe(s) + 2 Br^-(aq)\)

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The correct option that represents the constants of the reaction vessel after the N₂ and Cl₂ react as completely as possible is A.

The balanced chemical equation for the reaction is:

N₂(g) + 3Cl₂(g) → 2NCl₃(g)

This means that 1 mole of nitrogen gas and 3 moles of chlorine gas react to produce 2 moles of nitrogen trichloride gas.

The contents of the two vessels are:

Vessel 1: 1 mole of N₂

Vessel 2: 3 moles of Cl₂

When these two vessels are combined, the chlorine gas will react completely with the nitrogen gas to produce nitrogen trichloride gas. The remaining gas in the vessel will be nitrogen trichloride gas.

Which is:

[N₂] = 0 mol

[Cl₂] = 0 mol

[NCl₃] = 2 mol

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Complete question:

11 Question 3 A =N O=CI N2 +3 Cl2 2 NC13 The contents of two vessels are combined, as shown. The balanced chemical equation for the reaction that occurs is also shown. Which of the following represents the contents of the reaction vessel after the N2 and Cl2 react as completely as possible?

Answer:

The correct answer is 5.

CoS

Explanation:

Cobalt sulphide

is produced when the aqueous solutions of CoCl2 and Na2S are combined:

CoS(s) + 2NaCl = CoCl2(aq) + Na2S(aq) (aq)

In order to create cobalt sulphide, which is insoluble in water and precipitates out as a solid, the Co2+ ions in CoCl2 and the S2- ions in Na2S must interact. NaCl, one of the reaction's other byproducts, is soluble in water and stays in solution.

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Answer:

We can use stoichiometry to determine the amount of water produced when 29.3 g of LiOH reacts with CO2.

First, we need to convert 29.3 g of LiOH to moles:

moles of LiOH = mass/molar mass = 29.3 g / (6.941 g/mol + 15.999 g/mol + 1.008 g/mol) = 0.5 mol

From the balanced chemical equation, we see that 2 moles of LiOH react with 1 mole of CO2 to produce 1 mole of H2O. Therefore, we can say:

0.5 mol LiOH × (1 mol H2O / 2 mol LiOH) = 0.25 mol H2O

Now, we can use the molar mass of water to convert moles to grams:

mass of H2O = moles × molar mass = 0.25 mol × 18.015 g/mol = 4.504 g

Finally, we can use the density of water to convert grams to milliliters:

volume of H2O = mass / density = 4.504 g / 0.997 g/mL = 4.52 mL

Therefore, approximately 4.52 mL of water could form from 29.3 g of LiOH.

Weathering changes the chemical and physical nature of an element that is why it is often difficult to identify a highly weathered mineral.

The breakdown and alteration of rocks and minerals at or near the Earth's surface as a result of exposure to various weathering agents, such as water, wind and temperature changes is known as weathering.

Minerals can undergo physical changes as a result of weathering, such as being broken up into smaller pieces or having their color and texture altered. Additionally, it may result in chemical alterations such as the removal or addition of specific chemical components.

This may lead to the creation of brand-new minerals or the modification of already existing minerals into new ones.  Highly weathered minerals might not still possess the same physical and chemical characteristics as their unweathered counterparts.

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The products benzoin and meso-hydro benzoin melting points are same. Their temperature is 13341 degrees Celsius. Using the melting point method makes it challenging to directly determine. It would be necessary to determine the reaction's end product and then measure the melting point.

The thermal analysis that is most usually used to describe solid crystalline materials is melting point determination. It is used to identify solid crystalline chemicals and to assess their purity in research and development as well as in quality control in different industry areas.

The melting values of the products were between benzoin and meso-hydro benzoin 135 and 137, indicating that the molecule is likely benzoin because they are so similar to benzoin's melting points (135 to 137). Compared to hydroxybenzoic acid, benzoin has a melting point range that is closer to the finished product.

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Answer:

The atomic mass of element A, given that the first isotope has abundance of 18.18% and a mass of 147.99, is 131.43 amu

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

The atomic mass of the element A can be obtain as illustrated below:

Atomic mass = [(Mass of 1st × 1st%) / 100] + [(Mass of 2nd × 2nd%) / 100]

Inputting the given parameters, we have:

Atomic mass = [(147.99 × 18.18) / 100] + [(127.76 × 81.82) / 100]

Atomic mass = 26.90 + 104.53

Atomic mass = 131.43 amu

Thus, the atomic mass of element A obtained from the above calaculation is 131.43 amu

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Approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

To determine the number of molecules of HCl required to form 34.52 g of MgCl₂, we need to use the molar mass and stoichiometry of the balanced equation:

Mg(OH)₂(s) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

The molar mass of MgCl₂ is 95.21 g/mol.

First, we need to calculate the number of moles of MgCl₂ formed:

Moles of MgCl₂ = mass of MgCl₂ / molar mass of MgCl₂

Moles of MgCl₂ = 34.52 g / 95.21 g/mol

Moles of MgCl₂ = 0.363 mol

According to the balanced equation, the stoichiometric ratio between HCl and MgCl₂ is 2:1. Therefore, the moles of HCl required can be calculated as follows:

Moles of HCl = 2 * Moles of MgCl₂

Moles of HCl = 2 * 0.363 mol

Moles of HCl = 0.726 mol

To calculate the number of molecules, we need to use Avogadro's number, which is approximately 6.022 x 10^23 molecules/mol.

Number of molecules of HCl = Moles of HCl * Avogadro's number

Number of molecules of HCl = 0.726 mol * 6.022 x 10^23 molecules/mol

Number of molecules of HCl = 4.37 x 10^23 molecules

Therefore, approximately 4.37 x 10^23 molecules of HCl would be required to form 34.52 g of MgCl₂.

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1) Diesel has a higher viscosity than petrol.

2) Petrol is more flammable than diesel.

3) The formula will be C₁₀H₂₂.

4) The equation is; 2C8H18+25O2→16CO2+18H2O.

Depending on the precise composition and temperature, the viscosity of gasoline and diesel can change. In general, diesel is more viscous than gasoline. Higher viscosity fluids are thicker and flow more slowly than lower viscosity fluids because viscosity relates to the resistance of a fluid to flow.

Diesel is less flammable than gasoline. The lowest temperature at which gasoline can evaporate and turn into an ignitable combination in air is known as its flash point, and it is lower for gasoline. Compared to diesel fuel, petrol vapors are much more flammable and can ignite at lower temperatures.

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If ΔH is positive and entropy decreases, the reaction will be non-spontaneous, which means that it requires energy to proceed. The correct answer is D. Endothermic reaction.

What is an endothermic reaction?

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Answer:

D

Explanation:

Faults and problems in three routes of compound Br(bromide) synthesis, and the most feasible route, addressing competition, compatibility, and timing.

The main course includes a nucleophilic replacement of A with MeOH within the sight of HCl, trailed by parchedness with CS₂/Mel/heat. The fundamental issue with this course is the opposition between the replacement and end responses, which could prompt low yields of the ideal item.

The subsequent course includes a comparative nucleophilic replacement of A with MeOH within the sight of TsOH, trailed by a decrease with LiAlH₄. The fundamental issue with this course is the similarity between the acidic TsOH and the diminishing specialist, which could bring about the development of undesirable side items.

The third course includes an immediate buildup of A with ethyl oxalate within the sight of K₂CO₃, trailed by a decarboxylation with H₂SO₄. The fundamental issue with this course is the planning of the decarboxylation step, which could prompt the arrangement of undesirable side items because of overcompensation.

Generally speaking, the most possible course is by all accounts the first, with cautious advancement of the response conditions to limit the opposition among replacement and disposal responses.

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Answer:

To find the percentage by weight of oxygen in the gaseous compound, we need to first find the weight of the compound that is not oxygen. We can do this by subtracting the weight of the oxygen from the total weight of the compound: 50.0 grams - 30.0 grams = 20.0 grams.

Next, we need to divide the weight of the oxygen by the total weight of the compound and multiply the result by 100% to express the answer as a percentage: 30.0 grams / 50.0 grams * 100% = 60.0%.

Therefore, the percentage by weight of oxygen in the gaseous compound is 60.0%.

Explanation: