FILL IN THE BLANK. The lewis structure of pf3 shows that the central phosphorus atom has ________ nonbonding and ________ bonding electron pair(s)

Bone density can be affected by a mutation. This mutation can cause very dense bones that are almost unbreakable therefore the true statement about mutations in organism is that changes caused by mutations can be harmful, beneficial, or neutral to an organism which is therefore denoted as option B.

This is referred to as a change in the DNA sequence of an organism and it occurs when there are errors during cell division.

The changes caused have different effects on the organisms which could be harmful, beneficial, or neutral.

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At equilibrium, HCN is weaker Bronsted-Lowry acid, C₉H₇NH⁺ is stronger Bronsted-Lowry acid, C₉H₇NH is weaker  Bronsted-Lowry base and CN⁻ is stronger Bronsted-Lowry base.

CN⁻ + C₉H₇NH⁺ → HCN + C₉H₇N

Products are favored. So, CN⁻ accepted the proton instead of C₉H₇N and C₉H₇NH⁺ lost its proton instead of HCN.

 C₉H₇NH⁺ - stronger Bronsted-Lowry acid

 C₉H₇N - weaker  Bronsted-Lowry base

 HCN - weaker Bronsted-Lowry acid

 CN⁻ - stronger Bronsted-Lowry base

So, at equilibrium, HCN is weaker Bronsted-Lowry acid, C₉H₇NH⁺ is stronger Bronsted-Lowry acid, C₉H₇NH is weaker  Bronsted-Lowry base and CN⁻ is stronger Bronsted-Lowry base.

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The amount of heat released from the system as the pentane cools is 67.5kJ.

To determine the heat released from the system as the pentane cools, we need to consider the heat absorbed or released during each of the phase changes (vaporization and fusion) as well as the heat absorbed or released during the temperature change of the substance in each phase.

The heat absorbed or released during a phase change can be calculated using the enthalpy of vaporization (AHvap) or enthalpy of fusion (AHfus) and the number of moles of substance undergoing the phase change.

The heat absorbed or released during a temperature change can be calculated using the specific heat of the substance and the number of moles of substance and the temperature change.

In this case, we can first calculate the heat absorbed or released during the vaporization of the pentane:

Heat = ΔHvap * moles

Heat = 45.3 kJ/mol * 144.29 g / (58.12 g/mol) = 102.1 kJ

Next, we can calculate the heat absorbed or released during the cooling of the pentane vapor to the melting point:

Heat = specific heat * moles * temperature change

Heat = 168 J/mol K * 144.29 g / (58.12 g/mol) * (117 - (-84)) K = 68400 J = 68.4 kJ

Then, we can calculate the heat absorbed or released during the fusion of the pentane:

Heat = ΔHfus * moles

Heat = 120 kJ/mol * 144.29 g / (58.12 g/mol) = 270 kJ

Finally, we can calculate the heat absorbed or released during the cooling of the pentane from the melting point to the final temperature:

Heat = specific heat * moles * temperature change

Heat = 168 J/mol K * 144.29 g / (58.12 g/mol) * (-84 - (-84)) K = 0 J

The total heat released from the system can be calculated by adding up the heat absorbed or released during each of these processes:

Total heat = 102.1 kJ + 68.4 kJ + 270 kJ + 0 kJ = 440.5 kJ

The correct answer is therefore D) 67.5 kJ.

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In calorimetry, whatever happens in the system, the exact same change must occur to the surroundings is False.

What occurs in calorimetry?

Heat transfer rates to or from a substance are measured using calorimetry. Heat is transferred to do this using a calibrated item (calorimeter). Calculating the quantity of heat transmitted by the process under study requires using the temperature change recorded by the calorimeter.

The sealed calorimeter has thermometers that track temperature changes as a reaction takes place. This closed system functions similarly to a chemical reactor.

It's crucial to realize that in calorimetry difficulties, the material reacting is the "system," whilst the surroundings are made up of water and calorimetry. The heat exchanged between the "system" and "surroundings" is what is measured.

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The thing that the curves tell us about the distribution of dark matter in these galaxies is option B: Most of the dark matter is in the galaxies’ outskirts.

Beyond the visible borders of galaxies, the velocity of stars remains relatively constant, suggesting that there must be more matter present than what we can see as stars and gas.

When the impact of DE became noticeable, the rotation curves are shown to dip at a specific distance. r. These flat, v = constant, rotation curve data suggest that the amount of dark matter is increasing linearly with r up to great distances from the spiral galaxies' centers.

From the graph you can see they all pointing or going one direction, hence we can deduce that it is option B.

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Nuclides below the valley of stability can become more stable through  positron emission.

A positron emission tomography (puppy) test is an imaging check which can assist reveal the metabolic or biochemical feature of your tissues and organs. The puppy test uses a radioactive drug (tracer) to reveal both ordinary and abnormal metabolic pastime.

A positron is the anti-particle of a beta particle, and is emitted with the aid of a proton-rich nucleus. The collision of an electron and a positron yields two 0.511 MeV gamma rays. Positron gamma radiation can penetrate via inches of iron, concrete, timber, plastic, water, and many others.

They're converted into two high-electricity gamma rays moving immediately faraway from every other.

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Both E1 reaction and E2 reaction are more likely to produce 2-methyl-2-butene, instead of 2-methyl-1-butene.Therefore, it makes sense that "2-methyl-2-butene" is the major product while "2-methyl-1-butene" is the minor product because it is not as stable or substituted.

This has happened due of Zaitsev's Rule. Compared to 2-methyl-1-butene, 2-methyl-2-butene's double bond is more heavily replaced. Because it is more stable, 2-methyl-2-butene is more prevalent in the yield. The hydrogen needed to create 2-methyl-2-butene comes from the beta carbon that has undergone the most substitutions. According to Zaitsev's rule, the main product is the one produced as a result of hydrogen being lost during beta distillation, errors being made when weighing and measuring reagents and products, gathering the distillate above 45 C, and alkene products being lost because the alkenes were too close to a heat source. Future investigations should aim to reduce these flaws to obtain more precise results.

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The empirical formula for each compound are as follow,

a)[tex]Ag_2 O[/tex]

b)[tex]Co_3As_2O_8[/tex]

c)[tex]SeBr_4[/tex]

As per the data share in the above question are as follow,

The mass of their constituent elements are as follow,

a) Ag=1.651 g

   O=0.1224 g

b)Co =0.672 g

    As =0.569 g

    O=0.486 g

c) Se =1.443 g

    Br =5.841 g

We have to calculate the empirical formula for each compound.

a)Now Moles of (given mass upon Molar mass) [tex]Ag=\frac{1.651}{108} =0.015 \:moles[/tex]

Moles of [tex]O=\frac{0.1224}{8\times 2} =0.0077 \:moles[/tex]

Smallest mole value [tex]0.0077 \:moles[/tex]

Divide all component by smallest mass.

Therefore [tex]Ag \Rightarrow\frac{0.015}{0.0077} \rightarrow2\\\\O \Rightarrow\frac{0.0077}{0.0077} \rightarrow1[/tex]

Combine to get empirical formula,

[tex]Ag_2 O[/tex]

b)Now Moles of (given mass upon Molar mass) [tex]Co=\frac{0672}{59} =0.011 \:moles[/tex]

Moles of [tex]As=\frac{0.569}{75} =0.0076\: moles[/tex]

Moles of [tex]O=\frac{0.486}{16} =0.0304 \:moles[/tex]

Smallest mole value [tex]0.0076 \:moles[/tex]

Divide all component by smallest mass.

Therefore

[tex]Co \Rightarrow\frac{0.011}{0.0076} \rightarrow1.5\rightarrow1.5 \times 2\rightarrow3\\\\As \Rightarrow\frac{0.0076}{0.0076} \rightarrow1\rightarrow1 \times 2\rightarrow2\\\\O \Rightarrow\frac{0.0304}{0.0076} \rightarrow4\rightarrow4 \times 2\rightarrow8[/tex]

Combine to get empirical formula,

[tex]Co_3As_2O_8[/tex]

c)Now Moles of (given mass upon Molar mass) [tex]Se=\frac{1.443}{79} =0.018 \:moles[/tex]

Moles of [tex]Br=\frac{5.841}{80} =0.073 \:moles[/tex]

Smallest mole value [tex]0.018 \:moles[/tex]

Divide all component by smallest mass.

Therefore

[tex]Se \Rightarrow\frac{0.018}{0.018} \rightarrow1\\\\Br \Rightarrow\frac{0.073}{0.018} \rightarrow4[/tex]

Combine to get empirical formula,

[tex]SeBr_4[/tex]

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The mass percentage of C in morphine would be 4.21%.

The mass percentage of a composition in a compound is the mass of the composition relative to the mass of the entire compound. This can be mathematically expressed as:

Mass percentage = mass of component/mass of substance x 100%

In this case, we are looking for the mass percentage of C in [tex]C_{17}H_{19}NO_3[/tex].

Molar weight of C = 12 g/mol

Molar mass of  [tex]C_{17}H_{19}NO_3[/tex] = (12x17) + (1x19) + (14x1) + (16x3)

                                             = 285 g/mol

Mass percentage of C = 12/285

                                     = 4.21% to 2 decimal places.

In other words, the mass percentage of C in morphine is 4.21%.

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Molecules (II) and (IV) that is SF4 and H2S are polar.

In polar molecules, the electron density is unevenly distributed throughout the molecule, resulting in some regions of negative charge and some regions of positive charge. Molecular polarity depends on both individual bond polarities and molecular geometry, the latter of which can be predicted using his VSEPR theory.

Reason :

In molecule (I), molecule is symmetrical all polar bond cancel out polarity of each other so molecule is non-polar.

In molecule (II), molecule is asymmetrical, having a lone pair so molecule is polar.

In molecule (III), molecule is linear, bond polarity cancel out each other, so molecule is non-polar.

In molecule (IV), molecule is bent (asymmetric) having two lone pair, molecule is polar.

In molecule (V), no bond polarity, hence molecule is non-polar.

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A)The needed amount of phenacaine HCl is 0.6g and chlorobutanol is 0.3 g B) 40.8 ml is the required volume of 1.9% boric acid C) 0.7752 grams you do not have 1.9% boric acid solution and have to compound from boric acid powder

A)for calculating amount. of drug requieed Amount of phenacine Hcl SO = 1% w/v measure 19 in  100 ml

60/100×1g= 0.6 gm

Amount  of chlorobutanol = 0.5%  w/v  means 0.5g in 100 ml.

So 0.5g ⇒100ms

>> 60/100 ×0.5g = 0.39gm

b) so total volume

= Yol

At Vol. B

= 7.2ml + 12

= 19-2nd

So Volume Boric acid

= total volume of solution - volume of drug

= 60ml - 19.2ml

= 40.8gm

c) in this preparation. we need

40.8 ml = 0.4089 of Nac to =0.77529 grams boric acid powder to 40.8 ml

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1. The number of moles of KHP is found by dividing the mass of KHP by the molecular weight of KHP.

2. The number of moles of NaOH is found by multiplying the moles of KHP by the mole ratio of NaOH to KHP given by the above, balanced chemical reaction.

It will be presumed that more KHP is present in the solution if additional NaOH is added. As a result, the % KHP will be higher than anticipated. If the concentration of NaOH is doubled, just half the volume from the initial experiment is needed. To titrate NaOH with KHP, add NaOH from the burette to a predetermined amount of KHP. The mass and volume of KHP utilized to create the KHP solution are used to calculate the molarity of the KHP solution. The molarity of the NaOH is then determined using the information from the titration.

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From the given graph plotting the rate of photosynthesis versus the temperature, the rate of oxygen production is greatest at the point M.

Photosynthesis is the biochemical process of synthesising chemical energy by green plants  with the aid of light energy. The photosynthetic reaction involves the combination of carbon dioxide and water producing oxygen gas and glucose.

The chemical energy is stored in plants in the form of glucose which is then break down by the need of energy forming ATP.  The photosynthetic reaction is affected by the temperature and pressure of the atmosphere.

From the graph, it is clear that, the rate of photosynthesis is higher at the point M. Higher the rate of photosynthesis, higher will be the oxygen production. Hence, option B is correct.

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Answer:

8058.88

Explanation:

7.87 x 1024 = 8058.88

duh

b) 1.0×10−4 M1.0×10−4 M H2C2O4H2C2O4 will have the Highest pH among the solutions.

A weak acid is what oxalic acid is. It is not as strong as H3O+ ion (water). However, it is more potent than acids like acetic, sulfuric, nitrous, and benzoic. First of all, it is an organic substance, and organic substances are often not strong acids. Dicarboxylic acid is another name for oxalic acid.

Similar to how the weak acid in the example was calculated, a weak base's solution can also be calculated to determine its pH. However, the concentration of the hydroxide ion will be represented by the variable x.

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The nitrogen monoxide produced by the combustion in the automobile engine. the grams of the oxygen gas are used to form the 0.154 moles of the nitrogen dioxide is 2.46 g.

The reaction is given below :

2NO  +  O₂  --->  2NO₂

2 moles of the NO₂ produce by the 1 moles of O₂

moles of the  of NO₂ = 0.154 mol

moles of the  O₂ = (1/2) × 0.154

                            = 0.077 mol

molar mass of  O₂ = 32 g/mol

mass of  O₂ = moles × molar mass

                    = 0.077 × 32

                    = 2.46 g

The mass of oxygen gas needed to form the nitrogen dioxide is 2.46 g.

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Using xuv series, deduct about relationship between iodized energy and reactivity of metals that, The first ionization energy of a metal has little bearing on its reactivity.

Ionization energy, sometimes referred to as ionization potential, is the amount of energy needed to remove an electron from an isolated atom or molecule in physics and chemistry. Ionization energies are related with the removal of each subsequent electron, however they are most frequently employed when the first (and loosely held) electron is removed. In an electric discharge tube, when a quickly moving electron produced by an electric current collides with a gaseous atom of the element, causing it to eject one of its electrons, the ionization energy of a chemical element, given in joules or electron volts, is often measured. A hydrogen atom, which consists of an orbiting electron connected to a nucleus of one proton, requires one ionization energy of 1018 joules (13.6 electron volts) to completely expel an electron. The amount of an element's ionization energy depends on a number of factors, including the size of the atom, its electronic make-up, and the electric charge of its nucleus. Alkali metals are the simplest chemical elements to remove an electron from, whereas noble gases are the most difficult.

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The Balanced Eqn is 2CO+O2→2CO2.

A chemical reaction is a method wherein one or more materials, also called reactants, are transformed into at least one or more special substances, called merchandise. Materials are both chemical elements or compounds.

A few synthesis reactions can result in more than one product. Photosynthesis is one of the maximum important chemical reactions on earth. It allows plants and a few microbes to transform water and carbon dioxide gas into storable sugar and oxygen.

Chemical compounds are made from atoms of different elements, joined collectively through chemical bonds. A chemical synthesis normally involves the breaking of existing bonds and the formation of recent ones.

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Mechanical strength isn't a key factor when a wide range of metals and alloys need to be joined.

While many metal fabrication businesses employ a variety of methods, the majority rely on the three fundamental operations of cutting, bending, and assembly.

The following are some illustrations of fabrication or falsification: producing data artificially when it ought to be coming from a real experiment Unauthorized tampering with or fabrication of information, texts, pictures, audio, video, music, or other works.

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The reactant mole ratio for all reactions discovered using this method must be met in order to apply this method to other reactions. Furthermore, the reactant molar ratios for each reaction must be known in order to determine the mole ratio in which the principal reactants react.

To adjust the reaction temperature, water was added to reaction mixtures A through E. In this experiment, KCr,O- was chosen as the source of dichromate ions because it is a strong oxidising agent that is easily converted into dichromate ions.The mole ratio is the ratio of the mole amounts of any two compounds in an equitable chemical reaction. The balance chemical equation compares the ratios of the molecules required to complete the reaction. It calculates how much of one substance is required to produce a certain amount of another substance using the mole ratios between the different parts of the reaction.

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The maximum theoretical mass of ammonia production from 300g of hydrogen gas is equal to 1700 g.

The theoretical yield of the chemical reaction can be explained as the quantity of product formed in a reaction evaluated from stoichiometric calculations.

Given the reaction of the formation of ammonia from hydrogen gas:

3H₂ (s) + N₂  (g) →  2NH₃ (g)

The molecular mass of the ammonia =  17 g/mol

Given the mass of the hydrogen gas = 300 g

The number of moles of the hydrogen gas = 300/2 = 150 mol

From the above equation, 2 moles of ammonia are produced from 3 moles of hydrogen gas.

150 mol of hydrogen gas will produce ammonia = (2/3) ×150 = 100mol

The mass of ammonia is produced = 100 × 17 = 1700 g

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When the rates of the forward and reverse reactions are identical, equilibrium has occurred. At equilibrium, the concentrations of each reactant and product are both constant.

The relationship between a reaction's products and reactants at equilibrium is expressed by the equilibrium constant, K. A rise in pressure will cause the reaction to move to the right if there are more reactants than products by moles. It is untrue that a reaction with a higher equilibrium constant value proceeds more quickly than one with a lower equilibrium constant value. The equilibrium constant's value has no bearing on the reaction's rate. The rate constant affects the reaction's rate.

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The reagent are used in the chemical reaction to cause the chemical reaction or the test performed if one occurred.

The reagents are the compounds or the mixture that are added to the chemical reaction in order to start the chemical reaction. The reagent is not necessarily be consumed during the chemical reaction. the reagent is used to identify the compound or to measure the substance if it is present in the reaction.

The  reagent is  the  initiator , solvent , catalyst etc. the example of the reagents are : Tollen's reagent , Fehling's reagent , Fenton's reagent and collins reagent etc.

The question is incomplete, I answer the question in general according to my knowledge.

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 44.98 grams of al(s) are needed to produce 154.6 g of al2o3 .

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

mass obtained of an  Al₂O₃ (experimental yield) is 154..6 g.

Theoretical yield of an  Al₂O₃ is x g.

Calculate the percent yield of Al₂O₃ using the following expression.

%yield = (experimental yield / theoretical yield) × 100%

(154..6 / x) × 100% = 87.8%

theoretical yield= 176.08

molar mass of Al₂O₃ is 101.96 g/mol

mass of Al = 26.98 u*176.08/101.96= 44.98g

A bond's yield is a number that represents the return on investment. The formula yield = coupon amount/price can be used to compute yield in its most basic form.

Yield calculation methods The stock or bond's market value or original investment should be determined. Identify the profit the investment produced. Subtract the income from the market value. Add 100 to this amount.

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The formation of free radicals in the organic reaction takes place when the bond is cleavage homolytically. Free radicals are species with a single electron and thus, they are electron deficient in nature.

A free radical may be primary, secondary or tertiary depending upon the number of H-atoms attached to it. Further, the free radicals may be categorized as alkyl, allyl or benzyl radicals.

Primary free radicals: When radical is attached to two H-atoms.

Secondary free radicals: When radical is attached to one H-atoms.

Tertiary free radicals: When radical is not attached to any H-atoms.

Allylic free radicals: When radical is attached to two H-atoms and is in conjugation (resonance) with double bond.

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To calculate the standard enthalpy of reaction (ΔH∘rxn) for the given chemical reaction, we need to use the equation for enthalpy of reaction, which is:

ΔH∘rxn = Σ(ΔH∘f products) - Σ(ΔH∘f reactants)

where ΔH∘f is the standard heat of formation for a given compound, and the summation symbol (Σ) indicates that we need to sum the values for all of the products and reactants in the reaction.

In this case, we are given the standard heats of formation for the products and reactants in the reaction:

CO(g)+NH3(g)→HCN(g)+H2O(g)

We can use these values to calculate the enthalpy of reaction, as follows:

ΔH∘rxn = Σ(ΔH∘f products) - Σ(ΔH∘f reactants)

= (ΔH∘f HCN + ΔH∘f H2O) - (ΔH∘f CO + ΔH∘f NH3)

= (130.5 kJ + −241.8 kJ) - (−110.5 kJ + −45.90 kJ)

= (−111.3 kJ) - (−156.4 kJ)

= 45.1 kJ

Therefore, the standard enthalpy of reaction for the given chemical reaction is 45.1 kJ to three significant figures. This value represents the amount of heat that is absorbed or released when the reactants are converted into products, and it can be used to predict the energetics of the reaction under standard conditions.

Answer:

A. Protons and neutrons have the same mass; the mass of an electron is very close to zero

Explanation:

Answer: B

Explanation: 2 isotopes of boron are present.

Let percent abundance of B-10 is x

percent abundance of B-11 is (100-x)

Average atomic mass,      

 [tex]10.80=\frac{x*10+(100-x)*11}{100}[/tex]

        [tex]10.80=\frac{10x+1100-11x}{100}[/tex]

        [tex]1080=1100-x[/tex]

        [tex]x=20[/tex]

The correct answer is carboxylic acid. Option B.

In a similar reaction, a coupling agent such as DCC can be used to prepare amides by the reaction of carboxylic acids with amines. Simple amides can be made by reacting acid anhydrides with amines. Finally, amides can be formed from the direct reaction of carboxylic acids and amines.

Amides are generally prepared by coupling a carboxylic acid with an amine using a coupling agent or by first converting the carboxylic acid to a derivative. Alternative methods include Staudinger ligation, aminocarbonylation of aryl halides, and oxidative amidation of aldehydes. Amides can be hydrolyzed in the presence of aqueous acids.

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