An atom has seven valance electrons which element could have this type of atom

To draw the Lewis structure for C2H2, place the carbon atoms in the center connected by a double bond, connect each carbon atom to a hydrogen atom, and add two lone pairs on each carbon atom.

To draw the Lewis structure for C2H2 (whose skeletal structure is HCCH), we need to determine the total number of valence electrons. Carbon has 4 valence electrons, while hydrogen has 1 valence electron each. Since there are 2 carbon atoms and 2 hydrogen atoms, the total number of valence electrons is 6.

To draw the structure, we place the carbon atoms in the center, connected by a double bond. Each carbon atom is then connected to a hydrogen atom. The remaining two valence electrons are placed as lone pairs on each carbon atom.

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The number of moles can simply be calculated by the product of Molarity and Volume in L, therefore:

number of moles = (2.26 moles / L)* 0.0152 L

number of moles = 0.0344 moles

The molar mass of KOH is 56.11 g/mol, therefore the mass is:

mass = 0.0344 moles * 56.11 g/mol

mass = 1.93 g

Final answer:

To find moles and mass of KOH in a 15.2 ml sample of 2.26 M solution, you first calculate moles (0.034392 mol) using molarity and volume, then convert moles to grams using KOH's molar mass, resulting in 1.929 grams.

Explanation:

The task is to calculate the moles and mass of solute present in a 15.2 ml sample of a 2.26 M KOH solution. First, to find the number of moles of KOH in the sample, we use the formula moles of solute = molarity × volume of solution (in liters). Since the molarity is given as 2.26 M and the volume is 15.2 ml (which is equivalent to 0.0152 liters), the calculation is:

moles of KOH = 2.26 M × 0.0152 L = 0.034392 mol

Next, to find the mass of the solute, we multiply the number of moles by the molar mass of KOH. The molar mass of KOH is approximately 56.11 g/mol. Thus, the mass of KOH in the sample is:

mass of KOH = 0.034392 mol × 56.11 g/mol = 1.929 g

Therefore, the 15.2 ml sample of 2.26 M KOH solution contains approximately 0.034392 moles and 1.929 grams of KOH.

Final answer:

The volume occupied by a mixture of 0.522 mol of N2 and 0.522 mol of O2 gases at 0.56 atm and 42.7°C is calculated to be approximately 48.18 liters using the ideal gas law.

Explanation:

The question is asking to calculate the volume occupied by a mixture of gases using the ideal gas law. For a mixture of 0.522 mol of N2 and 0.522 mol of O2 gases at 0.56 atm and 42.7°C, we first need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature, which gets us 315.85 K. We then use the ideal gas law equation PV = nRT, where P is the pressure in atm, V is the volume in liters, n is the amount of gas in moles, R is the ideal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹), and T is the temperature in Kelvin. Since both gases are ideal and in the same conditions, there is no need for separate calculations; we can combine the moles of the gases as the total amount is the sum of the individual moles of N2 and O2.

The total moles of gases is 0.522 mol + 0.522 mol = 1.044 mol. Placing these values into the ideal gas law equation and solving for V (volume), we have V = (nRT)/P. The calculated volume is then:

V = (1.044 mol x 0.0821 L·atm·K⁻¹·mol⁻¹ x 315.85 K) / 0.56 atm = 48.18 L

The volume occupied by the mixture is approximately 48.18 liters.

In the chemical formula of water,the lack of subscript after O indicates that there is one oxygen atom in the compound.

Chemical  formula is a way of representing the number of atoms present in a compound or molecule.It is written with the help of symbols of elements. It also makes use of brackets and subscripts.

Subscripts are used to denote number of atoms of each element and brackets indicate presence of group of atoms. Chemical formula does not contain words. Chemical formula in the simplest form is called empirical formula.

It is not the same as structural formula and does not have any information regarding structure.It does not provide any information regarding structure of molecule as obtained in structural formula.

There are four types of chemical formula:

1)empirical formula

2) structural formula

3)condensed formula

4)molecular formula

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In each box of periodic table, atomic number which is equal to number of protons and electrons, mass number and symbol of element is displayed.

Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.

It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.

Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.

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To prepare 350 mL of 0.100 M solution from a 1.50 M solution, we simply have to use the formula:

M1 V1 = M2 V2

So from the formula, we will know how much volume of the 1.50 M we actually need.

1.50 M * V1 = 0.100 M * 350 mL

V1 = 23.33 mL

So we need 23.33 mL of the 1.50 M solution. We dilute it with water to a volume of 350 mL. So water needed is:

350 mL – 23.33 mL = 326.67 mL water

Steps:

1. Take 23.33 mL of 1.50 M solution

2. Add 326.67 mL of water to make 350 mL of 0.100 M solution

Final answer:

To prepare 350 mL of a 0.100 M C12H22O11 solution from a 1.50 M C12H22O11 solution, measure 23.33 mL of the 1.50 M solution and dilute it with solvent to reach a total volume of 350 mL.

Explanation:

To prepare 350 mL of a 0.100 M C12H22O11 solution starting with 3.00 L of a 1.50 M C12H22O11 solution, we can use the equation C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Plugging in the values, we have (1.50 M)(3.00 L) = (0.100 M)(350 mL). Solving for V1, we get V1 = (0.100 M)(350 mL) / (1.50 M) = 23.33 mL.

Therefore, you should measure 23.33 mL of the 1.50 M C12H22O11 solution and dilute it with enough solvent (such as water) to reach a total volume of 350 mL.

To find the percentage of water vaporized, use the ideal gas law to calculate the initial and final amount of water in the flask. Then, subtract the final amount from the initial amount to find the mass of water vaporized. Finally, divide the mass of water vaporized by the initial mass of water and multiply by 100 to find the percentage.

To find the percentage of water vaporized, we need to calculate the initial and final amount of water in the flask. We know that the initial mass of water is 1.20 g. Using the ideal gas law, we can calculate the amount of water vapor at 65°C and the pressure of 187.5 mmHg. Then, we can subtract the final amount of water vapor from the initial amount to find the mass of water vaporized. Finally, we can divide the mass of water vaporized by the initial mass of water and multiply by 100 to find the percentage:

By following these steps, you should be able to calculate the percentage of water vaporized.

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Actually this depends on what kind of titration is being done. If we are titrating Ba(OH)2 using an acid, it actually does not matter that some Pb(OH)2 is formed because the acid will still neutralize even those OH- ions that combined with lead Pb.

But other than that, generally we can say that the calculated concentration can be affected because lead (II) chloride would co-precipitate with Ba(OH)2 to form Pb(OH)2

Final answer:

The contamination of a Ba(OH)₂ sample with lead(II) chloride before a titration can affect the calculated concentration of the sample due to the common ion effect. This effect decreases the solubility of lead(II) chloride, altering the equilibrium conditions and potentially leading to an underestimation of the Ba(OH)₂ concentration.

Explanation:

Suppose a small amount of lead(II) chloride got into your Ba(OH)₂ sample before your titration, contaminating it. This would impact your calculated concentration for the sample due to the common ion effect. The presence of extra chloride ions from lead(II) chloride can lead to a decrease in the solubility of lead(II) chloride. This phenomenon occurs because the solution now has an increased concentration of a common ion, in this case, chloride ions which both lead(II) chloride and barium chloride (potentially formed during titration) share. Consequently, the solubility of lead(II) chloride decreases, leading to a decrease in the concentration of lead(II) ions in the solution. This change in ion concentration can ultimately affect the accuracy of the titration results by altering the expected equilibrium conditions, potentially leading to an underestimation of the concentration of Ba(OH)₂ in the sample.

Answer: B. 15

Explanation:

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

Atomic number is defined as the number of protons or number of electrons that are present in an atom.

Atomic number = Number of electrons = Number of protons

Given : number of protons = 8

number of electrons = 8

number of neutrons = 7

Hence, mass number = Number of protons + Number of neutrons = 8+ 7 = 15.

chemical weathering

The increase in size or large that is physically visible is called growth. In one-hour bamboo plant will grow 0.984 inches tall.

Height is the lateral growth or increased size of an object or person that can be calculated in units like meters (m), centimeters (cm), inches, etc.

If in 24 hours (1 day), the bamboo plant grows 60 cm then in 1 hour it will grow,

[tex]\dfrac{60\;\rm cm}{24\;\rm hour}\times 1\;\rm hour = 2.5\;\rm cm[/tex]

We know, 1 cm is equivalent to 2.54 cm so, 2.5 cm is equivalent,

[tex]\dfrac{2.5\;\rm cm}{21.54\;\rm cm}\times 1\;\rm inch = 0.984 \;\rm inch[/tex]

Therefore, in a one-hour bamboo plant will grow 0.984 inches.

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If the specific rotation of (r)-2-butanol is −13.5° then the specific rotation of (s)-2-butanol is simply the opposite in sign 13.5°. In a 3:1 mixture of (s):(r) 2-butanol, 75% is (s) while 25% is r, so the specific rotation of the mixture is:

mixture rotation = 0.75 * (13.5°) + 0.25 * (- 13.5°)

mixture rotation = 6.75°

the answer is that the changes are weather

To find the limiting reactant, compare the amount of product expected for the complete reaction of each reactant. The reactant yielding the lesser amount of product is the limiting reactant.

The limiting reactant in a chemical reaction is the reactant that is completely consumed and therefore limits the amount of product that can be formed. To determine the limiting reactant, you need to compare the amount of product that would be formed from each reactant based on the reaction's stoichiometry. Whichever reactant produces the lesser amount of product is the limiting reactant.

For the given reaction: 2Li(s) + F2(g) → 2LiF(s)

1. For the initial quantities of 1.0 g Li and 1.0 g F2: Determine the moles of each reactant and calculate the moles of product formed for each reactant. The reactant that produces the lesser amount of product is the limiting reactant.

2. Repeat the same steps for the initial quantities of 10.5 g Li and 37.2 g F2, and for 2.85×10^3 g Li and 6.79×10^3 g F2.

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In the reaction 2Li(s) + F₂(g) → 2LiF(s), the limiting reactant for each set of initial reactant quantities was determined using Approach 2. For the given amounts of 1.0 g Li and 1.0 g F₂, 10.5 g Li and 37.2 g F₂, and 2850 g Li and 6790 g F₂, F₂ was found to be the limiting reactant in all three cases.

Finding the Limiting Reactant

For the reaction 2Li(s) + F₂(g) → 2LiF(s), we will determine the limiting reactant for each set of initial reactant quantities using Approach 2 (The Product Method).

Balance the chemical equation: The given chemical equation is already balanced: 2Li(s) + F₂(g) → 2LiF(s)

Convert the given information into moles:

Molar masses: Li = 6.94 g/mol, F₂ = 38.00 g/mol

For 1.0 g Li and 1.0 g F₂:

Moles of Li = 1.0 g / 6.94 g/mol = 0.144 moles

Moles of F₂ = 1.0 g / 38.00 g/mol = 0.0263 moles

For 10.5 g Li and 37.2 g F₂:

Moles of Li = 10.5 g / 6.94 g/mol = 1.51 moles

Moles of F₂ = 37.2 g / 38.00 g/mol = 0.979 moles

For 2850 g Li and 6790 g F₂:

Moles of Li = 2850 g / 6.94 g/mol = 410.81 moles

Moles of F₂ = 6790 g / 38.00 g/mol = 178.68 moles

Calculate the product formed:

2 moles of Li react with 1 mole of F₂ to form 2 moles of LiF.

For each set of initial reactants, calculate the product:

1.0 g Li and 1.0 g F₂: Li is in excess (0.144 > 0.0263 × 2). Therefore, F₂ is the limiting reactant.

10.5 g Li and 37.2 g F₂: Li is in excess (1.51 > 0.979 × 2). Therefore, F₂ is the limiting reactant.

2850 g Li and 6790 g F₂: Li is in excess (410.81 > 178.68 × 2). Therefore, F₂ is the limiting reactant.

In conclusion, in all three cases, the limiting reactant is F₂.

Answer:

C.option 3

Explanation:

We are given that Your boss tell you that she needs a decision by the end of the day about the machine you want to purchase for your new operation .

You have four choices , and your boss reminds you to be cost-efficient as possible but she wants the best  efficiency .

Efficiency of machine  : It tell us how much input energy converted into useful work.

Efficiency of machine= Output energy[tex]\div[/tex]input energy

In option 1  machine cost = $5000

The rate of  supplied energy of machine =1900 W

The rate of useful work= 1786 W

Efficiency=[tex]\frac{1786}{1900}=0.94[/tex]

In option 2

Machine cost =$45,00

The rate of  supplied energy of machine= 1450 W

The rate of useful work=1300 W

Efficiency=[tex]\frac{1300}{1450}=0.897[/tex]

In option 3

Machine cost=$14750

The rate of  supplied energy of machine=1950 W

The rate of useful work=1833 W

Efficiency=[tex]\frac{1833}{1950}=0.94[/tex]

In option 4

Machine cost=$4750

The rate of  supplied energy of machine=2000 W

The rate of useful work= 1822 W

Efficiency=[tex]\frac{1822}{2000}=0.911[/tex]

In option 1 and 3 cost are different but the efficiency of machine in option 3  and option  1 is same. But the cost of option 3 is smaller than the cost of machine in option 1. Hence, option 3 is best .

Answer: C.option 3

The maximum percent of the product should be 100 percent, so it was not possible to produce a higher than 100 percent yield like 105 percent mentioned in the question. If this happens, then it will be against the law of conservation of energy, that is, mass cannot be created.  

The enhanced mass may have got produced due to impurities. One of the reason is if one has rinsed the container and has left some amount of water. The water is not considered as a substrate, but it may enhance the weight of the result. There is also a probability that some reaction has taken place in the air, and generated liquid or solid substance, which would have increased the weight.  

While it is not possible to truly obtain more than 100% of a product from a chemical reaction, the presence of impurities, absorption of substances, incomplete reactions, apparatus contamination, or measurement errors can lead to an reported percent yield that exceeds 100%.

The reported percent yield of sand of 105% is indeed greater than 100%, which suggests that more product has been collected than was originally present. While it may seem impossible to collect more sand than was initially used, there are several possible explanations for this occurrence that do not involve calculation mistakes:

1. Presence of Impurities: The extra product could be due to the presence of impurities in the starting materials or the environment. These impurities could contribute to the mass of the final product, leading to an apparent percent yield greater than 100%.

 2. Absorption of Moisture or Other Substances: Sand, being a porous material, can absorb water or other substances from the air. If the product sand absorbed moisture or other chemicals during the experiment or while being stored, this would increase its mass, resulting in a percent yield over 100%.

3. Incomplete Reaction: If the reaction to produce sand was not carried to completion, some of the starting materials might remain unreacted. These unreacted materials could be mistakenly included in the mass of the product, inflating the yield.

 4. Apparatus Contamination: During the experimental procedure, the apparatus used might have contributed some of its mass to the product. This could happen if there was any abrasion or if small pieces of the apparatus broke off and were collected with the sand.

 5. Error in Initial Mass Measurement: While calculation mistakes are not to be considered, it is possible that the initial mass of the reactants was underestimated. This could occur due to an improperly calibrated balance or human error in recording the mass.

6. Instrumental Error: The balance used to measure the mass of the reactants and products could have instrumental errors that lead to inaccurate measurements. This could result in an incorrect percent yield calculation.

7. Human Error: Other human errors, such as misreading the balance or incorrectly recording data, could also lead to an inflated percent yield.

In conclusion,  It is important for the lab group to critically evaluate their experimental procedure and measurements to identify the source of the discrepancy.

Final answer:

The question about lactate dehydrogenase cannot be answered without specific information on the enzyme's kinetics such as turnover number or assay conditions. Enzyme activity is determined by several factors including enzyme concentration, substrate concentration, and environmental conditions.

Explanation:

The question posed about lactate dehydrogenase (LDH) does not provide the specific information required to determine how many molecules of lactate one molecule of LDH can act on in a period of 21 minutes. The activity of lactate dehydrogenase on lactate molecules is a property that can be measured in terms of enzyme kinetics which includes factors like enzyme concentration, substrate concentration, temperature, and pH.

These factors would typically be reported in a study as part of the enzyme's turnover number or kcat, which is the number of substrate molecules an enzyme can convert per second when the enzyme is fully saturated with the substrate. Without this information or data from a specific assay, we cannot provide a precise answer to how many molecules of lactate can be acted upon by lactate dehydrogenase in 21 minutes.

Further data is required regarding the specific activity, kcat, or details about the conditions under which the enzyme is acting to answer this question accurately. These are essentials in the field of enzymology which is a subdivision of biochemistry dealing with the properties and functions of enzymes.

Actually since Bromine is located at the 1 Carbon, so we can say that this is a primary alkyl halide and which undergoes SN2 or E2 reactions. This reaction is a bimolecular, single step process because it is a primary.

The substitution product formed will be 1-ethoxybutane (main product) and sodium bromide (side product).

The structure of the substitution product from the reaction of 1-bromohexane with sodium ethoxide is hexanol (CH3(CH2)4CH2OH), which occurs through an SN2 mechanism.

The student has asked for the structures of the substitution products expected when 1-bromohexane reacts with sodium ethoxide (NaOCH2CH3). This reaction involves a nucleophilic substitution process where sodium ethoxide acts as the nucleophile and displaces the bromide ion (Br-) from 1-bromohexane.

For the reaction with sodium ethoxide, the expected major substitution product is hexanol. This reaction is an example of an SN2 mechanism where the nucleophile attacks the electrophilic carbon bearing the leaving group (in this case, Br-) from the opposite side, leading to an inversion of configuration at that carbon atom if it is chiral.

The overall reaction can be described as follows:

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The pH of acid is between 0-7 on pH scale while for base pH range is from 7-14. Therefore, the pH of  0.000081M  HClO₄  is 4.09. pH is a unitless quantity.

pH is a measurement of amount of hydronium ion H₃O⁺ in a given sample. More the value of hydronium ion concentration, more will be the solution acidic.

On subtracting pH from 14, we get pOH which measures the concentration of hydroxide ion in a given solution. pH depend on the temperature. At room temperature pH scale is between 0 to 14. pH of neutral solution is 7.

The concentration of  HClO₄ is 0.000081M or 8.1x10⁻⁵M

Concentration of H₃O⁺ in HClO₄ =0.000081M or 8.1x10⁻⁵M

Mathematically,

pH=-log[H⁺]

Substituting the values

pH=-log[0.000081]

pH of HClO₄=4.09

Thus the pH of  0.000081M HClO₄ is 4.09.

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Final answer:

When solid potassium nitrate is put into water, it dissociates into its constituent ions: K+ and NO3-. The compound potassium nitrate is a strong electrolyte.

Explanation:

When solid potassium nitrate (KNO3) is put into water, it undergoes dissociation into its constituent ions. The balanced chemical equation for this reaction is:

KNO3 (s) → K+ (aq) + NO3- (aq)

This reaction shows that potassium nitrate is a strong electrolyte as it completely ionizes in water, producing free moving ions and allowing for electrical conductivity.

To determine the percentage of MES that is protonated, we use the Henderson-Hasselbalch equation. By substituting the given values, we find that approximately 85% of the MES buffer is protonated at a pH of 6.85.

To find out the percentage of MES buffer that is protonated, we need to apply the Henderson-Hasselbalch equation, which relates the pKa value of an acid to the pH of the solution, via the following relation: pH = pKa + log10([A-]/[HA]).

In this case, the pKa value is 6.1 and pH given is 6.85. Let the concentration of (protonated) HA is [HA], and (deprotonated) A- is [A-].

To find the percentage of MES that is protonated, we need to rearrange and solve the equation for [HA]/[A-]:

[HA]/[A-] = 10^(pH - pKa) = 10^(6.85 - 6.1) = 5.6234

The fraction of MES that is protonated is [HA]/([HA]+[A-]). So the percentage is [HA]/([HA]+[A-]) * 100 = 5.6234 /(1+5.6234) * 100 = 84.88%

Therefore, approximately 85% of the MES buffer is protonated at pH 6.85.

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The percent yield of the alcohol fermentation reaction can be calculated by using the stoichiometry of the reaction and the ideal gas law. This involves converting the given mass of glucose to moles, calculating the expected moles of CO₂, and comparing this to the actual moles of CO₂ collected.

In your question, you're interested in the percent yield of alcohol fermentation - a biological process where yeast converts glucose to ethanol and CO₂. The starting amount of glucose and the amount of collected CO₂ are given, both of which can be used to calculate percent yield. Use the molecular weight of glucose to convert its initial given mass to moles. From there, you can calculate expected moles of CO₂ using the stoichiometry of the proposed alcohol fermentation reaction: 1 mole of glucose yields 2 moles of CO₂.

Furthermore, the actual moles of CO₂ collected can be calculated using the ideal gas law (PV=nRT), where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. Finally, percent yield can be calculated by taking the actual yield (from the collected volume of CO₂) divided by the theoretical yield (from the stoichiometry of the reaction), and multiplying by 100% to obtain a percentage.

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Apart from the gold, a ring can be described by its physical properties like size, style, or shape, its craftsmanship, its visual appeal, emotional significance, the presence of other elements like gemstones, and its condition.

There are several ways to describe a gold ring beyond the material from which it is made, such as gold. You could describe the physical properties of the ring, such as its size, the style of ring, or the shape (is it a band, solitaire, or cluster style?). You can discuss the craftsmanship, indicating if it's handmade or mass-produced. Another aspect could be the visual appeal - is it shiny, elegant, or understated? The ring might have emotional associations - perhaps it's a family heirloom or a symbol of love. It can also be described based on the presence of other elements, like gemstones. Finally, you can describe its condition (new, antique, worn out, well-maintained).

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