Boston, MA and Barcelona, Spain have very similar latitudes (they are both a similar distance from the equator.) Why is the climate of Boston so different from the climate of Barcelona?

Answer:

A. Yes, the calculated density in agreement with the tabulated value.

B. 19.28 mL of volume of cyclohexane should be used.

C. 742.20 is the mass of the sphere of lead.

Explanation:

A.

Volume of the liquid = V = 25.0 mL

Mass of the liquid = m = 21.95 g

Density of the liquid = d

[tex]d=\frac{m}{V}[/tex]

[tex]=\frac{21.95 g}{25.0 mL}=0.878 g/mL[/tex]

Density mentioned in the report book = d' = 0.878 g/mL

d' = d = 0.878 g/mL

Yes, the calculated density in agreement with the tabulated value.

B.

Volume of the liquid cyclohexane= V = ?

Mass of the liquid cyclohexane= m = 15.0 g

Density of the liquid cyclohexane = d = 0.7781 g/mL

[tex]d=\frac{m}{V}[/tex]

[tex]V=\frac{15.0 g}{0.7781 g/mL}=19.28 mL[/tex]

19.28 mL of volume of cyclohexane should be used.

C.

Diameter of the ball = d = 5.0 cm

Radius of the ball = r = 0.5 × d = 2.5 cm

Volume of sphere ,V= [tex]\frac{4}{3}\pi r^3[/tex]

[tex]V = \frac{4}{3}\times 3.14\times (0.25 cm)^3=65.45 cm^3[/tex]

Volume of the spherical lead ball = V  

Mass of the  spherical lead ball= m = ?

Density of the  spherical lead ball = d = [tex]11.34 g/cm^3[/tex]

[tex]d=\frac{m}{V}[/tex]

[tex]m=d\times V=11.34 g/cm^3\times 65.45 cm^3=742.20 g[/tex]

742.20 is the mass of the sphere of lead.

A. The tabulated density of benzene  [tex]15^0 C[/tex] is 0.878 g/mL. B. 19.28 mL of volume of cyclohexane should be used. C. The mass of the lead sphere is approximately 743.5 g.

A. To determine if the calculated density of the liquid matches the tabulated value for benzene, we need to calculate the density using the provided data and then compare it to the handbook value. The density is calculated by dividing the mass of the liquid by its volume.

Given:

Mass of the liquid = 21.95 g

The volume of the liquid = 25.0 mL

Calculated density = [tex]Mass / Volume = 21.95 g / 25.0 mL = 0.878 g/mL[/tex]

The tabulated density of benzene  [tex]15^0 C[/tex] is 0.878 g/mL. Since the calculated density matches the tabulated value, the liquid in the bottle is likely benzene.

B. To find the volume of cyclohexane required for the experiment, we can use the formula: Volume = Mass / Density

Given:

Mass of cyclohexane needed = 15.0 g

Density of cyclohexane at [tex]25^0C[/tex] = 0.7781 g/mL

Volume of cyclohexane = 15.0 g / 0.7781 g/mL = 19.28 mL

Therefore, 19.28 mL of cyclohexane should be used for the experiment.

C. The mass of the lead sphere can be calculated using the formula for the volume of a sphere and the density of lead:

The volume of a sphere = [tex](4/3)\pi r^3[/tex]

Density = Mass / Volume

Given:

The diameter of the lead sphere = 5.0 cm

Radius of the lead sphere = Diameter / 2 = 5.0 cm / 2 = 2.5 cm

Density of lead = [tex]11.34 g/cm^3[/tex]

The volume of the lead sphere =[tex](4/3)\pi (2.5 cm)^3 =20.94375\pi cm^3[/tex]

Mass of the lead sphere = Density — Volume = [tex]11.34 g/cm^3 - 20.94375\pi cm^3= 11.34 g/cm^3- 65.45902 cm^3= 743.5 g[/tex]

Therefore, the mass of the lead sphere is approximately 743.5 g.

Greenhouse gases trap energy in the atmosphere.

Explanation:

Greenhouses gases trap the radiation, especially long infrared wavelengths, that emanate from the earth's surface after solar radiation hits the earth’s surface.  These infrared wavelengths are electromagnetic radiation that are significant in heat transfer. Mostly a higher percentage of this radiation from the earth’s surface escapes into space. Trapping this radiation in the atmosphere causes warming of the atmosphere - a phenomenon called greenhouse effect. This is why an increase in greenhouse gases causes an increase in global temperatures and subsequently drives climate change.

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Answer: The molecular formula for the given organic compound is [tex]C_4H_8O_2[/tex]

Explanation:

We are given:

Percentage of C = 54.5 %

Percentage of H = 9.1 %

Percentage of O = 36.4 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 54.5 g

Mass of H = 9.1 g

Mass of O = 36.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{54.5g}{12g/mole}=4.54moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{9.1g}{1g/mole}=9.1moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{36.4g}{16g/mole}=2.28moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.28 moles.

For Carbon = [tex]\frac{4.54}{2.28}=1.99\approx 2[/tex]

For Hydrogen  = [tex]\frac{9.1}{2.28}=3.99\approx 4[/tex]

For Oxygen  = [tex]\frac{2.28}{2.28}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 4 : 1

Hence, the empirical formula for the given compound is [tex]C_2H_{4}O_1=C_2H_4O[/tex]

Mass of empirical formula = [tex]C_2H_4O[/tex]  = 2(12) + 4(1) + 16 = 44 g/eq.

Now we have to determine the molar mass of compound by using ideal gas equation.

[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]

where,

P = pressure of gas = 1.00 atm

V = volume of gas = 120 mL = 0.120 L

T = temperature of gas = [tex]100^oC=273+100=373K[/tex]

w = mass of gas = 0.345 g

M = molar mass of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the above formula, we get:

[tex]PV=\frac{w}{M}RT[/tex]

[tex](1.00atm)\times (0.120L)=\frac{0.345g}{M}\times (0.0821L.atm/mol.K)\times (373K)[/tex]

[tex]M=88.04g/mol[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

We are given:

Mass of molecular formula = 88.04 g/mol

Mass of empirical formula = 44 g/mol

Putting values in above equation, we get:

[tex]n=\frac{88.04g/mol}{44g/mol}=2[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_2H_4O=(C_2H_4O)_n=(C_2H_4O)_2=C_4H_8O_2[/tex]

Thus, the molecular formula for the given compound is [tex]C_4H_8O_2[/tex]

The empirical formula of the given compound is [tex]\bold {C_2H_4O}[/tex]. The empirical formula is the smallest whole-number ratio of the compound.

Assume the mass of the compound is 100 g. So, the percentages given are taken as mass.

Mass of C = 54.5 g  = 4.54 moles

Mass of H = 9.1 g  = 9.1 moles

Mass of O = 36.4 g = 2.28 moles

To formulate the empirical formula, Calculate the molar ratio by dividing moles by the smallest number,

For the mole ratio, we divide each value of the moles by the smallest number of moles, we get.

For Carbon = 2

For Hydrogen  =4

For Oxygen  = 1

Therefore, the empirical formula of the given compound is [tex]\bold {C_2H_4O}[/tex].

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Answer:A

Explanation:

The solar system consist of the sun, the planets, stars and other objects. The chemical composition of the Sun consist mainly of Hydrogen and helium.

The sun is the largest object in the Solar system, it comprises nearly all the matter in the Solar System, Also the largest planet after the Sun are Jupiter and Saturn are giant planets forming almost the remaining matter of the solar system.

Like the Sun, the mass of Jupiter and Saturn are composed of roughly 98% hydrogen and helium with 2% of all the other elements combined.

Answer:

108.04g

Explanation:

Looking at the balanced chemical equation theoretically, we can see that 3 moles of chlorine yielded 3 moles of HCl. This means that they have equal mole ratio.

Now let’s get the actual reactive moles. The number of moles is obtained by dividing the mass by the molar mass.

The molecular mass of chlorine gas is (2 * 35.5) = 72g/mol

The number of moles is thus 210.3/71 = 2.96 moles

Since the number of moles are equal, the number of moles of HCl produced too is 2.96 moles.

Now to get the mass of HCl produced, we multiply the number of moles by the molar mass of HCl. The molar mass of HCl is 36.5g/mol

The mass is thus 2.96 * 36.5 = 108.04g

Answer:

(A) An investor will exercise the option to put the bond if yields rise significantly

Explanation:

A put option on the bond is a mechanism to allow the buyer of the bond the ability to compel the lender to repay the principal on the bond. The put option offers the buyer of the bond the ability to collect the principal of the bond anytime they choose until maturity for any purpose.

Recall that once the price drops (that is, the yield increases), put options are exercised. If the yield significantly increased, the put choice on a municipal bond is executed.

Answer:

The correct answer is No, it would boil at 109.13°C

Explanation:

This question can be solved by knowing the boiling point elevation formula and the fact that ethylene glycol dissolves in water without dissociation

The boiling point elevation formula is given by

ΔT = i × [tex]K_{b}[/tex] ×[tex]m_{solute}[/tex]

Where [tex]K_{b}[/tex] = 0.51 °C/mole

i = Vant't Hoff factor

m = molality of the solution

When ethylene glycol, C2H6O2, (antifreeze) enters into solution in water it disociates into

C2H6O2 (aq) ---> 2OH(-1)(aq) + C2H4(+2)(aq)

Thus one mole of C2H6O2 dissociates into two moles of hydroxyl ions and one mole of C2H4(+2) ion

Hence the Van't Hoff factor, i, = 3

Therefore the mass of the mole

Therefore ΔT = 3 × 0.51 × 17.9 = 27.387 K

However Ethylene formula = (CH2OH)2 it dissolves in water without dissociation

Therefore i = 1

and ΔT = 1 × 0.51 × 17.9 = 9.129 ≅ 9.13

Hence at the boiling point of the water with antifreeze dissolved in it it

Boiling point of water + Boiling point elevation = 100 + 9.13 = 109.13 °C

The water will not boil until it reaches 109.13 °C

The coolant mixture will not boil at 108°C. The calculated boiling point elevation suggests it will boil at approximately 109.16°C, based on the molality and the boiling point elevation constant for water.

To determine whether the coolant mixture in your car will boil at 108°C, we can use the concept of boiling point elevation. The boiling point of a solution increases when a solute is added to a solvent due to the colligative properties of the solution. Using the molality provided (17.9m), and knowing the boiling point elevation constant (Kb) for water is approximately 0.512 °C/m, we can calculate the boiling point elevation.

ΔTb = i * Kb * m

Where ΔTb is the boiling point elevation, i is the van't Hoff factor (i = 1 for ethylene glycol as it does not dissociate in solution), Kb is the ebullioscopic constant for water, and m is the molality of the solution.

ΔTb = 1 * 0.512 °C/m * 17.9m = 9.16 °C

The normal boiling point of water is 100°C, so adding the elevation to the normal boiling point gives us:

100°C + 9.16°C = 109.16°C

Therefore, the solution will not boil at 108°C because it would boil at approximately 109.16°C.

Answer:

True is the correct answer to the above question.

Explanation:

Answer:

"True" is the correct answer to this question.

Explanation:

The program to the given question as follows:

Program:

public class data //defining class

{

  public static void main (String [] aw)//defining the main method

  {

String x="OH"; //defining string variable x and assign value

System.out.println("assign value: "+x); //print value

x = new String("OH"); //defining instance variable and assign value

System.out.println("assign value by creating instance: "+x); //print value

  }

}

Output:

assign value: OH

assign value by creating instance: OH

Explanation of the program:

In the above java program, a class data is defined, inside the class the main method is declared, In this main method a string variable "x" is defined that holds a value "OH", then we the print function to print this variable value.

In the next line, An instance of variable x is created, which holds a value "OH" in its parameter. In this question, both are correct because both hold the same value.

Answer:

Law of definite proportion

Explanation:

As per law of definite proportion, ratio of elements present in a compound is always fixed irrespective of the  source, amount and method of preparation.

In the given case, the hydrogen and oxygen react with each other to form a compound. The ratio of oxygen and hydrogen is fixed which is 8 : 1 and this ratio does not change upon changing amount of oxygen and hydrogen.

So, the this experiment support the law of definite proportion.

Therefore, the correct option is option 4

The atomic mass of element X is approximately 107.8682 amu with an accuracy of .001%.

To calculate the atomic mass of element X, we need to use the given information about the isotopes of element X.

We are given that element X consists of two naturally occurring isotopes: 107X with an abundance of 53.84% and an isotopic mass of 106.9051 amu, and 109X with an isotopic mass of 108.9048 amu.

To calculate the atomic mass of element X, we can use the formula:

Atomic mass = (abundance of isotope 1 x mass of isotope 1) + (abundance of isotope 2 x mass of isotope 2)

Plugging in the values for element X, we get:

Atomic mass = (0.5384 x 106.9051 amu) + (0.4616 x 108.9048 amu)

Atomic mass ≈ 107.8682 amu

Therefore, the atomic mass of element X is approximately 107.8682 amu to an accuracy of .001%.

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To calculate the atomic mass of element X, multiply the mass of each isotope by its abundance and sum them up. The atomic mass of element X is 106.22288 amu.

To calculate the atomic mass of element X, we need to consider the abundance and isotopic mass of its two naturally occurring isotopes. The atomic mass is calculated by multiplying the mass of each isotope by its abundance and summing them up.

For isotope 107X with an abundance of 53.84%, we multiply its mass (106.9051 amu) by its abundance (0.5384). For isotope 109X with an abundance of (100% - 53.84% = 46.16%), we multiply its mass (108.9048 amu) by its abundance (0.4616).

Finally, we add these two values together to get the atomic mass of element X to an accuracy of .001%. The calculation is as follows:

(106.9051 amu * 0.5384) + (108.9048 amu * 0.4616) = 106.22288 amu

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Answer:

Lead (Pb)

Explanation:

To find the atomic mass of the element, we need to take into consideration all the naturally occurring isotopes. We then find the atomic mass by multiplying the natural abundance of the isotopes by their mass for all of the isotopes and summing them together.

1. 1.390/100 * 203.97304 = 2.835225256

2. 24.11/100 * 205.97447 = 49.660444717

3. 22.09/100 * 206.97590 = 45.72097631

4. 52.41/100 * 207.97665 = 109.000562265

We then add all of these masses together:

109.000562265 + 45.72097631+ 49.660444717 + 2.835225256 = 207.217208548

To 4 sf = 207.2 amu

Element is Lead (Pb)

Answer:

w/v% = gm per 100 ml

75gm X gm 7500

------- = ------ == --------

1500ml 100ml 1500X

X=5%

Answer: False

Explanation:

From definition dynamic equilibrium is a state of balance between continuing processes. When the rate of evaporation equals the rate of condensation the system has reached a dynamic equilibrium.

However, it is possible, to disturb a system that is in dynamics equilibrium by changing conditions of the system. In fact anything that changes the thermodynamic state of the system will disturb the system and it will no longer be at equilibrium.

An increase in temperature to the system will favour evaporation more than condensation. Molecules with higher kinetic energy will escape the system.

A reducing agent does get oxidized as it reacts in a process known as a redox reaction. During this reaction, the reducing agent loses electrons, effectively donating them to another substance. Disproportion reactions are also possible, where the same substance gets oxidized and reduced.

Yes, the statement is true: A reducing agent gets oxidized as it reacts. In redox reactions, the substance that is oxidized loses electrons and is therefore referred to as the reducing agent. For example, in the reaction, 'Al(s) + NiO(s) --> Al2O3(s) + Ni(s)', aluminum (Al) is the reducing agent as it gets oxidized from 0 to +3 oxidation state while reducing nickel oxide (NiO).

Redox reactions involve the transferring of electrons from one atom (which gets oxidized) to another (which gets reduced). In the process, the reducing agent is oxidized because it essentially donates its own electron to the other substance.

There are also cases of disproportion reactions where the same substance gets oxidized and reduced. These reactions are very interesting in the context of oxidation-reduction chemistry.

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Answer:

2,347.8 grams

Explanation:

The freezing point depression Kf of water = 1.86° C / molal.

To still freeze at -12° C, then the molality of the solution 12/ 1.86 = 6.45 moles

The molecular weight of sorbitol (C6H14O6)is:

6 C = 6 ×12 = 72

14 H = 14 × 1 = 14

6 O = 6 × 16 = 96

...giving a total of 182

 So one mole of sorbitol has a mass of 182 grams.

Since there are 2 kg of water,  2 × 6.45 moles = 12.9 moles can be added to the water to get the 12° C freezing point depression. 

Therefore

grams = moles × molar mass

12.9 moles × 182 grams / mole = 2,347.8 grams of sorbitol can be added and still freeze

Answer:

33.3 kg of air

Explanation:

This is a problem of conversion unit.

Density is mass / volume

Therefore we have to calculate the volume in the room, to be multiply by density. That answer will be the mass of air.

Volume of the room → 9 ft . 11 ft . 10 ft = 990 ft³

Density is in g/L, therefore we have to convert the ft³ to dm³ (1 dm³ = 1L)

990 ft³ . 28.3 dm³ / 1ft³ = 28017 dm³ → 28017 L

This is the volume of the room, if we replace it in the density formula we can know the mass of air in g.

1.19 g/L = Mass of air / 28017 L

Mass of air = 28017 L .  1.19 g/L → 33340 g of air

Finally, let's convert the mass in g to kg → 33340 g . 1kg / 1000 g = 33.3 kg

Answer:

Lowers the actual yield

Explanation:

Answer:

option c, 0.1 M [tex]AlCl_3[/tex]

Explanation:

Addition of non-volatile solute to a solvent decreases its vapour pressure which results in decrease in melting point.

Decrease in melting point is known as depression in freezing point. the depression in freezing point is related with molality and no. of ions as follows:

[tex]\Delta T_f = imK_f[/tex]

Where, i is von't Hoff factor, m is molatilty and ΔTf  is depression in freezing point.

As, in the given case concentration of all the solution is same, therefore, depression in freezing point will depend upon the no. of ions produced by the ionization of the salts in the aqueous solution.

In case of NaCl, the no. of ions produced will be 2.

Therefore, value of i will be 2

In case of [tex]MgCl_2[/tex], the no. of ions produced will be 3.

Therefore, value of i will be 3

In case of [tex]AlCl_3[/tex], the no. of ions produced will be 4.

Therefore, value of i will be 4.

More, the value of van't Hoff factor, more will be depression in freezing point.

Therefore, assuming the given solution to be aqueous, the solution expected to have lowest freezing point is 0.1 M  [tex]AlCl_3[/tex].

The correct option is b. 0.1 M MgCl2 would be expected to have the lowest freezing point.

To understand why 0.1 M MgCl2 has the lowest freezing point, we need to consider the colligative properties of solutions, specifically the freezing point depression. The freezing point depression is directly proportional to the number of particles (ions or molecules) in the solution. When a solute is dissolved in a solvent, it dissociates into ions, and the more ions it dissociates into, the greater the freezing point depression.

Let's analyze each of the given solutions:

a. 0.1 M NaCl: Sodium chloride (NaCl) dissociates into two ions in solution, Na+ and Cl-. Therefore, the total number of particles in solution is 2 times the concentration of NaCl, which is 0.2 moles of particles per liter.

b. 0.1 M MgCl2: Magnesium chloride (MgCl2) dissociates into three ions in solution, Mg2+ and 2Cl-. Therefore, the total number of particles in solution is 3 times the concentration of MgCl2, which is 0.3 moles of particles per liter.

c. 0.1 M AlCl3: Aluminum chloride (AlCl3) can dissociate into four ions in solution, Al3+ and 3Cl-. However, AlCl3 is not fully dissociated in aqueous solution due to its Lewis acid behavior and the formation of complex ions such as AlCl4-. Assuming complete dissociation for simplicity, which is not the case in reality, the total number of particles in solution would be 4 times the concentration of AlCl3, which is 0.4 moles of particles per liter.

Since MgCl2 dissociates into the most number of ions compared to NaCl and AlCl3 (assuming complete dissociation for AlCl3), it will have the greatest effect on freezing point depression. Therefore, 0.1 M MgCl2 is expected to have the lowest freezing point among the given options.

It is important to note that in reality, the actual freezing point depression of AlCl3 would be less than calculated here due to its incomplete dissociation and complex ion formation. This further supports the conclusion that MgCl2 would have the lowest freezing point, as AlCl3 would not reach the expected freezing point depression based on complete dissociation.

Answer:

Lithium and Sodium

Explanation:

Losing and electron in natural setting is characterizes of elements in group one. These are elements known as the alkaline earth metals. They are the most electropositive elements on the periodic table.

These elements ionize by losing an electron yin their outermost shell to attain the configuration of the nearest noble gas. These elements are usually found in combined and rarely seen in uncombined state principally due to their very reactive nature.

Sodium naturally would ionize by losing one electron. Other elements capable of this even at a better rate because they are more electropositive are potassium and lithium. Both are also group one alkaline metals

Answer : The mass of nitrogen gas collected is, 290.9 grams

Explanation :

To calculate the mass of nitrogen gas we are using ideal gas equation:

[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]

where,

P = pressure of nitrogen gas = 1 atm

V = volume of nitrogen gas = 250 L

T = temperature of nitrogen gas = [tex]20^oC=273+20=293K[/tex]

R = gas constant = 0.0821 L.atm/mole.K

w = mass of nitrogen gas = ?

M = molar mass of nitrogen gas = 28 g/mole

Now put all the given values in the ideal gas equation, we get:

[tex](1atm)\times (250L)=\frac{w}{28g/mole}\times (0.0821L.atm/mole.K)\times (293K)[/tex]

[tex]w=290.9g[/tex]

Therefore, the mass of nitrogen gas collected is, 290.9 grams.

Answer:

The net charge of 1.3 g nugget of pure gold after 1.68% of its electrons are removed is 559 C

Explanation:

When an atom gains electrons it becomes negatively charged. Conversely, when it looses electrons the atoms becomes positively charged thus

To solve this question, we rely on the relationship between the nmber of particles present in a given mass of an atom, Avogadro's number and number of moles, n

The given variables are

mass of pure gold nugget = 1.30 g

Quantity of electrons removed = 1.68% of electrons present in the gold sample

Molar mass of gold = 197 g/mol

Avogadro's number = 6.02 × 10²³ atoms/mole  

qc = one electron charge = -1.06 × 10⁻¹⁹ C/electron

Electrical charge of gold nugget = 0 C

Number of electrons in one gold atom = 79 electrons

Solving for the number of prticles or gold atoms in 1.3 grams of gold we get

n mass/(molar mass) = 1.3/197 moles of gold =  0.0066 moles

number of particles in 0.0066 moles of gold N = n×[tex]N_{A}[/tex] = 0.0066 × 6.02 × 10²³  = 3.97 × 10²¹ atoms

since 79 electrons are present per particle we have

3.97 × 10²¹ × 79 = 3.14 × 10²³ electrons

quantity of elecrtrons removed = 1.68% of  3.14 × 10²³ electrons =1.68/100 × 3.14 × 10²³ electrons = 0.0168 × 3.14 × 10²³ electrons = 5.3 × 10²¹ electrons

The net charge of 5.3 × 10²¹ electrons = 5.3 × 10²¹ electrons × -1.06 × 10⁻¹⁹ C/electron =

5.59 × 10² C = 559 C

Final answer:

After 1.68% of its electrons are removed, a 1.30 g nugget of pure gold will have a net positive charge of approximately +842 C, calculated based on the initial number of electrons and the charge per electron.

Explanation:

To calculate the net charge of a 1.30 g nugget of pure gold after 1.68% of its electrons are removed, we first need to understand the composition and characteristics of gold. The atomic mass of gold (Au) is approximately 197 g/mol, and each atom has 79 protons and, when neutral, 79 electrons. The number of gold atoms in the nugget can be calculated using its mass and the atomic mass of gold.

First, calculate the number of moles of gold in the nugget:
Moles of gold =
(1.30 g) / (197 g/mol) = 0.0066 mol. The number of gold atoms = 0.0066 mol ×
(6.022 × 1023 atoms/mol) ≈ 3.97 × 1021 atoms. Since each gold atom has 79 electrons when neutral, the total number of electrons initially is approximately 3.13 × 1023 electrons.

To find out how many electrons 1.68% represents:
1.68% of the total electrons = 1.68 / 100 × 3.13 × 1023 ≈ 5.26 × 1021 electrons.

The charge of one electron is approximately -1.6 × 10-19 C. Therefore, removing 5.26 × 1021 electrons means the nugget would have a net positive charge equal to:
5.26 × 1021 × 1.6 × 10-19 C ≈ 8.42 × 102 C. Thus, the net charge of the gold nugget after removing 1.68% of its electrons would be approximately +842 C.

Answer:

10

Explanation:

Binary search's runtime is proportional to log (base two) of the number of list elements.

Answer:

The answer to your question is Gallium-71 = 70.9202 amu

Explanation:

Gallium atomic weight = 69.7

Gallium-69 = 68.9 amu    abundance = 60.4%

Gallium-71 =     x                abundance = 39.6%  

To solve this problem just write an equation and solve it for the mass of gallium-71.

Equation

Gallium = Gallium-69(abundance) + Gallium-71(abundance)

Substitution

69.7 = (68.9)(0.604) + Gallium-71(0.396)

69.7 = 41.6156 + Gallium-71(0.396)

Gallium-71(0.396) = 69.7 - 41.6156

Gallium-71(0.396) = 28.0844

Gallium-71 = 28.0844/0.396

Gallium-71 = 70.9202 amu

Answer:

a) 2

b) 1/4

c) 9

Explanation:

a) for a first order reaction in reactant A

r initial = k*[A initial]

then if the concentration is doubled [A final ]= 2*[A initial]  , then

r final = k*[A final ] = 2* k*[A initial] = 2*r initial

then the velocity changes by a factor of 2

b) for a second order reaction in reactant B

r initial = k*[B initial]²

then if the concentration is halved: [B final ]= [B initial]/2  , then

r final = k*[B final ]²  = k*( [B initial]/2 )²  =k* [B initial]² /4 = r initial /4

then the velocity changes by a factor of 1/4

c) for a second order reaction in reactant C

r initial = k*[C initial]²

then if the concentration is tripled : [C final ]= 3* [C initial]  , then

r final = k*[C final ]²  = k*( 3*[C initial] )²  =k* [C initial]² *9 = 9 * r initial  

then the velocity changes by a factor of 9

For first-order reactions, doubling the concentration of the reactant doubles the reaction rate. For second-order reactions, halving the concentration reduces the rate by a factor of four, while tripling the concentration increases the rate by a factor of nine.

(a) For a reaction that is first order in reactant A, if [A] is doubled, the rate of the reaction will also double. This is because the rate of a first-order reaction is directly proportional to the concentration of the reactant. When [A] is doubled, the rate is doubled as well.

(b) For a reaction that is second order in reactant B, if [B] is halved, the rate of the reaction will decrease by a factor of four. This is because the rate of a second-order reaction is directly proportional to the square of the concentration of the reactant. When [B] is halved, the concentration is squared and the rate is decreased by a factor of four.

(c) For a reaction that is second order in reactant C, if [C] is tripled, the rate of the reaction will increase by a factor of nine. Similar to the previous case, the rate of a second-order reaction is directly proportional to the square of the concentration of the reactant. When [C] is tripled, the concentration is squared and the rate is increased by a factor of nine.

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Answer:

Yes, this is true. The reason is that the flower transpires and sucks the water in and distributes it as much as it can. You can also flip it upside down and hang it with petals down , allowing the liquid to enter the flower and then retaining color for longer periods of time and having more color.

Explanation:

Answer:

If you add more solute in a saturated solution, it will have no effect. Additional solute does not dissolve in a saturated solution.

Explanation:

If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.

Answer: The partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm

Explanation:

Assuming ideal gas behavior, the equation follows:

PV = nRT

We know that:

[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]

Rearranging the above equation:

[tex]M=\frac{dRT}{P}[/tex]

where,

d = density of gas mixture = 2.2 g/L

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature = [tex]26^oC=[26+273]K=299K[/tex]

P = pressure of the mixture = 0.80 atm

M = average molar mass of mixture

Putting values in above equation, we get:

[tex]M_{avg}=\frac{2.2g/L\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 299}{0.80atm}\\\\M_{avg}=67.5g/mol[/tex]

We know that:

Molar mass of [tex]NO_2[/tex] = 46 g/mol

Molar mass of [tex]N_2O_4[/tex] = 92 g/mol

Let the mole fraction of [tex]NO_2[/tex] be 'x' and that of [tex]N_2O_4[/tex] be '(1-x)'

For average molar mass calculation:

[tex]M_{avg}=M_{NO_2}\chi_{NO_2}+M_{N_2O_4}\chi_{N_2O_4}[/tex]

Putting values in above equation:

[tex]67.5=46x+92(1-x)\\\\x=0.533[/tex]

Mole fraction of [tex]N_2O_4[/tex] = (1 - x) = (1 - 0.533) = 0.467

To calculate the partial pressure, we use the equation given by Raoult's law, which is:

[tex]p_{A}=p_T\times \chi_{A}[/tex]

We are given:

[tex]p_T=0.80atm\\\chi_{NO_2}=0.533[/tex]

Putting values in above equation, we get:

[tex]p_{NO_2}=0.80atm\times 0.533\\\\p_{NO_2}=0.426atm[/tex]

We are given:

[tex]p_T=0.80atm\\\chi_{N_2O_4}=0.467[/tex]

Putting values in above equation, we get:

[tex]p_{N_2O_4}=0.80atm\times 0.467\\\\p_{N_2O_4}=0.374atm[/tex]

Hence, the partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm

Answer:

The peptide will definitely migrated towards cathode (negative terminal)

Explanation:

The positively and negatively charged side chains of proteins cause them to behave like amino acids in an electrical field; that is, they migrate during electrophoresis at low pH values to the cathode (negative terminal) and at high pH values to the anode (positive terminal). The isoelectric point, the pH value at which the protein molecule does not migrate, is in the range of pH 5 to 7 for many proteins.

Answer : The precipitate will not be formed when these solutions are mixed.

Explanation :

The chemical equation for the reaction of calcium chloride and lithium sulfate follows:

[tex]CaCl_2(aq)+Li_2SO_4(aq)\rightarrow 2LiCl(aq)+CaSO_4(s)[/tex]

We are given:

Concentration of calcium ion = [tex]2.1\times 10^{-5}M[/tex]

Concentration of sulfate ion = [tex]4.75\times 10^{-5}M[/tex]

[tex]K_{sp}=2.4\times 10^{-5}[/tex]

The salt produced is calcium sulfate.

The equation follows:

[tex]CaSO_4(s)\rightleftharpoons Ca^{2+}(aq)+SO_4^{2-}(aq)[/tex]

The expression of [tex]Q_{sp}[/tex] (ionic product) for above equation follows:

[tex]Q_{sp}=[Ca^{2+}]\times [SO_4^{2-}][/tex]

Putting values of the concentrations in above expression, we get:

[tex]Q_{sp}=(2.1\times 10^{-5})\times (4.75\times 10^{-5})\\\\Q_{sp}=9.9\times 10^{-10}[/tex]

There are 3 conditions:

As, the [tex]K_{sp}>Q_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.

Hence, the precipitate will not be formed when these solutions are mixed.

Answer:I beams can withstand greater Moment compared to other beams like rectangular beam when given the same stress.

Rectangular beams are better for casting concrete beams, and for making beams of high and low moment of inertia and its centriod is easy to understand.

Explanation: I beams are beams usually used in metal work and in making beams of high moments of inertia.

Rectangular beams are beams whose centroid ( the geometric center of a regular rectangular beam) are easily identifiable and they can be built to give high and low moment of inertia. This is one of the advantage of using rectangular beams.

The advantage of use of I beam over rectangular beam has been the wide spread and high moment of inertia.

In the construction industry, for the support to the buildings and the structure, beams of different shapes has been used. I beam are the steel or metal beams that have high functionality.

I beam has been made with the shape of rolled joist. They have been installed in the building because of their high moment of inertia.

I beam has been working with the spread of moment of inertia in the structure, that has been able to bend and withstand the pressure more than compared to the rectangular cross-section with concentrated inertia.

Learn more about I beam, here:

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