starches and glycogen are carbohydrates polymers comprised of which monomer

a) nucleotides
b) fatty acids
c) amino acids
d) monosaccharides.

1) The gold will occupy a volume of 2.99 dm³.

[tex]\text{Volume = 17 200 g} \times \frac{\text{1 cm}^{3}}{\text{5.75 g}} = \text{2990 cm}^{3} = \textbf{2.99 dm}^{3}\\[/tex]

2) The density of the olive oil is 0.850 g/cm³.

[tex]\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{425 g}}{\text{500 cm}^{3}}= \textbf{0.850 g/cm}^{3}[/tex]

3) The mass of the silver is 10.4 kg.

[tex]\text{Mass} = \text{987 cm}^{3} \times \frac{\text{10.5 g} }{\text{1 cm}^{3 }} = \textbf{10 400 g = 10.4 kg}\\[/tex]

Metals conduct heat from one to another end.....this property will help you

Heat is used to measure a mineral's thermal conductivity and specific heat to determine its metallic nature. By comparing these values with known metallic minerals and observing properties like color and luster, one can identify if a mineral is metallic.

Heat can be utilized to determine if a mineral has a metallic nature by examining its thermal conductivity and specific heat. Metallic minerals generally conduct heat more efficiently than non-metallic minerals due to the movement of free electrons. To quantify this, the specific heat of a mineral could be measured by transferring a known quantity of heat to the mineral and observing the change in its temperature.

For example, consider a scenario where a piece of metal at a known temperature is placed in water at a different temperature. By measuring the resulting temperature change in the water, and knowing the mass and specific heat of the water, the specific heat of the metal can be calculated. This value can then be compared against known values for metallic minerals to help identify the mineral in question.

If the measured specific heat is close to known values for metals such as gold or lead, additional observations such as color and luster can help refine the identification. For instance, a silver/gray color, when paired with an appropriate specific heat value, suggests a lead mineral over gold, which has a yellowish color.

Hey There!

Ideal gas equation: PV = nRT where P is pressure, V is volume, n is moles of gas, R is molar gas constant and T is temperature.

Since P, R and T are constant, V/n = constant

V1 = 2.00 L, V2 = 3.30 L

n1 = (2.00 g)/(4.00 g/mol) = 0.5 mol, n2 = ?

V1/n1 = V2/n2

2.00/0.5 = 3.30/n2

n2 = 0.825 mol

Moles of He added = n2 - n1

= 0.825 - 0.5 = 0.325 mol

Mass of He added = moles x atomic mass

= 0.325 * 4.00 = 1.30 g of He

The amount of helium added to the cylinder, while keeping the pressure and temperature constant and increasing the volume from 2.00L to 3.70L, is approximately 1.70 grams.

The problem you've presented involves the concept of an ideal gas law which states that the number of moles of gas is directly proportional to the volume it occupies, provided that temperature and pressure remain constant. Considering that you are keeping the pressure and temperature constant, and you have increased the volume from 2.00 L to 3.70 L, the mass of helium will also increase proportionally.

To find out how much helium was added, we first must determine the initial number of moles of helium using its given mass (2.00 g) and the molar mass of helium (around 4 g/mol). That is, initial moles of helium = 2.00 g / 4 g/mol = 0.5 mol. Since number of moles is directly proportional to volume, we can set up a proportion: initial volume/initial moles = final volume/final moles (2.00 L / 0.5 mol) = (3.70 L / x mol), which yields a solution of roughly 0.925 mol for 'x', the final number of moles.

 Subtracting the initial moles (0.5 mol) from the final moles (0.925 mol), we find that around 0.425 mol of helium has been added. The mass of the added helium equals the moles of added helium multiplied by the molar mass, i.e., 0.425 mol * 4 g/mol = 1.70 g.

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Hey there!

Stock solution:

Concentracion bromide = 0.128 M

initial solution in volumetric flask =  450 µl

So , moles of bromide present:

450  µl in liters :

1 µl  =   = 1*10⁻⁶ liters

450 * ( 1*10⁻⁶ )  = 0.00045

0.128 * 0.00045 =>  57.6 * 10⁻⁶ moles

Now volume final is 25 mL , in liters : 0.025 L ou 25*10⁻³

so  new  bromide concentration:

57.6*10⁻⁶ / 25*10⁻³=> 2.304*10⁻³ M

To find the bromide concentration in the diluted solution, use the formula for dilution C1V1 = C2V2. Your values are C1= 0.128 M, V1= 0.45 ml, and V2= 25 ml. Solve the equation for C2 to obtain the required bromide concentration.

The bromide concentration in the diluted solution can be calculated using the formula for dilution: C1V1 = C2V2. Where C is concentration, V is volume, 1 refers to initial conditions, and 2 refers to conditions after dilution.

Here, C1 is 0.128 M (the initial concentration of bromide in the stock solution), V1 is 450 μl (the volume of the aliquot), V2 is 25 ml (the volume of the diluted solution in the volumetric flask). Inserting these values into the formula, we need to find C2 (unknown concentration of diluted solution).

Therefore, C2 = C1V1 / V2.

Before doing calculations, make sure that all volumes are in the same unit. In this case, convert μl to ml: 1μl = 0.001 ml, so 450 μl = 0.45 ml.

The calculation becomes: C2 = (0.128 M * 0.45 ml) / 25 ml, which, when calculated, provides the concentration of bromide in the diluted solution.

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correct answer is B) equal to the number of protons

answer D)

explanation:

Bases turn the red litmus to Blue

while aciDs turn blue litmus to reD.

Answer: The correct answer is Option d.

Explanation:

Litmus paper is defined as the indicator which is used to determine the nature of the solution, whether it is acidic or basic.

There are 2 types of litmus paper:

Hence, the correct answer is Option d.

First you can take cotton balls, and make a model.

So first make a circle of cotton balls to resemble the eye of a hurricane.

Hurricanes are not round so, when your creating the outside if the eye make it took like the hurricane has bands.

One you create the model, you can simply write things you know about the hurricane, etc.

Examples of thing's you can say:

-The peak of hurricane season is September.

-Hurricane season starts in June.

-September is one of the most active month for hurricanes.

Hey there!:

absorbance = log 100 - log Transmitance

absorbace = 0.85

log 100 = 2

- log transmitance = absorbace / log 100

0.85 / 2= 0.425

transmitance =  10 ^ ( - 0.425 )

transmitance = 0.376

Fatty acids are carboxylic acids with a carboxylate group and a hydrocarbon chain. Saturated fatty acids have no double bonds and are fully 'saturated' with hydrogen, while unsaturated fatty acids contain one or more double bonds and have fewer hydrogen atoms.

Differences and Similarities Between Saturated and Unsaturated Fatty Acids

Fatty acids are carboxylic acids that serve as building blocks for various types of lipids. All fatty acids have a carboxylate group (-COOH) attached to a hydrocarbon chain. The primary difference between saturated and unsaturated fatty acids lies in the hydrocarbon chain's bond types. In saturated fatty acids, all the carbons are connected by single covalent bonds, and each carbon atom is 'saturated' with hydrogen atoms. This means that there are no double or triple bonds, allowing for the maximum number of hydrogen atoms to be attached to the carbon skeleton. Stearic acid is an example of a saturated fatty acid. On the other hand, unsaturated fatty acids contain one or more double bonds within the hydrocarbon chain. These double bonds reduce the number of hydrogen atoms attached to the carbon skeleton. They are called unsaturated because they do not contain the maximum amount of hydrogen possible.

D the belguim people atacked using it

milligrams. kilograms are too big.

Answer:

B) Milligrams

Explanation:

just did the test on usatestprep

B) 83 mL

Find the moles of NaOH by using Molarity=moles/volume(L) so Moles = molarity x volume.

Then use the mol ratio to go from moles of NaOH to moles of H2SO4 by dividing the moles of NaOH by 2. Now you're in moles of H2SO4 so again use M = moles/Volume to find the volume. Volume = moles/M.

(The volume is in Liters in the equation but you can use the mL on this problem because your conversions cancel each other out.)

Answer:

B Plato

Explanation:

1 mole has 6.02*10^23 molecules in it.

1 nickel (II) chloride molecule, NiCl2, has 1 Ni atom in it.

so 1 mole of nickel (II) chloride molecule has 1 mole of Ni atom in it.

so 100 moles of nickel (II) chloride molecule has 100*6.02*10^23

= 6.02*10^25 Ni atom in it.

Nickel (II) chloride has the chemical formula of NiCl2. So 1 mole of NiCl2 contains 1 mole of Ni atoms.

1 mole contains 6.02*10^23 atoms. So 100 moles of Nickel (II) chloride contains 6.02*10^25 Ni atoms.

[tex]A_r = 62.9296 \; \text{amu} \times 69.15 \% + 64.9278 \; \text{amu} \times 30.85 \%\\\phantom{A_r} = 63.5460 \; \text{amu}[/tex]

The idea is to sum up the product of atomic mass and abundance for each of the isotope- e.g. 62.9296 and 69.15% for X-63- to find the average of isotope atomic mass weighted regarding their abundance, which is by definition the relative atomic mass of the element.

Answer : The average atomic mass of an element X is, 63.546 amu

Solution : Given,

Mass of isotope X-63 = 62.9296 amu

% abundance of isotope X-63 = 69.15% = 0.6915

Mass of isotope X-64 = 64.9278 amu

% abundance of isotope X-64 = 30.85% = 0.3085

Formula used for average atomic mass of an element X :

[tex]\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})[/tex]

[tex]\text{ Average atomic mass of an element X}=\sum[(62.9296\times0.6915)+(64.9278\times 0.3085)][/tex]

[tex]\text{ Average atomic mass of an element X}=63.546amu[/tex]

Therefore, the average atomic mass of an element X is, 63.546 amu

BaSO₄ is relatively harmless, but BaS is highly toxic.

BaSO₄ is quite insoluble (240 µg/100 mL). It is a mild irritant in cases of skin contact and inhalation. However, it is safe enough that health professionals ask patients to drink a suspension of BaSO₄. The Ba is opaque to X-rays, so it makes the stomach and intestines more visible to radiographers.

BaS is soluble (7.7 g/100 mL). It reacts slowly with water and more rapidly in the acid conditions of the stomach to release H₂S.

BaS + 2HCl ⟶ BaCl₂ + H₂S

An H₂S concentration of 60 mg/100 mL can be fatal within 30 min.

Don’t eat barium sulfide!

A) deposition is the processes where particles of rock or laid down in sections with heavier sediments building up first

Answer:

The correct answer is option A, that is, deposition.

Explanation:

The erosion, weathering, and deposition refer to the procedures, which works spontaneously on or close to the surface of the Earth. With time, these procedures lead to the generation of sedimentary rocks.  

Deposition refers to the procedure where the rocks particles are laid down in segments with the heavier sediments forming up initially. It is the geological phenomenon in which the soil, sediments, and rocks are supplemented to a land mass or landform. Water, ice, wind, and gravity carry the previously weathered surface material that at the expense of adequate kinetic energy in the fluid gets deposited, forming up the sediment layers.  

I agree with what you've chosen for question two and three.

I would choose C for the fourth question and argue that you shall also go with C for question five.

One of the main ideas about repetition is to have multiple Trials- or independent experiments- for a single independent variable using identical experimental setup. The table presented in option C is the only one among the four choices that allots spaces for measurements from multiple independent experiments.

Hypotheses are predictions of the outcome of a particular experiment. A hypothesis is correct only if it accurately describes the outcome of an experiment. The hypothesis, not the data, would have to be revised if the two doesn't fit. The researcher could keep developing and testing new hypotheses until arriving at a correct one.

Additionally, I would prefer option B over option D in the first question- responses to that question can depends on what your teacher says about the validity of different sources, given that neither B nor D are reliable as sources that one would like to cite in a paper.

Given:

Density of iron:7.874g/mL

Mass of iron : 7.75 gms

Now we know that

Volume = Mass/density

Substituting the given values in the above formula we get

Volume of iron= 7.874/7.75= 1.016 mL

Volume of iron = 0.01016 dL

The volume of a 7.75 gram iron sample, with a density of 7.874 g/mL, is approximately 0.9843 mL, which converts to approximately 0.009843 dL.

To calculate the volume of an iron sample with a mass of 7.75 grams, given the density of iron is 7.874 g/mL, use the density formula:

Density = Mass/Volume

By rearranging the formula to solve for volume, we get:

Volume = Mass/Density

Substitute in the known values:

Volume = 7.75 g / 7.874 g/mL

Volume ≈ 0.9843 mL

To convert milliliters to deciliters (dL), use the conversion factor that 1 dL = 100 mL:

Volume ≈ 0.9843 mL × (1 dL / 100 mL)

Volume ≈ 0.009843 dL

Therefore, the volume of the sample of iron in deciliters is approximately 0.009843 dL.

Hey There!

Assuming the part of the round flask containing liquid to be perfect sphere, we can calculate it's volume.

Given diameter of sphere = 24 cm  

Therefore radius of the sphere, r= 24/2= 12 cm

Volume of the sphere of diameter 24 cm = (4/3)*pi*r³ = 7,238.23 cm3

Conversion of cm³ to L

1 L= 1 dm³

1 dm = 10 cm

Therefore 1L= 1 dm³ = 1000 cm³

So the volume of liquid in the round flask = 7.24 L( round of to nearest 0.01 L)

The volume of a round flask can be calculated using the formula: Volume = 4/3 * π * radius³. The calculated volume should then be converted from cubic meters to liters, and rounded to the nearest 0.01 liters.

The volume of a round flask or sphere can be calculated using the mathematical formula: Volume = 4/3 * π * radius³. Assuming you have the radius of the flask, you plug it into this formula, calculate the volume in cubic meters, and then convert to liters (since 1 cubic meter equals 1000 liters). If you are given the diameter, remember to divide it by 2 to find the radius. Once you have your answer, round to the nearest  0.01 liters as per the question's instructions.

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In a Directly proportional graph, a straight line will be observed between two variables and inversely proportional graph no straight line will be observed between two variables.

So, from this, we can conclude that in a :

Directly proportional graph, a straight line will be observed between two variables, and inversely proportional graph no straight line will be observed between two variables.

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The equilibrium concentration of NO is [tex]\boxed{{\text{0}}{\text{.0018 M}}}[/tex].

Further Explanation:

Chemical equilibrium is a stage where the rate at which forward reaction proceeds becomes equal to the rate at which the backward reaction occurs. At equilibrium, the formation of product from reactant gets balanced out by the formation of reactants from products so there is no change in concentrations of both reactants and products.

The expression for a general equilibrium is,

[tex]a{\text{A}} + b{\text{B}} \rightleftharpoons c{\text{C}} + d{\text{D}}[/tex]

Here,

A and B are the reactants.

C and D are the products.

a and b are the stoichiometric coefficients of reactants.

c and d are the stoichiometric coefficients of products.

The expression for the equilibrium constant for the general reaction is as follows:

[tex]{K_{\text{c}}}=\dfrac{{{{\left[ {\text{C}} \right]}^c}{{\left[ {\text{D}} \right]}^d}}}{{{{\left[ {\text{A}} \right]}^a}{{\left[ {\text{B}} \right]}^b}}}[/tex]  

Here,

[tex]{K_{\text{c}}}[/tex] is the equilibrium constant.

[C] is the concentration of C.

[D] is the concentration of D.

[A] is the concentration of A.

[B] is the concentration of B.

The given reaction occurs as follows:

[tex]2{\text{NO}}\left( g \right) \rightleftharpoons {{\text{N}}_2}\left( g \right) + {{\text{O}}_{\text{2}}}\left( g \right)[/tex]

The initial concentration of NO is 0.175 M. Let x to be the change in concentration at equilibrium. Therefore, the concentration of NO becomes 0.175-2x at equilibrium. The concentration of both [tex]{{\text{N}}_{\text{2}}}[/tex] and [tex]{{\text{O}}_{\text{2}}}[/tex] become x at equilibrium.

The expression of [tex]{K_{\text{c}}}[/tex] for the above reaction is as follows:

[tex]{K_{\text{c}}}=\dfrac{{\left[ {{{\text{N}}_2}} \right]\left[{{{\text{O}}_2}} \right]}}{{{{\left[ {{\text{NO}}} \right]}^2}}}[/tex]     …… (1)                                                                          

Substitute x for [tex]{{\text{N}}_{\text{2}}}[/tex], x for [tex]{{\text{O}}_{\text{2}}}[/tex], 0.175-2x for [NO] and [tex]2.4 \times {10^3}[/tex] for [tex]{K_{\text{c}}}[/tex] in equation (1).

[tex]2.4 \times {10^3} = \dfrac{{\left( {\text{x}}\right)\left({\text{x}} \right)}}{{{{\left( {0.175 - 2x}\right)}^2}}}[/tex]                                                        …… (2)

Solve for x,

[tex]{\text{x}} = 0.0866[/tex]  

Or,

[tex]{\text{x}} = 0.0884[/tex]  

The value of [tex]{\text{x}} = 0.0884[/tex] is rejected as it makes concentration of NO negative that is not possible. So the value of x is 0.0866.

The concentration of NO at equilibrium is calculated as follows:

[tex]\begin{aligned}\left[ {{\text{NO}}} \right]&= 0.175 - 2\left( {0.0866} \right)\\&= {\text{ 0}}{\text{.0018 M}}\\\end{aligned}[/tex]  

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: chemical equilibrium, reactants, products, concentration, A, B, C, D, a, b, c, d, kc, equilibrium constant, 0.0018 M, NO, N2, O2.

The problem involves calculating equilibrium concentrations using the law of mass action and the given equilibrium constant. By setting up the equilibrium constant equation and solving the resulting quadratic equation, we can determine the required equilibrium concentration.

This Chemistry problem involves calculating equilibrium concentration using the Law of Mass Action and the given equilibrium constant (Kc). Let's denote the change in concentration of NO as '-2x' (since two moles of NO are consumed), and the change in concentration of N2 and O2 as '+x' (since one mole of each is produced).

At equilibrium, the concentration of NO would be 0.175 - 2x, N2 would be x and O2 would be x. The equilibrium constant expression (Kc) for this reaction would be Kc = [N2][O2] / [NO]². Substituting the equilibrium constant value (Kc=2.4×10³), and the equilibrium concentrations into the Kc expression, we get 2.4 x 10³ = x² / (0.175 - 2x)².

Solving this quadratic equation, we can find the value of 'x' and thus, the equilibrium concentration of NO (which would be 0.175 - 2x). Note that if 'x' is small compared to 0.175, it can be neglected to simplify the calculation. This assumption should then be verified. Law of Mass Action, equilibrium constant, and equilibrium concentration are key concepts involved in the solution.

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Answer: Option (B) is the correct answer.

Explanation:

Reaction between the given reactants will be as follows.

   [tex]Fe(C_{2}H_{3}O_{2})_{2} + KI \rightarrow K(C_{2}H_{3}O_{2})_{2} + FeI_{2}[/tex]

Therefore, we can see that the above reaction results in the formation of potassium acetate and iron(II) iodide.

Out of which iron(II) iodide [tex](FeI_{2})[/tex] is the precipitate.

When solutions of iron(II) acetate and potassium iodide are mixed, the precipitate formed is FeI2, Iron (II) Iodide(B). This happens because of a double displacement reaction, where the iodide ions from potassium iodide and iron(II) ions from iron(II) acetate combine to form a precipitate.

When solutions of iron(II) acetate, which is Fe(C2H3O2)2, and potassium iodide (KI) are mixed, the resulting compound is FeI2 or known as Iron (II) Iodide. This is due to a chemical reaction known as a double displacement reaction where the iodide ions (I-) from the potassium iodide (KI) solution and the iron(II) ions (Fe2+) from the iron(II) acetate solution will combine and precipitate out of the solution. The resulting compound, FeI2, is insoluble in water and hence forms the precipitate. Therefore, the correct answer to the question is option B: FeI2.

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The cladogram of the kingdom Animalia shows the relationship of selected animals based on their shared anatomical features. The black numbered squares represent features that are critical for forming each branch in the animal kingdom. Square one represents a dorsal nerve cord, which means the animals in all the branches have a dorsal nerve cord.

Which numbered square would represent the trait "mammary glands

Answer - C. 5

The earth's surface will split due to the rising currents of magma. When two oceanic plates drift apart a ridge is formed. When two continental plates drift apart a ridge push is formed.

The Magma rises, spreads outwards,cools and forms new rock and the older rocks are pushed away.

b. dispersion

Definition of Dispersion by Mimiwhatsup: a mixture in which fine particles of one kind of  substance is scattered throughout another substance.

The weakest force of molecular attraction is dispersion.

(Option B)

The dispersion force is also known as London dispersion forces is a weak intermolecular force.

These dispersion forces arise due to temporary fluctuations in electron distribution within molecules, creating temporary dipoles.

These temporary dipoles induce similar dipoles in neighboring molecules, resulting in attractive forces between them.

Dispersion forces are present in all molecules, regardless of their polarity or the presence of other types of bonds or interactions.

However, they tend to be weaker compared to other intermolecular forces, such as dipole-dipole interactions and hydrogen bonds.

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The answer is D. 3.15 × 10−4

Now the dissolution of methylamine can be represented as follows:

CH3NH2 + H2O → CH3NH3OH  

The product formed dissociates and it is represented as given below

CH3NH3OH ↔ CH3CH3+ + OH-  

Now pH = 11.4 ,

Then pOH = 2.6  

[OH-] = 10^-2.6  

[OH-] = 2.51*10^-3  

Considering the below equation again

CH3NH3OH ↔ CH3CH3+ + OH-  

We can calculate Kb = [CH3NH3] [ OH-] / [CH3NH3OH]  

where Kb is the equilibrium constant.

Now [CH3NH3] = [OH-] = 2.51 × 10^-3

and [CH3NH3OH] = 0.020M (given)

Substituting these values we get  

Kb = ( 2.51 × 10^-3)² / 0.02  

Kb = 6.31 × 10^-6 / 0.02  

Kb = 3.15 × 10^-4

[tex]390 \; \text{K}[/tex]

Explanation

The rate of a chemical reaction is directly related to its rate constant [tex]k[/tex] if the concentration of all reactants in its rate-determining step is held constant. The rate "constant" is dependent on both the temperature and the activation energy of this particular reaction, as seen in the Arrhenius equation:

[tex]k = A \cdot e^{-E_A/( R\cdot T)[/tex]

where

Taking natural logarithms of both sides of the expression yields:

[tex]\ln k = \ln A - {E_a}/ ({R \cdot T})[/tex]

[tex]k_2 = 4.50 \; k_1[/tex], such that

[tex]\ln k_2 = \ln 4.5 + \ln k_1[/tex]

[tex]\ln A- {E_a}/ ({R \cdot T_2}) = \ln k_2 \\\phantom{\ln A-{E_a}/ ({R \cdot T_2})} = \ln 4.5 + \ln k_1\\ \phantom{\ln A- {E_a}/ ({R \cdot T_2})} = \ln 4.5 +\ln A- {E_a}/ ({R \cdot T_1})[/tex]

Rearranging gives

[tex]-{E_a}/ ({R \cdot T_2}) = \ln 4.5- {E_a}/ ({R \cdot T_1})[/tex]

Given the initial temperature [tex]T_1 = 355 \; \text{K}[/tex] and activation energy [tex]E_A = 49.40 \; \text{kJ} \cdot \text{mol}^{-1}[/tex]- assumed to be independent of temperature variations,

[tex]- {49.40 \; \text{kJ} \cdot \text{mol}^{-1}}/ ({8.314 \times 10^{-3} \; \text{kJ} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \cdot T_2}) \\= \ln 4.5- {49.40 \; \text{kJ} \cdot \text{mol}^{-1}}/ ({355 \; \text{K}\cdot 8.314 \times 10^{-3} \; \text{kJ} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}})[/tex]

Solve for [tex]T_2[/tex]:

[tex]-(8.314 \times 10^{-3}/ 49.40) \; T_2 = 1/ (\ln 4.5 - 49.40 / (355 \times 8.314 \times 10^{-3})[/tex]

[tex]T_2 = 390 \; \text{K}[/tex]

Given:

Ea(Activationenergy):49.4kJ/molecule

k1: the rate constant of the first reaction

k2 : rate constant of the second reaction.

T2: Temperature of the second reaction.

T1: Temperature of the first reaction.

k2/k1=4.55

Now by Arrhenius equation we get

log(k2/k1)=[Ea/(2.303xR)] x[(1/T1)-(1/T2)]

Where k1 is the rate constant of the first reaction.

k2 is the rate constant of the second equation.

T2 is the temperature of the second reaction measured in K

T1 is the temperature of the first reaction measured in K

Ea is the activation energy kJ/mol

R is the gas constant measured in J/mol.K

Now substituting the given values in the Arrhenius equation we get:

log(k2/k1)=[Ea/(2.303xR)] x[(1/T1)-(1/T2)]

log(4.55)=[Ea/(2.303xR)] x[(1/T1)-(1/T2)]

0.66=[49.4/(2.303x8.314x10^-3)]x[(1/355)-(1/T2)]

0.66= 2579.75x [(1/355)-(1/T2)]

0.000256= (T2-355)/355T2

0.0908T2-T2= -355

0.9092T2=355

T2=390.46K

If the density of an object is affected by its mass, then osmium will have the highest density and hydrogen will have the lowest density

Density = mass/volume or D = m/V

If we hold V constant, we can write D = km. where k = 1/V.

Thus, D ∝ m. As the mass of a fixed volume of a substance increases, the density increases.

The mass of H₂ in 1 cm³ 0.080 g. The mass of Os in 1 cm³ is 22.6 g.

However, there are substances with even higher densities. For example, the material at the centre of a neutron star is so dense that 1 cm³ of the material has a mass of 10¹² kg!

The order of the electronegativity for the given elements is:

Fluorine > Phosphorous > Copper > Aluminium > Barium > Francium

Electronegativity measures the tendency of an element to attract electrons of bonded pairs towards itself. The value of electronegativity of an atom is with respect to another bonded atom.

When two atoms are bonded with each other through a bond and one of them is more electronegative than another bonded atom. Then the electron density of the bond will lie closer to the more electronegative atom.

Firstly, Fluorine is the most electronegative element in the modern periodic table. Aluminium and phosphorous are both present in the 2nd period. As we know, the electronegativity increases as we move from left to right in the periodic table.

Therefore phosphorous is more electronegative than aluminium. Copper is more electronegative than aluminium as the value of the electronegativity for copper is 1.92 and for aluminium is 1.6.

Barium and Francium lie at the bottom of the periodic table, Therefore they are the least electronegative. But the electronegativity of the Braium is more than that of Francium.

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