Meteorology is best defined as the study of

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

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 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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Now ,

C + O2  → CO2

According to above equation, 1 mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.Thus this implies that 12 g of carbon reacts with 32 g of O2 to produce 44 g of CO2.

No of moles = mass of the substance/molecular mass of the substance.

In this case 1.2 g of carbon reacts with "x "g of O2 to produce 4.4 g of CO2.

No of moles of carbon in this case = 1.2÷ 12 = 0.1 moles.

No of moles of carbon dioxide formed = 4.4÷44 =0.1 moles

Thus already discussed above, 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide. Hence to produce 0.1 mole of CO2 ,0.1 mole of carbon needs to react with 0.1 mole of oxygen.

Also number of moles of O2 = mass of O2÷ molar mass of O2

Substituting number of moles of O2 as 0.1 we get

mass of O2(x)  = Number of moles of O2 × Molar mass of O2

Mass of O2 (x) = 0.1 × 32= 3.2 g

Thus mass of 3.2 g O2 reacts with 1.2 g of CO2 to produce 4.4 g of CO2.

In the reaction between carbon and oxygen to form carbon dioxide, the amount of oxygen reacted can be calculated by subtracting the mass of carbon used from the mass of carbon dioxide formed. In this case, approximately 3.2 g of oxygen reacted with 1.2 g of carbon to produce 4.4 g of carbon dioxide.

In the reaction between carbon and oxygen to form carbon dioxide, the amount of oxygen reacting can be calculated using stoichiometry. The balanced chemical equation for this reaction is C + O2 -> CO2. According to this equation, each mole of carbon reacts with one mole of oxygen to produce one mole of carbon dioxide.

Knowing that the molar mass of carbon is about 12.0 g/mol and carbon dioxide is approximately 44.0 g/mol, we can figure out the amount of oxygen that reacted. If the experiment has resulted in 4.4 g of carbon dioxide from 1.2 g of carbon, the amount of oxygen that participated may be calculated as follows: the difference between the mass of carbon dioxide formed (4.4 g) and the mass of carbon used (1.2 g) gives the mass of oxygen. Therefore, 4.4 g - 1.2 g equals 3.2 g of oxygen.

This shows that in the given reaction, approximately 3.2 g of oxygen reacted with 1.2 g of carbon to produce 4.4 g of carbon dioxide.

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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.  

[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

milligrams. kilograms are too big.

Answer:

B) Milligrams

Explanation:

just did the test on usatestprep

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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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.

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.

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!

Answer;

Compounds of carbon and hydrogen

Explanation;

An organic compound is any of a large class of chemical compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen.

The primary difference between organic compounds and inorganic compounds is that organic compounds always contain carbon while most inorganic compounds do not contain carbon. Additionally, nearly all organic compounds contain carbon-hydrogen or C-H bonds.

Organic compounds includes nucleic acids, fats, sugars, proteins, enzymes and hydrocarbon fuels. All organic molecules contain carbon, nearly all contain hydrogen, and many also contain oxygen.

Answer:

Compounds of carbon and hydrogen

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.

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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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.

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.

[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

Answer:

Monosaccharides!

Explanation:

monosaccharies are the building blocks of all carbohydrates! Starch and glycogen are carbs that consist of multiple monos.

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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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

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:

48 grams of CO₂ contains approximately 6.56 x 10²³ carbon atoms, whereas 12 grams of pure carbon contains exactly 6.022 x 10²³ carbon atoms. Hence, 48 grams of CO₂ has more carbon atoms than 12 grams of pure carbon.

Comparing Carbon Content in CO₂ and Pure Carbon

To determine which sample contains more carbon atoms, we first need to consider their molar masses and the mass of carbon in each. The atomic mass of carbon is approximately 12 u, and one mole of a substance contains Avogadro's number of particles, which is roughly 6.022 x 10²³ particles/mole.

For CO₂ (molar mass approximately 44 g/mol), 48 grams corresponds to approximately 48 g / 44 g/mol = 1.09 moles of CO₂. Since each mole of CO₂ contains one mole of carbon atoms, there would be 1.09 moles of carbon atoms in 48 grams of CO₂.

For pure Carbon, 12 grams corresponds directly to 12 g / 12 g/mol = 1 mole of carbon since its molar mass is 12 g/mol. This equals 1 mole, or 6.022 x 10²³ carbon atoms.

Given that 1 mole of a substance contains 6.022 x 10²³ atoms:

48 grams of CO₂ has 1.09 moles x 6.022 x 10²³ atoms/mole = 6.56 x 10²³ carbon atoms.

12 grams of pure carbon has 1 mole x 6.022 x 10²³ atoms/mole = 6.022 x 10²³ carbon atoms.

Therefore, 48 grams of CO₂ has more carbon atoms than 12 grams of pure carbon.

correct answer is B) equal to the number of protons

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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D the belguim people atacked using it

Hey there!:

1 mole of  C6H12O6 ------------------ 6 moles of C

3.0 moles of C6H12O6 ------------- ??

3.0 *6 / 1 =>

18.0 moles of C

Hope that helps!

Answer:

18.0

Explanation:

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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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!

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.

Given:

Density:8.92g/cm3

Volume:25.8 mL

Now we know that

Mass = density X volume

Substituting the given values in the above equation we get:

Mass = 8.92 x 25.8= 230.136 g

= 230136 mg

Final answer:

To find the mass of a copper sample in milligrams, calculate the mass using density and volume (mass = density × volume), resulting in 230.016 g, and convert to milligrams to get 230016 mg.

Explanation:

The question asks to find the mass of a sample of copper in milligrams, given its density is 8.92 g/cm3 and it occupies a volume of 25.8 mL. First, we need to recall that density is defined as mass over volume (d = m/v), which means mass can be calculated as m = d × v. Given the density of copper (8.92 g/cm3) and the volume of the sample (25.8 mL, which is equal to 25.8 cm3 because 1 mL = 1 cm3), we can calculate the mass in grams and then convert it to milligrams.

To calculate the mass in grams: mass = 8.92 g/cm3 × 25.8 cm3 = 230.016 g. To convert this into milligrams (remembering that 1 g = 1000 mg), we multiply by 1000, resulting in 230016 mg.

Therefore, the mass of the copper sample is 230016 milligrams.