What is the pH of 0.0050 M NH3 (Kb = 1.8 x 10-5)
11.70
2.30
3.52
10.48

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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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.

The answer is 125 gms.

Answer:

first half of 1000gram=1000/2=500g

second half of 1000gram=500/2=250g

third half of 1000gram=250/2=125 g

Explanation:

in simple

third half of 1000 g=1000/(2³)=125gram

Control or controlled group is the part of the experiment where conditions are kept the same.

milligrams. kilograms are too big.

Answer:

B) Milligrams

Explanation:

just did the test on usatestprep

[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

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

[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

Carolus Linnaeus is responsible.

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 is  A - HCl is the titrant and D-NaOH is the analyte.

In a titration process, the solution of unknown concentration considered is the analyte. During titration, a standard solution(titrant) is added to an analyte until the equivalence point is achieved.

So in this case as the concentration of NaOH is not known,NaOH is the analyte.

Titrant is the solution whose concentration  is known to us and is added to an analyte until the equivalence point is reached. In this case since we know the concentration of HCl, HCl is the titrant.

We have that for the Question "Calculate the activation energy for the reaction" it can be said that The  activation energy is

From the question we are told

Generally the equation for the Activation energy  is mathematically given as

[tex]10(\frac{kr}{ki})=\frac{Ea}{2.303R}(\frac{1}{t_1}-\frac{1}{t_2}})\\\\Therefore\\\\10(\frac{0.0815}{0.0796})=\frac{Ea}{2.303*8.314}(\frac{1}{1010}-\frac{1}{1220}})\\\\Ea=\frac{0.01024}{8.9*10^{-6}}/mol\\\\[/tex]

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The activation energy for the reaction is 96.8 kJ/mol.

To calculate the activation energy for the reaction, we can use the Arrhenius equation: k = Ae-Ea/RT, where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. We are given the rate constants at two different temperatures, so we can set up two equations:

k1 = A e-Ea/(R × T1)

k2 = A e-Ea/(R × T2)

Dividing the second equation by the first equation, we get:

(k2/k1) = (A e-Ea/(R ×T2))÷(A e-Ea/(R ×T1))

Simplifying, we get:

eEa/R = (k2/k1) ×(T2/T1)

Taking the natural logarithm of both sides, we can solve for the activation energy:

Ea = -R × ln((k2/k1) × (T2/T1))

Using the given values, we can plug in and calculate the activation energy:

Ea = -8.314 J/mol ×K × ln((0.0815 m-1 × s-1)/(0.0796 m-1 × s-1) × (947 + 273)/(737 + 273)) = 96.8 kJ/mol

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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

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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Answer: Inertia is based only on an object’s mass.

Explanation:

Any physical object's resistance towards any change in its velocity is known as inertia.

Therefore, more is the mass of an object, the more inertia will be there.

Thus, we can conclude that inertia is based only on an object’s mass and not on velocity.

The correct answer is option A which is mass.

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:

B. because the little number below the H stands for how many atoms there are, and b has 1 more Hydrogen atom than A does.

(I think)

ans is B) 2 moles of CH₄

because it has 4 Hydrogen atoms per molecule, 1 more than NH3 has.

The weather would be around the same maybe dropping 5 degrees.

Barometric pressure is the measurement of air pressure in the atmosphere.

A barometer that has a high reading — meaning high pressure — and is stable, indicates good weather.

If one of the station models were to show that the barometric pressure was steadily dropping and the current weather condition showed a slight drizzle with the temperature of 65 degrees then would be around the same maybe dropping 5 degrees.

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Given:

E = 7.3 × 10–17 Hz                                                                                      

 h= 6.63 × 10–34 J•s

Now E = hf

where E is the energy of the photon                                                          

h is the Planck's constant                                                                          

f is the frequency of the photon

Substituting the values in the equation we get                                        

E= 7.3 × 10^-17 × 6.63 × 10^-34                                                                  

E= 4.8399 × 10^-50  J.                                                                                                      

The energy of the photon, to the nearest tenths place, is 4.8399 × 10⁻⁵⁰ J.

Energy of the photon can be calculated as :

E = hυ, where

E = energy of the photon = to find?

h = plank's constant = 6.63 × 10⁻³⁴ J•s (given)

and υ = frequency = 7.3 × 10⁻¹⁷ Hz

Now putting all these values in the above equation, we get

E = (6.63 × 10⁻³⁴) × (7.3 × 10⁻¹⁷)

E = 4.8399 × 10⁻⁵⁰ J.

Hence, 4.8399 × 10⁻⁵⁰ J is the energy of the photon.

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It would be D. Enzymes breaking down the food. Enzymes are used to speed up chemical reactions

Answer: Option (D) is the correct answer.

Explanation:

A change that does not bring any difference in chemical composition of a substance are known as physical change.

For example, shape, size, mass, volume, density, etc of a substance are all physical properties.

Whereas a change that brings difference in chemical composition of a substance is known as chemical change.

For example, precipitation, reactivity, toxicity etc are chemical property.

During break down of food, salivary amylase breaks down starch into simple sugars. In stomach the enzyme pepsin converts protein into peptones in presence of acidic medium. Small intestine receives both intestinal and pancreatic juices (chemical substances) and the final digestion of fats, proteins and sugars occurs here.

Thus, we can conclude that example of a chemical change during digestion is enzymes breaking down food.

Answer:

The application of X-rays in the hospital.

Explanation:

An X-ray is a painless and brisk method generally used in order to generate inside images of the body. The X-rays are generally carried out in the X-ray departments in the hospital under the guidance of trained specialists known as radiographers, however, it can also be performed by other healthcare professionals.

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.

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.

Aluminium most often has an oxidation state of +3 because it has three valence electrons that are removed relatively easily.

Al is in Group 13, so it has three valence electrons.

To get a complete octet, it must either lose three electrons or gain five.

It is easier to remove three electrons, so Al most often has an oxidation state in its compounds.

Final answer:

Aluminum typically has a +3 oxidation state because it loses all three valence electrons in its outer shell during oxidation, forming an Al³+ ion. This oxidation state is maintained in both ionic and covalent aluminum compounds to achieve stability and comply with electrical neutrality in compounds like Al₂O₃.

Explanation:

Aluminum commonly exhibits an oxidation state of +3 because it has three valence electrons in its outer shell (ns²np¹ configuration). When aluminum atom is oxidized, it tends to lose all three of these electrons, resulting in an Al³+ ion with a +3 oxidation state. Transition metals, in comparison, have multiple oxidation states because they can lose electrons from both their s and d orbitals. This is not as straightforward for aluminum, which has only p orbital electrons to lose.

While in many compounds of aluminum, such as AlF3 and Al₂(SO₄)₃, the oxidation state is represented as ionic, some aluminum-containing compounds are covalent. However, in both cases, aluminum adopts a +3 oxidation state to achieve stability. In aqueous solutions, aluminum salts dissociate to release [Al(H₂O)₆]³+ cations, demonstrating aluminum's charge even when coordinated. Furthermore, during redox reactions, aluminum is oxidized by losing three electrons to give a +3 charge, which complies with the electrical neutrality essential in ionic compounds like Al₂O₃.

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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1 mole has mass = 56 g  

0.28 mol has mass = 56 x 0.28 = 15.68 g

Hope this helps!

Final answer:

To calculate the mass of 0.28 moles of iron, multiply the number of moles by the molar mass of iron, which is 55.85 g/mol, resulting in a mass of 15.638 grams.

Explanation:

To find the mass of 0.28 moles of iron, we need to use the molar mass of iron. The molar mass of iron (Fe) can be found on the periodic table and it is approximately 55.85 grams per mole.

Step 1: Molar Mass of Iron

First, we establish the molar mass of iron is 55.85 g/mol.

Step 2: Calculate the Mass

Next, we multiply the number of moles by the molar mass of iron to calculate the mass.

Mass = moles × molar mass

Mass = 0.28 moles × 55.85 g/mol

Mass = 15.638 grams

Hi!

Why is second ionization energy larger than first ionization energy?

The reason second ionization energy is larger is because it takes more energy to remove an electron from a poistively charged ion than it does from an neutral atom.

Ionization energy is the energy required to remove an electron from a neutral atom. When an electron is removed from an atom, the neutral atom becomes positively charged. When another electron is attempted to be removed from the same atom,this requires more energy than the first one, as the attractive force between the electrons and protons are greater owing to the presence of the extra proton. In order to break this force we need more energy to remove the second electron. Hence the second ionization energy is greater than the first ionization energy.

Answer:

Monosaccharides!

Explanation:

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

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.

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: