Draw the Lewis Structure for NaCl

Answer:

1. Percentage by weight = 0.5023 = 50.23 %

2. molar fraction =0.153

Explanation:

We know that

Molar mass of HClO4 = 100.46 g/mol

So the mass of 5 Moles= 5 x 100.46

       Mass (m)= 5 x 100.46 = 502.3 g

Lets assume that aqueous solution of HClO4  and the density of solution is equal to density of water.

Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.

We know that

Mass = density x volume

Mass of 1000 ml  solution = 1 x 1000 =1000     ( density = 1 gm/ml)

            m'=1000 g

1.

Percentage by weight = 502.3 /1000

Percentage by weight = 0.5023 = 50.23 %

2.

We know that

molar mass of water = 18 g/mol

mass of water in 1000 ml = 1000 - 502.3 g=497.9 g

So moles of water = 497.7 /18 mole

moles of water = 27.65 moles

So molar fraction = 5/(5+27.65)

molar fraction =0.153

Answer:

b- The heat capacity ratio increases but output temperature don’t change

Explanation:

The heat capacity is the amount of energy required to raise the temperature of a body, by 1 degree. On the other hand, the specific heat capacity is the amount of heat required to raise the temperature of a of unit mass of a material by 1 degree.

Heat capacity is an extensive property meaning its value depends on the amount of material. Specific heat capacity is found by dividing heat capacity by the mass of the sample, thus making it independent of the amount (intensive property). So if the specific heat capacity increases and the mass of the sample remains the same, the heat capacity must increase too. Because of that options c and d that say that heat capacity reamins same are INCORRECT.

On the other hand, in which has to be with options a and b both say that the heat capacity increases which is correct, but about the output temperatures what happens is that if we increase the specific heat capacity of both fluids that are involved in a process of heat exchange in the same value, the value of the output temperatures do not change so only option a is CORRECT.

To round 129.752416 to the requested number of significant figures:
- For 3 significant figures, round down to 129.7.
- For 4 significant figures, round up to 129.8.
- For 6 and 7 significant figures, keep the given digits.

To round 129.752416 to the requested number of significant figures:

Answer:

Because for dilute and aqueous solutions the mass of solvent will be a very close value to the volume of solution.

Explanation:

Molar concentration is defined as:

[tex][M]=\frac{molessolute}{volumesolution}[/tex]

And molal concentration is defined as:

[tex][m]=\frac{molessolute}{kgsolvent}[/tex]

And:

Msolution=Msolute+Msolvent

For dilute solutions, we have small amounts of solute, then we have:

Msolution=Msolute+Msolvent, and as the mass of  solute is very small: Msolution≅Msolvent

If the solution is also aqueous (water as solvent), and considering that the density of water is around 1 gm/cm3 or 1 kg/m3:

Msolvent≅Msolution≅Vsolution

Therefore, if we look to the molar and molal equations, we have the same numerator in both (moles of solute) and nearby numbers for the denominator, giving to the molar and molal concentration close values.  

Answer:

Part a:

Isobars: Ca-41, K-41 and Ar-41

Part b:

Number of proton

Part c:

Incorrect statement

Explanation:

Part a:

Nuclei with same mass number are called isobars.

Given:

Ca-40, Ca-41, K-41 and Ar-41

Ca-41, K-41 and Ar-41 have equal mass numbers, so these are isobars.

Part b:

Atomic number of Ca = 20

Atomic no. = Number of proton.

So, Ca-40 ans C-41 have same number of proton or in other words proton count is common in both.

Part c:

Number of neutrons in Ca-41

Atomic number = 20

Mass number = 41

Number of neutron = Mass number -atomic number

                                = 41 -20 = 21

Number of neutrons in K-41

Atomic number = 19

Mass number = 41

Number of neutron = Mass number -atomic number

                                = 41 -19 = 22

Number of neutrons in Ar-41

Atomic number = 18

Mass number = 41

Number of neutron = Mass number -atomic number

                                = 41 -18 = 23

So, the statement is incorrect.

The isobars among the examples given are Ca-41, K-41, and Ar-41. Ca-40 and Ca-41 are isotopes of calcium with different neutron counts. The statement about Ca-41, K-41, and Ar-41 having the same number of neutrons is incorrect because they have different atomic numbers.

Nuclei with the same mass number but different atomic numbers are known as isobars. This means that while they have the same mass number (sum of protons and neutrons), they belong to different elements (different atomic numbers). Consequently, Ca-41, K-41, and Ar-41 are isobars since they all have a mass number of 41 but belong to different elements with atomic numbers 20 (calcium), 19 (potassium), and 18 (argon), respectively. Nevertheless, Ca-40 is not an isobar to these because it has a different mass number.

Isotopes are atoms with the same atomic number but differing numbers of neutrons, resulting in different mass numbers. In this case, Ca-40 and Ca-41 are isotopes because they are both calcium atoms (same atomic number of 20) with different mass numbers.

The commonality between Ca-40 and Ca-41 is they are isotopes of calcium and thus share the same atomic number and chemical properties, despite having different mass numbers due to different neutron counts.

The statement 'Atoms of Ca-41, K-41, and Ar-41 have the same number of neutrons' is incorrect because while these isotopes have the same mass number, they have different numbers of protons. To calculate the number of neutrons in these isotopes, you subtract the atomic number from the mass number and because the atomic numbers are different, the resulting neutron numbers will also differ.

Answer:

W = 1.8 KJ

Explanation:

turbine adiabatically: Q = 0

∴ Lost work:

∴ To = 300 K

ideal gas:

⇒ S2 - S1 = (9/2)*(8.314) Ln (383.5/494.8) - (8.314) Ln (1.7/6.3)

⇒ S2 - S1 = 1.36 J/mol.K

entropy generated in the sourroundings:

∴ Q = ΔU = Cv*ΔT

∴ Cv = 3/2*R = 12.5 J/mol.K.....ideal gas

∴ ΔT = 494.8 - 383.5 = 111.3 K

⇒ Q = 12.5 * 111.3 = 1391.25 J/mol

⇒ ΔSo = 1391.25/300 = 4.638 J/mol.K

⇒ W = ( 300 K ) * [ 1.36 J/mol.K + 4.638 J/mol.K ]

⇒ W = 1799.25 J/mol * 1 mol * ( KJ/1000J )

⇒ W = 1.799 KJ ≅ 1.8 KJ

Answer:

Mass of sea food = 30.98 Kg

Mass of sea food in pound = 68.31 lbs

Explanation:

Salmon, crab and oysters all are sea food.

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

Mass of salmon = 22 kg

Mass of crab = 5.5 kg

Mass of oysters = 3.48 kg

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

                             = 22 + 5.5 + 3.48

                             = 30.98 Kg

1 Kg = 2.205 lbs

Therefore, 30.98 kg = 30.98 × 2.205

                                 = 68.31 lbs

Answer:

E) More neutrons in the nucleus

Explanation:

Isotope -

The atoms which have same number of protons but different number of neutrons.

And since , atomic number = protons number

and , mass number = proton + neutrons .

Hence ,

The atomic number will not change , but the mass number will change .

Hence , the correct option is more neutrons in the nucleus .

To convert cm³ or mm³ to m³, one must divide by 1,000,000 or 1,000,000,000 respectively. Thus, 2.7 cm³ is equal to 2.7 x 10^-6 m³ and 2.0 mm³ is equal to 2.0 x 10^-9 m³.

To convert cubic centimeters (cm³) and cubic millimeters (mm³) to cubic meters (m³), you need to know the unit conversions. One square meter is equal to 1,000,000 cubic centimeters and 1,000,000,000 cubic millimeters.

This means you can convert 2.7 cm³ to meters by dividing by 1,000,000, yielding an answer of 0.0000027 m³ (to two significant figures, or 2.7 x 10^-6 m³).

Similarly, 2.0 mm³ can be converted to meters by dividing by 1,000,000,000, resulting in an answer of 0.000000002 m³ or 2.0 x 10^-9 m³.

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The conversion from cm3 and mm3 to m3 is done by multiplying the original number by 1e-6 for cm3 and 1e-9 for mm3. The results for the examples given are 2.7 cm3 equals 2.7e-6 m3, and 2.0 mm3 equals 2.0e-9 m3.

The task is to convert measurements from one unit (cubic centimeters or cubic millimeters) to another (cubic meters). It's important to understand the relevant conversion factors:

1 cm3 = 1e-6 m3

1 mm3 = 1e-9 m3

Applying these to your examples we get:

2.7 cm3 = 2.7 * 1e-6 m3 = 2.7e-6 m3

2.0 mm3 = 2.0 * 1e-9 m3 = 2.0e-9 m3

So these are the conversions using two significant figures.

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Answer:

Fischer projection is the method used for representing a three-dimensional organic molecule as a two dimensional molecule.  

This method can be used for determining the D- and L- configuration of the organic molecules.

In the Fisher projection of carbohydrate molecule, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the right-hand side, then the carbohydrate is said to have D-configuration.

Whereas, if the hydroxyl group attached to the last stereocenter of the molecule is placed on the left-hand side, then the carbohydrate is said to have L-configuration.

Final answer:

To determine if a carbohydrate is D or L in a Fischer projection, look at the hydroxyl (-OH) group on the penultimate carbon; if it's to the right, it's a D-sugar, and if it's to the left, it's an L-sugar. This classification does not directly relate to the sugar's optical activity but its stereochemistry relative to glyceraldehyde.

Explanation:

How to Determine if a Carbohydrate is D or L Using a Fischer Projection

When examining a Fischer projection of a monosaccharide, you can determine whether it is a D-sugar or an L-sugar by looking at the orientation of the hydroxyl (-OH) group on the penultimate carbon (second-last carbon) in the chain. The rule is straightforward: if the -OH group on this carbon is to the right side of the Fischer projection, the sugar is designated as a D-sugar. Conversely, if the -OH group is to the left side, the sugar is an L-sugar.

This method of classification is based on the relative configuration to glyceraldehyde, where D-glyceraldehyde has the -OH on the right at the chiral center farthest from the carbonyl group, hence all D-sugars follow this pattern. L-sugars are the mirror images (enantiomers) of the D-sugars, with their -OH groups flipped to the opposite side.

It's important to note that the D/L configuration does not directly correlate with the optical activity of the sugar (++/--) but rather describes its stereochemistry relative to glyceraldehyde. The D/L nomenclature is fundamental in distinguishing the stereochemistry of sugars and their derivatives.

Answer:

9 L

Explanation:

According to the question , the given reaction is -

2NO(g) + O₂(g)------->2NO₂(g)

Since ,

At STP ,

One mole of a gas occupies the volume of 22.4 L.

Hence , as given in the question -

9 L of NO , i.e .

22.4 L = 1 mol

1 L = 1 / 22.4 mol

9 L = 1 / 22.4  * 9 L = 0.40 mol

From the chemical reaction ,

The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .

Hence ,  

2 moles of NO will produce 2 moles of NO₂.

Therefore ,

0.40 mol of NO will produce 0.40 mol of NO₂.

Hence , the volume of NO₂ can be calculated as -

1 mol = 22.4 L

0.40 mol = 0.40 * 22.4 L = 9 L

To find the volume of NO2 produced, calculate the number of moles of NO reacted using the ideal gas equation. Convert moles to volume using the molar volume of gases at STP.

To find the volume of NO2 produced, we first need to determine the number of moles of NO reacted. Using the ideal gas equation, we can calculate the number of moles:

n = PV/RT = (1 atm)(9 L)/(0.0821 L atm/(mol K))(273 K) = 0.408 mol

Since the reaction has a 2:2 mole ratio, 0.408 mol of NO will produce 0.408 mol of NO2. To convert moles to volume at STP, we can use the molar volume of gases, which is 22.4 L/mol:

Volume of NO2 = 0.408 mol * 22.4 L/mol = 9.1392 L

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Answer: [tex]41.4\times 10^{6}[/tex]

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s preceding the first integers are never significant.

All zero’s after the decimal point are always significant.

Engineering notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form times 10 raise to the power.  It is similar to the scientific notation but in engineering notation, the powers of ten are always multiples of 3.

The engineering notation written in the form:

[tex]a\times 10^b[/tex]

where,

a = the number which is greater than 0 and less than 999

b = an integer multiple of 3

Now converting the given value of 41,405,000 into engineering notation, we get [tex]41.4\times 10^{6}[/tex]

Therefore the scientific notation with 3 significant figures is [tex]41.4\times 10^{6}[/tex]

Answer:

You need 8,53 L of ammonia

Explanation:

Global reaction of remotion of nitrogen oxide with ammonia is:

4 NH₃ + 6 NO ⇒ 5 N₂ + 6 H₂O

This balanced equation shows that 4 NH₃ moles reacts with 6 NO moles.

With 100% yield and temperature and pressure constants it is possible to apply Avogadro's law. This law is an experimental gas law relating the volume of a gas to the amount of substance of gas present. The formula is:

[tex]\frac{V_{1} }{n_{1} } = \frac{V_{2} }{n_{2} }[/tex]

Where:

V₁ is the NO volume =  12,8L

n₁ are NO moles = 6

n₂ are NH₃ moles = 4

V₂ is NH₃ volume, the unknown.

Thus, V₂ are 8,53 L of ammonia

I hope it helps!

Answer:

N₂ is the limiting reactant

Explanation:

The balanced reaction between N₂ gas and H₂ gas is:

N₂ + 3H₂ → 2NH₃

In order to determine the limiting reactant, we have to calculate the number of moles of each rectant, using their molecular weight:

Lastly, we multiply the number of moles of N₂ by 3, and the number of moles of H₂ by 1; due to the coefficients in the balanced reaction. Whichever number is lower, belongs to the limiting reactant.

N₂ => 10.7

H₂ => 50.0

Thus N₂ is the limiting reactant

Infrared spectroscopy is a technique used to identify functional groups in molecules by measuring absorption frequencies. It helps determine the presence or absence of specific groups in a molecule.

Infrared spectroscopy is a technique used to identify different functional groups in molecules based on their characteristic absorption frequencies. It involves passing infrared light through a sample and measuring the frequencies that are absorbed. Different functional groups have different absorption frequencies, allowing scientists to determine the presence or absence of specific groups in a molecule. For example, if a sample absorbs in the carbonyl frequency range but not in the alkyne range, it can be inferred that the molecule contains a carbonyl group but not an alkyne.

Answer:

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

[tex]r = k [A]^{x} [B]^{y}[/tex]

where:

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

[tex]v(t) = -\frac{d[A]}{dt} = k [A]^{n}[/tex]

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

Rate Law:                                    rate = k

Concentration-time Equation:   [A]=[A]o - kt

where

For first-order kinetics, we have:

Rate Law:                                        rate= k[A]

Concentration -Time Equation:      ln[A]=ln[A]o - kt

where:

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use:                        [A]=[A]o - Kt

we replace the data:   5 = 100 - K (60)

we clear K:                 K = [100 - 5 ] (mg/L) /60 (s)  = 1, 583 [mg/L.s]

First-order kinetics

we use:                                  ln[A]=ln[A]o - Kt

we replace the data:               ln(5)  = ln(100) - K (60)

we clear K:                                   K = [ln(100) - ln(5)] /60 (s)  = 0,05 [1/s]

Answer : The  value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

[tex]2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)[/tex]

First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].

[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]

[tex]\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}][/tex]

where,

[tex]\Delta H^o[/tex] = enthalpy of reaction = ?

n = number of moles

[tex]\Delta H_f^0[/tex] = standard enthalpy of formation

Now put all the given values in this expression, we get:

[tex]\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)][/tex]

[tex]\Delta H^o=-1035.6kJ=-1035600J[/tex]

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction [tex](\Delta S^o)[/tex].

[tex]\Delta S^o=S_f_{product}-S_f_{reactant}[/tex]

[tex]\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}][/tex]

where,

[tex]\Delta S^o[/tex] = entropy of reaction = ?

n = number of moles

[tex]\Delta S_f^0[/tex] = standard entropy of formation

Now put all the given values in this expression, we get:

[tex]\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)][/tex]

[tex]\Delta S^o=-153J/K[/tex]

Now we have to calculate the Gibbs free energy of reaction [tex](\Delta G^o)[/tex].

As we know that,

[tex]\Delta G^o=\Delta H^o-T\Delta S^o[/tex]

At room temperature, the temperature is 500 K.

[tex]\Delta G^o=(-1035600J)-(500K\times -153J/K)[/tex]

[tex]\Delta G^o=-959100J=-959.1kJ[/tex]

Therefore, the value of [tex]\Delta G^o[/tex] for the reaction is -959.1 kJ

Answer:

pH of the solution = 4.80

Explanation:

pKa = 8.3

Concentration of tris acid = 8.3

[tex]pH = \frac{1}{2} \times p_{ka} - \frac{1}{2} logC[/tex]

[tex]pH = \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(0.05)[/tex]

     [tex]= \frac{1}{2} \times 8.3 - \frac{1}{2}\times log(5\times 10^{-2})[/tex]

     [tex]= 4.15 - \frac{1}{2}\times (-1.30103)[/tex]

    [tex]= 4.15 - (-0.650515)[/tex]

   [tex]= 4.15 + 0.650515[/tex]

   = 4.80

pH of the solution = 4.80

Answer:

Explanation:

The metallic properties of steel would strongly not favor its use in construction of aircraft.

Steel is an alloy of carbon and iron. It is denser and generally weighs more than equivalent amount of aluminum. This would imply that the weight of aircrafts would be much more heavier than usual. Weight is a very key component in construction of aircraft. The lighter the mass, the faster propellers can move the craft and lesser amount of energy used. The weight would be a major problem.

Although alloying iron and carbon improves the resistant of the steel to corrosion, it still cannot be compared to that of aluminium. We would build airplanes that would not be durable for so long and that can readily rust on frequent usage.

If airplanes were made of steel, they would be heavier, requiring more fuel and more powerful engines, reducing efficiency and payload capacity. Aluminum's lightness, toughness, and resistance to corrosion are pivotal for modern airplane performance and cannot be matched by steel without compromising efficiency and increasing operating costs.

If airplanes had to be made of steel instead of aluminum, there would be significant consequences for the aviation industry. Steel is much heavier than aluminum, which means that airplanes would be considerably heavier as well. This increase in weight would lead to a need for more powerful engines to provide the necessary lift, which could result in higher fuel consumption and lower efficiency. Moreover, the added weight would decrease the payload capacity of the airplane, as the plane's own structure would constitute a larger portion of its maximum takeoff weight.

The major advantages of aluminum in airplane construction include its lightness, toughness, and resistance to corrosion. These properties are crucial for ensuring that airplanes are able to carry substantial payloads, achieve adequate fuel efficiency, and maintain structural integrity over time. The use of aluminum, a metal that meets these critical needs, is therefore one of the key reasons why modern airplanes are able to achieve the performance levels that they do today. Utilizing steel, despite its strength, would compromise these benefits and could alter the economics and environmental impact of air travel.

Based on conversion ratios, 47 mg of sodium would be:

1g is 1,000 grams.

47 mg would be:

= 0.47 / 1,000

= 0.047 grams

16 ounces is 453.6 grams so 0.047 grams would be:

= (0.047 x 16) / 453.6

= 0.00166 ounces.

1 pound is 453.6 grams so 0.047 grams would be:
= 0.047 / 453.6

= 0.000104 pounds

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To convert 47 mg of sodium to grams, divide by 1000. To convert grams to ounces, use the conversion factor 1 oz = 28.35 g. To convert grams to pounds, use the conversion factor 1 lb = 453.6 g.

To convert 47 mg to grams, divide by 1000:

47 mg = 47/1000 = 0.047 grams

To convert grams to ounces, use the conversion factor:

0.047 g x (1 oz/28.35 g) = 0.00166 oz

To convert grams to pounds, use the conversion factor:

0.047 g x (1 lb/453.6 g) = 0.0001035 lb

Answer:

OH-

Explanation:

H2O can act as base or acid.

When H2O acts as a base the conjugate acid is H3O+

When H2O acts as an acid the conjugate base is OH-

Answer:

The number 10,847,100 in Scientific Notation is [tex]1.0847x10^{7}[/tex]

Explanation:

Scientific notation is an easy form to write long numbers and it is commonly used in the scientific field. To write a long number in a shorter way it is necessary to 'move' the decimal point to the left the number of positions that are necessary until you get a unit. Then you write the number and multiplied it by 10 raised to the number of positions you moved the decimal point. In this case, it is necessary to move the decimal point 7 positions so, we multiply the number by 10 raised to 7.

Answer:

the "d" option

Explanation:

 Cómo se pronuncia

To calculate the flow I need the volume and time to comply with the formula:

S = V / t

V = volume

t = time

In this problem I have volume and density as data.

There is no way to calculate the time so the correct answer is the "d" option

Explanation:

A mixture is material which is composed up of two or more substances and these substances are physically combined. The identities of each specie in the mixture are retained.

A solution is a type of the homogeneous mixture which is composed of two or more substances. The specie which is dissolved into the another substance is known as solute and in which it is dissolved is known as a solvent.

A compound is the substance which is formed when two or more elements are bonded together chemically.

A molecules are electrically neutral and is a combination of of two or more atoms which are held together by means of chemical bonds.

Acids are the species which furnish hydrogen ions in the solution or is capable of forming bonds with electron pair species as they are electron deficient species.

Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species.

Answer:

Try to remember that 1 nanometer is 1 x 10^9, so calculate 6.23 x 10^-3 m  x 1 x 10^9. Your answer is 6.23 x 10^-3 m = 6.23 x 10^6 nm

Explanation:

Answer:

6.23 x 10^-3 m = 6.23 x 10^6 nm

Explanation:

Answer: The moles of carbon dioxide gas formed is 12.33 moles.

Explanation:

We are given:

Moles of oxygen gas = 12.33 moles

The chemical equation for the reaction of carbon and oxygen to produce carbon dioxide follows:

[tex]C+O_2\rightarrow CO_2[/tex]

By stoichiometry of the reaction;

1 mole of oxygen gas produces 1 mole of carbon dioxide gas.

So, 12.33 moles of oxygen gas will produce = [tex]\frac{1}{1}\times 12.33=12.33mol[/tex] of carbon dioxide gas.

Hence, the moles of carbon dioxide gas formed is 12.33 moles.

Final answer:

An interpretation of the results of many tests is called a theory. It is a well-substantiated explanation of an aspect of the natural world, based on extensive research and experimentation.

Explanation:

An interpretation of the results of many tests is called a theory. A theory is a well-substantiated explanation of an aspect of the natural world. It is a scientific explanation that has been repeatedly tested and supported by many experiments.

For example, the theory of evolution is a well-supported explanation of how species have evolved over time. It is based on extensive research and experimentation.

It is important to note that a theory in science is different from the everyday use of the word, which often refers to a guess or speculation. In science, a theory is a well-tested and supported explanation.

Answer:

[tex]\Delta H_{rxn}[/tex] for the given reaction is -890.3 kJ

Explanation:

[tex]\Delta H_{rxn}=\sum n_{i}.\Delta H_{f}(product)_{i}-\sum n_{j}.\Delta H_{f}(reactant)_{j}[/tex]

where [tex]n_{i}[/tex] and [tex]n_{j}[/tex] represents number of moles of i-th product and j-th reactant in balanced reaction respectively.

Hence [tex]\Delta H_{rxn}=[1mol\times \Delta H_{f}(CO_{2})_{g}]+[2mol\times \Delta H_{f}(H_{2}O)_{l}]-[1mol\times \Delta H_{f}(CH_{4})_{g}]-[2mol\times \Delta H_{f}(O_{2})_{g}][/tex]

so, [tex]\Delta H_{rxn}=[1mol\times -393.5kJ/mol]+[2mol\times -285.8kJ/mol]-[1mol\times -74.8kJ/mol]+[2mol\times 0kJ/mol]=-890.3 kJ[/tex]

So, Correct answer is -890.3 kJ

Answer:

Molarity of a solution that contains 3.11 mol of NaNO3 is 1,24 M

Explanation:

We understand molarity as the number of moles of solute that are contained in 1 L of solution, then if in a solution of 2.50 L we have 3.11 moles, it remains to calculate how many moles do we have in 1 liter.

2,50 L .......... 3,11 moles

1 L .................. x

X = ( 1 L x 3,11 moles) / 2,50 L = 1,24

Explanation:

Here it is given that carbon is sample 2 = 25.9 g

No. of moles of carbon will be calculated as follows.

         No. of moles of carbon = [tex]\frac{\text{mass carbon}}{\text{molar mass carbon}}[/tex]

                                                 = [tex]\frac{1.47 g}{12.01 g/mol}[/tex]

                                                 = 0.1224 mol

It is also given that mass of hydrogen = 0.123 g

Hence, calculate number of moles of hydrogen as follows.

           No. of moles of hydrogen = [tex]\frac{\text{mass hydrogen}}{\text{molar mass hydrogen}}[/tex]

                                                      = [tex]\frac{0.123 g}{1.008 g/mol}[/tex]

                                                      = 0.122 mol

Therefore, [tex]\frac{\text{moles of carbon}}{\text{moles of hydrogen}}[/tex]

                              = [tex]\frac{0.1224 mol}{0.122 mol}[/tex]

                                = 1.003

Therefore, calculate the number of moles of hydrogen as follows.

       No. of moles of hydrogen = [tex]\frac{\text{mass hydrogen}}{\text{molar mass hydrogen}}[/tex]

                                          = [tex]\frac{2.17 g}{1.008 g/mol}[/tex]

                                          = 2.1528 mol

Hence, calculate the moles of carbon as follows.

      Moles of carbon = [tex]\text{moles hydrogen} \times \frac{\text{moles of carbon}}{\text{moles hydrogen}} [/tex]

                                   = [tex]2.1528 mol \times 1.003[/tex]

                                  = 2.16 mol

Mass of carbon = moles carbon × molar mass carbon

                           = (2.16 mol) × (12.01 g/mol)

                           = 25.9 g

Thus, we can conclude that 25.9 g of carbon is expected in the sample.

Final answer:

The expected amount of carbon in Sample 2 would be 25.93 grams, calculated using the constant carbon-to-hydrogen mass ratio determined from Sample 1.

Explanation:

The question requires determining how much carbon is expected in Sample 2, given the known amounts of carbon and hydrogen in Sample 1 and the amount of hydrogen in Sample 2. We assume that the compound is the same in both samples, meaning the ratio of carbon to hydrogen must be constant. First, we find the ratio of carbon to hydrogen in Sample 1:

Sample 1: 1.47 g C / 0.123 g H = 11.95 g C/g H

Using this ratio, we calculate the expected amount of carbon in Sample 2:

Expected carbon in Sample 2: 2.17 g H x 11.95 g C/g H = 25.93 g C

Therefore, we would expect Sample 2 to contain 25.93 grams of carbon, assuming it is composed of the same pure compound as Sample 1.