Which is the next logical step in balancing the given equation?

3CO(g) + Fe2O3(s) Fe(s) + 3CO2(g)

A.Place the coefficient 2 in front of iron(III) oxide.
B.Place the coefficient 2 in front of iron(III) oxide.
C.Place the coefficient 2 in front of elemental iron.
D.Replace the coefficient 3 in carbon monoxide with 6.

somebody help pleaseeee ):

To classify each molecule as an aliphatic or an aromatic hydrocarbon, we need to determine if they are alkanes, alkenes, or alkynes.

To classify each molecule as an aliphatic or an aromatic hydrocarbon, we need to determine if they are alkanes, alkenes, or alkynes.

By examining the structure of each molecule, we can identify their classification as aliphatic or aromatic hydrocarbons.

Answer:

i honestly dont know but ill try

Explanation:

hope this helps its yes

Final answer:

Determining bond polarity involves identifying the direction of the polarity based on electronegativity differences. The atom with higher electronegativity gains a partial negative charge, defining the bond's polarity. The H-F bond is typically the most polar due to a large difference in electronegativity.

Explanation:

The question involves identifying the direction of polarity in chemical bonds and determining which bond is the most polar. Polarity arises when there is a difference in electronegativity between the atoms forming a bond. The atom with higher electronegativity will have a partial negative charge (δ-) and the other a partial positive charge (δ+). To predict the most polar bond, compare the electronegativity values of the atoms involved; the greater the difference, the more polar the bond.

Examples:

Generally, bonds involving atoms like fluorine, oxygen, and nitrogen with other less electronegative atoms tend to be highly polar due to the significant electronegativity differences. Thus, among common bonds, the bond between hydrogen and fluorine (H-F) would be expected to be the most polar because of the large electronegativity difference between the two atoms.

In a 1.70-g dose of lithium carbonate, which contains 18.8% lithium, there is approximately 0.3196 g of lithium.

The question is asking for the amount of lithium in a 1.70-g dose of lithium carbonate (Li2CO3) which is known to contain 18.8% lithium. The calculation is fairly straightforward once we understand that the percentage given is by mass. Hence, to find out the lithium content, we simply multiply the total mass of the dose by the percentage of lithium.

Step 1. Convert the lithium percentage to a decimal by dividing 18.8 by 100, giving us 0.188.
Step 2. Multiply the mass of the lithium carbonate dose (1.70 g) by the decimal from Step 1. The calculation is: 1.70 g × 0.188 = 0.3196 g.

The result indicates that in a 1.70-g dose of lithium carbonate, there is approximately 0.3196 g of lithium.

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To find the mass of lithium in a 1.70-g dose of lithium carbonate, multiply the mass by the percentage of lithium content (18.8%). This calculation results in 0.3196 grams of lithium.

Step-by-Step Solution

Mass of lithium = (percentage of lithium / 100) * mass of lithium carbonate

Mass of lithium = (18.8 / 100) * 1.70 g

Mass of lithium = 0.3196 g

Therefore, there are 0.3196 grams of lithium present in a 1.70-g dose of lithium carbonate.

Answer:

0.0365

Explanation:

Given number,

0.036549,

In the above number there are 5 significant figures (i.e. 3, 6, 5, 4 and 9)

When we round a decimal number to 3 significant figure we check the fourth significant number after decimal,

if it is 5 or more than 5 then the number is rounded next significant figure it rounded to previous significant figure,

Here, the fourth significant number after decimal = 4

Thus, 0.036549 ≈ 0.0365 ( rounded to 3 significant figure )

Explanation:

Atomic number of copper is 29 and it is a d-block element.The electronic configuration of copper is as follows.

Electronic configuration of copper is [tex][Ar]4s^1 3d^{10}[/tex]

According to the Aufbau principle, in the ground state of an atom or ion the electrons fill atomic orbitals of the lowest energy levels first, before filling the higher energy levels.

But completely filled d-shell is more stable in nature. Hence, one electron from the 4s shell goes into the d-orbital so the copper atom can attain stability.

So, when a copper atom loses an electron then it goes out from the 4s shell first.

Final answer:

The observed electron configuration of copper (Cu) is [Ar] 3d10 4s1.

Explanation:

The observed electron configuration of Cu is [Ar] 3d10 4s1.

Copper has an anomalous electron configuration because it is a transition metal. The half-filled 3d manifold with one 4s electron is more stable than a partially filled d-manifold and a filled 4s manifold. This leads to the anomalous electron configuration of Cu.

This deviation from the normal electron design happens on the grounds that copper has a higher dependability when its 3d sublevel is totally loaded up with 10 electrons, as opposed to having 9 electrons in the 3d sublevel and 2 during the 4s sublevel. This peculiarity is known as the "half-filled or completely filled sublevel soundness."

If the ion is negatively charged, anion, it means that it gains two electrons from its surroundings. The number of electrons of an argon (neutral atom) is 18. Two of the electrons in our anion is from the surrounding, so originally it only has 16 electrons.

In the periodic table, the 16th element is sulphur. The chemical symbol of sulphur is S. Thus, the anion is the negatively charged sulphur with chemical symbol of S2-. 

During buildup in reflux condenser, the greatest tallness is most of the way up to the condenser. In the event that dissolvable gathers higher than this most extreme tallness, the majority of the dissolvable may keep running over into the response cup or might be the response will begin prior, And if the dissolvable don't go up than the greatest stature, the response does not occur to begin prior.  

Further Explanation:  

Reflux:  

Heartburn is otherwise called acid reflux, corrosive acid reflux, or pyrosis. It happens when a portion of the acidic stomach substance return up into the throat. Heartburn makes a consuming agony in the lower chest zone, frequently in the wake of eating.  

reason for a reflux device:  

Reflux is a procedure including the buildup of vapors and the arrival of this condensate to the framework from which it began. It is utilized in mechanical and research center refining processes. It is likewise utilized in science to supply vitality to responses over an extensive stretch of time.  

Reflux is utilized:

To keep the dissolvable from dissipating, a reflux condenser is utilized. This is a glass segment with a subsequent segment encompassing it through which cool water streams. As vapor from the bubbling dissolvable ascents into inside section of the reflux condenser, it is cooled by the coat of water outwardly and consolidates.

Subject: chemistry  

Level: college

Keywords: Reflux, reason for a reflux device, Reflux is utilized.

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Answer: All substances are composed of elements.

Explanation: Matter is anything which has mass and occupies space.

The substances can be made up element alone like Helium (He).

They can be made up of molecules of elements also like hydrogen [tex](H_2)[/tex]

They can also be made up of molecules of compounds like carbon dioxide [tex](CO_2)[/tex]

They can be made up of mixtures like salt solution in water.

Since all the mixtures, compounds and molecules consists of elements. Therefore, all the substances are composed of elements.

Answer:

Kr < Ar < Ne < N (N is fastest and Kr is slowest)

Explanation:

According to graham law of diffusion, rate of diffusion of a gas is inversely proportional to the square root of molar mass of the gas.

                                        [tex]R\propto \frac{1}{\sqrt{M}}[/tex]

Where R is rate of diffusion and M is molar mass

Molar masses of given gases are - Ar(40 g/mol), Ne(20 g/mol), N(14 g/mol) and Kr(84 g/mol)

So molar mass order : Kr > Ar > Ne > N

Hence rate of diffusion order: Kr < Ar < Ne < N (N is fastest and Kr is slowest)

I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y, for the following reason: 

If you look at the combining masses of X and Y in each of the two compounds, 

The first compound contains 0.25g of X combined with 0.75g of Y 
so the ratio (by mass) of X to Y = 1 : 3 

The second compound contains 0.33 g of X combined with 0.67 g of Y 
so the ratio (by mass) of X to Y = 1 : 2 

Now, you suppose to prepare each of these two compounds, starting with the same fixed mass of element Y ( I will choose 12g of Y for an easy calculation!) 

The first compound will then contain 4g of X and 12g of Y 
The second compound will then contain 6g of X and 12g of Y 

The ratio which combined the masses of X and the fixed mass (12g) of Y
= 4 : 6 
or 2 : 3 

So, the ratio of MOLES of X which combined with the fixed amount of Y in the two compounds is also = 2 : 3 

The two compounds given with the plausible formula must therefore contain the same ratio.

To identify plausible sets of formulas for two compounds containing elements X and Y, use the given masses to calculate the ratios and apply the law of multiple proportions to determine simple whole-number ratios for potential empirical formulas.

The question is regarding the identification of plausible sets of formulas for two compounds containing generic elements X and Y based on the amounts of element X in both compounds. By using the mass data provided (0.25 g and 0.33 g of X), along with the total sample mass (1.00 g for each compound), we can find the respective amounts of element Y in each compound (0.75 g and 0.67 g of Y, respectively).

Then, by taking the ratios of the masses of X to Y for each compound, we can determine simple whole-number ratios, which can suggest possible empirical formulas for the compounds. These formulas are based on the law of multiple proportions, which states that if two elements form more than one compound between them, then the ratios of the masses of the second element that combine with a fixed mass of the first element will be ratios of small whole numbers.

Final answer:

To find the mass of HCl in a 45.0 mL aqueous solution with a specific density and concentration, the volume is converted to mass using the given density, and then the mass of HCl is calculated based on its percentage concentration, resulting in 19.93 g of HCl.

Explanation:

The question asks to find the mass of HCl contained in 45.0 mL of an aqueous HCl solution with a density of 1.19 g cm–3 and a concentration of 37.21% HCl by mass. First, we convert the volume of the solution to mass using the given density. Calculating the mass of the solution: 45.0 mL × 1.19 g/mL = 53.55 g. Next, to find the mass of HCl, we use the percentage concentration of HCl in the solution. Calculating the mass of HCl: 53.55 g × 0.3721 = 19.93 g. Therefore, the mass of HCl in the 45.0 mL solution is 19.93 g.

Final answer:

The material, with a density of 0.67g/cm³, will float on water since its density is less than water's density of 1.0g/cm³. This is determined by dividing its mass by its volume and comparing it to water's density.

Explanation:

The question asks whether a new material with a mass of 2.0g and volume of 3.0cm3 will float on water, which has a density of 1.0g/cm3. To determine whether this material will float, we calculate its density using the formula density = mass / volume. Plugging in the given values, we find the density of the material to be 2.0g / 3.0cm3 = 0.67g/cm3.

Since the density of the material (0.67g/cm3) is less than the density of water (1.0g/cm3), the material will float on water. This is explained by the principle of buoyancy, which states that an object will float in a fluid if its density is less than the density of the fluid.

There are 194 moles of phosphorus in 48.5 moles of P4O10 because each molecule of P4O10 contains 4 phosphorus atoms.

To calculate the number of moles of phosphorus (P) in 48.5 moles of P4O10, we must understand the molar relationship between P4O10 and P. The molecule P4O10 has a composition of 4 phosphorus atoms to 10 oxygen atoms. This means that in every mole of P4O10, there are 4 moles of phosphorus atoms because the subscript '4' in P4 indicates that there are 4 phosphorus atoms in each molecule.

Using this stoichiometric ratio, the calculation is straightforward:

Number of moles of P = 4  imes Number of moles of P4O10

Number of moles of P = 4  imes 48.5 moles

Number of moles of P = 194 moles

Therefore, 48.5 moles of P4O10 contains 194 moles of phosphorus (P).

At pH 3.75, the ratio of [tex]\frac{[A^-]}{[HA]}\right)[/tex] for formic acid ([tex]pK_a[/tex] = 3.75) is 1.

To find the ratio of the concentrations of the conjugate base [A⁻] to the acid [HA] for formic acid (HCOOH) at pH 3.75, we can use the Henderson-Hasselbalch equation:

Given that the pH is 3.75 and the [tex]pK_a[/tex] of formic acid is also 3.75, we can set up the equation as follows:

Subtract 3.75 from both sides:

To solve for the ratio [tex]\frac{[A^-]}{[HA]}\right)[/tex], we need to exponentiate both sides with base 10:

Therefore:

This result makes sense because when the pH of the solution is equal to the [tex]pK_a[/tex] of the acid, the concentrations of the acid and its conjugate base are equal. Consequently, the ratio is exactly 1.

It is theoretically possible to have 5.0 × 10⁻²⁵ g of Al, but this is realistically impossible because it is smaller than the mass of a single aluminum atom.

The question asks if it's possible to have 5.0 × 10⁻²⁵ g of aluminum (Al), given that the atomic mass of Al is 26.98154 g/mol. One mole of Al has a mass in grams that is numerically equivalent to its atomic mass. Since the mass of 1 mol of Al is 26.98 g, having a sample of 5.0 × 10⁻²⁵ g is theoretically possible but practically not feasible since this mass represents far less than a single atom of Al. The smallest practical amount of Al one can have is the mass of one atom, which is much more than 5.0 × 10⁻²⁵ g. Hence, 5.0×10⁻²⁵ g of Al is not a meaningful physical quantity because it is significantly less than the mass of one atom, and one cannot have a fraction of an atom.