When 2.50 g of copper reacts with oxygen the copper oxide product has a mass of 2.81 g. what is the simplest formula of the copper oxide?

Mark him brainliest, please. Don't be afraid to ask how. I know how to.

Answer:

Its CCCCCCCC

Explanation:

your welcome please give 5 starts and like bc this is right i  got 100% in the final text over the second semester

To balance this chemical equation, we have to make sure that equal number of elements are located on the left side and the right side by placing coefficients:

CCl4  +  O2   <-- --> COCl2  +  Cl2

Balancing this equation:

2CCl4(g)+ O2(g) ⇌ 2COCl2(g) + 2Cl2(g)

Answer:

2,1,2,2

To balance the equation ccl4(g)+o2(g)⇌cocl2(g) + cl2(g), the coefficients should be: 1,2,1,1.

To balance the equation ccl4(g)+o2(g)⇌cocl2(g) + cl2(g), we need to ensure that the number of each atom is the same on both sides of the equation.

Starting with the carbon atoms, we have 1 on the left side and 1 on the right side. So, the coefficient for CCl4 remains 1.

Next, we have 2 chlorine atoms on the left side and 2 on the right side. So, the coefficient for Cl2 remains 1.

Finally, we have 2 oxygen atoms on the right side and 0 on the left side. To balance the oxygen atoms, we need a coefficient of 2 for O2.

Therefore, the balanced equation is: CCl4(g) + 2O2(g) ⇌ COCl2(g) + Cl2(g).

a) 0 is the work of expansion accompanying the complete combustion of 1.0 g of glucose to carbon dioxide b) -81.12 J is the work of expansion.

When a force is applied to an item and the object is moved in the direction of the force, the energy transfer known as "work" takes place. The fundamental idea of work has applications in both physics and chemistry, particularly in the context of thermodynamics.The expansion or compression of a gas, which can take place in a variety of processes including isothermal (constant temperature), adiabatic (no heat exchange), and isobaric (constant pressure) processes, is frequently linked to work in thermodynamics.

a) w = -Pext(final pressure- initial pressure)

       = -Pext(nfRT/Pf- niRT/Pi)

since initial and final number of moles is same

Work of expansion is zero.

b)moles of water is calculated as:

=1g glucose/180 g/mol×6 mol water/1 mol glucose

=[tex]3.33 \times 10^{-2}[/tex] mol of water

W =- nRΔT

   = [tex]3.33 \times 10^{-2}[/tex]×8.31×293.15

   =-81.12 J

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Final answer:

Coal is the type of rock formed in a plant-rich swamp. It forms over millions of years as dead plants are buried under sediment and subjected to pressure and heat, resulting in a carbon-rich sedimentary rock.

Explanation:

The type of rock that would be formed in a plant-rich swamp is coal. Through a process spanning millions of years, dead plants that accumulated at the bottom of such swamps were buried under layers of sand, water, and clay. With the absence of oxygen and the presence of pressure and heat over time, the organic materials gradually transformed into a carbon-rich sedimentary rock known as coal. This transformation takes place in environments where stagnant water, low levels of nitrogen and oxygen, and organic matter coalesce, creating the perfect conditions for coal formation.

Coal is not just a rock; it's a major fossil fuel that has played a critical role in the development of human civilization by providing energy for warmth, manufacturing, and electricity. However, the creation of coal reserves also highlights the long-term carbon sequestration abilities of ancient ecosystems that thrived millions of years ago.

To calculate the vapor pressure of a benzene solution with camphor, use Raoult's Law considering the negligibility of camphor's vapor pressure and determine the mole fraction of benzene.

Calculating Vapor Pressure of a Solution

To calculate the vapor pressure of a solution containing 28.2 g of camphor (C10H16O) dissolved in 94.8 g of benzene, we can use Raoult's Law. This law states that the partial vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. The formula is P1 = X1 * P1o, where P1 is the vapor pressure of the solvent in the solution, X1 is the mole fraction of the solvent, and P1o is the vapor pressure of the pure solvent.

First, we need to calculate the mole fraction of benzene in the solution. To do that, we must find the moles of both camphor and benzene using their respective molar masses (152.23 g/mol for camphor and 78.11 g/mol for benzene). After calculating the moles, we can find the mole fraction of benzene (X1) by dividing the moles of benzene by the total moles of both compounds. We can then apply the equation above to find the new vapor pressure of the benzene in the solution (vapor pressure of a solution).

Note that this calculation assumes camphor has negligible vapor pressure, which is a reasonable assumption given it is a low-volatility solid. The answer will be in mmHg because that is the unit given for the vapor pressure of pure benzene.

The correct answer is...

D. Only the hydrogen needs to be balanced. There are equal numbers of aluminum and chlorine.

Answer:

A substance with precise melting and freezing point.

Explanation:

A pure substance is one that has a precise melting and freezing point

Like water, pure water boils at 100C while impure water will boil at a temperature less than that.

Impurity also affects the refractive index and specific gravity of a substance

The compounds in group c are a mixture of both ionic and molecular acids. Ionic compounds tend to dissociate in water, while molecular compounds do not dissociate in water.

Based on the information provided, we can determine the nature of the compounds in group c. KCl and MgCl2 are ionic compounds because they consist of a metal (K and Mg) and a nonmetal (Cl). On the other hand, NC13, ICl, PCl5, and CCl4 are molecular compounds because they only contain nonmetals (N, C, I, P, and Cl). So, the compounds in group c can be classified as both ionic (KCl and MgCl2) and molecular acids (NC13, ICl, PCl5, and CCl4).

Regarding dissociation, ionic compounds tend to dissociate into ions when dissolved in water. So, we would expect KCl and MgCl2 to dissociate into K+ and Cl- ions. However, molecular compounds usually do not dissociate into ions in water. Instead, they remain as intact molecules. Therefore, we would not expect NC13, ICl, PCl5, and CCl4 to dissociate when dissolved in water.

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The determination involves analyzing the compounds' composition. Molecular compounds do not dissociate into ions, while ionic compounds and molecular acids do. If the compounds in Group C are acids, they will dissociate in water.

To determine if all four compounds in group C are molecular, ionic, or molecular acids, we need to analyze their composition and behavior in water. Molecular compounds typically consist of nonmetals and do not dissociate into ions in water. Ionic compounds are formed from metals and nonmetals and dissociate into cations and anions when dissolved in water. Molecular acids, such as HCl, are covalent but dissociate in water to produce H+ ions and anions.

Given this, if the compounds are molecular acids, you would expect them to dissociate in water.

For every mole of P4O10, there are 4 moles of phosphorus atoms. To find the total moles of P, you multiply the moles of P4O10 by 4, resulting in 78.5 mol  imes 4 = 314 mol of phosphorus.

To calculate the number of moles of phosphorus (P) in 78.5 mol of P4O10, we have to understand the composition of the compound P4O10. The molecular formula shows that one molecule contains 4 atoms of phosphorus. Hence, for every mole of P4O10, there are 4 moles of phosphorus atoms.

By multiplying the moles of P4O10 by 4, we can find out the total moles of phosphorus. Therefore:

So, 78.5 mol of P4O10 contains 314 mol of phosphorus (P).

The mass, in grams, of Ag is produced from reaction of 31.75 g of Cu is 107.9g

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

Given,

Mass of Cu = 31.75g

Molar mass of Cu = 63.5 g/mol

Moles of Cu = 31.75 / 63.5

= 0.5 moles

From the reaction, 1 mole of Copper gives 2 moles of silver.

Moles of Ag = 2 × 0.5 = 1 mole

Mass of Ag = moles × molar mass

= 1 × 107.9

= 107.9g

Therefore, the mass, in grams, of Ag is produced from reaction of 31.75 g of Cu is 107.9g

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MW H2 = 2

MW 2 = ?

rate 1 = 1/12.1

rate 2 = 2.42

aopply graham law

Rate1/rate2 = sqrt(MW2/MW1)

(12.1/2.42)^2 = MW2/2

MW = 2(12.1/2.42)^2 = 50 g/mol

Therefore, the molar mass of the unknown gas is 50 g/mol

According to Graham's Law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. By comparing the rates of effusion of H2 gas and the unknown gas, we can calculate the molar mass of the unknown gas. The molar mass of the unknown gas is approximately 0.404 g/mol.

Graham's Law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Let's use the given information to calculate the molar mass of the unknown gas.

According to the question, an equal volume of H2 effuses in 2.42 min, while the unknown gas effuses in 12.1 min.

Applying Graham's Law, we can set up the following ratio:

(Rate of H2 effusion) / (Rate of unknown gas effusion) = √(Molar mass of unknown gas) / √(Molar mass of H2)

Substituting the given values, we have:

2.42 / 12.1 = √(Molar mass of unknown gas) / √(2.02 g/mol)

Simplifying this equation, we find:

Molar mass of unknown gas = (2.42 / 12.1) * (2.02 g/mol)

Calculating this, we get:

13 kilojoules

Given:

A synthesis of ammonia in which the volume contracts from 7.4 l to 4.5 l at a constant external pressure of 44 atm.

Question:

Calculate the work in kilojoules done during the synthesis of ammonia. Express your answer using two significant figures.

The Process:

Let us calculate the change in volume and also convert units.

[tex]\boxed{ \ \Delta V = V_2 - V_1 \ }[/tex]

[tex]\boxed{ \ \Delta V = 4.5 - 7.4 \ }[/tex]

Therefore ΔV = - 2.9 L.

Work done during expansion and contraction at constant pressure.

[tex]\boxed{ \ W = p \ \Delta V \ }[/tex]

Let us find out the work done.

[tex]\boxed{ \ W = 4,457,200 \times -2.9 \times 10^{-3} \ }[/tex]

W = -12,925.88 joules

W = -13,000 joules to 2 significant figure.

Thus, the work in kilojoules done during the synthesis of ammonia is 13 kilojoules.

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Notes

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Notes:

Final answer:

The balanced equation for photosynthesis is 6CO2 + 6H2O + light energy → C6H12O6 + 6O2, affirming the conservation of atoms but not molecules.

Explanation:

Photosynthesis Chemical Reaction

The balanced chemical equation for the photosynthetic conversion of CO2 and H2O into glucose (C6H12O6) and O2 is:

6CO2 + 6H2O + light energy → C6H12O6 + 6O2

b. Yes, the number of each type of atom on either side of the arrow is the same, maintaining the law of conservation of mass.

c. The number of molecules on either side of the equation is not the same because there are different numbers of reactant and product molecules, but the number of atoms is conserved.

Answer:

2Cu(s)+S(s)→Cu2S(s)

Explanation:

The chemical equation for the reaction between solid copper  and solid sulfur to form solid copper(i) sulfide is 2Cu(s) + S(s) -> Cu₂S(s)

The chemical equation for the reaction between solid copper and solid sulfur to form solid copper(i) sulfide can be written as follow:

Solid copper + Solid sulfur -> Solid copper(i) sulfide

Cu(s) + S(s) -> Cu₂S(s)

There are 2 atoms of Cu on the right side and 1 atom on the left. It can be balanced by writing 2 before Cu as shown below:

2Cu(s) + S(s) -> Cu₂S(s)

Thus, the balance equation for the reaction is 2Cu(s) + S(s) -> Cu₂S(s)

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A compound is a pure substance because its molecule cannot be broken down into simpler particles by physical means.

Answer:

A compound is a pure substance because its molecule cannot be broken down into simpler particles by physical means.

Explanation:

To determine if it's worth the time and trouble for Richard and Diane to make their own alum from discarded aluminum cans, we can compare the cost of making alum with the cost of buying it from the grocery store. To calculate the cost of making alum, we need to determine the cost of sulfuric acid. By calculating the amount of alum that can be made from the given quantity of cucumbers, we can compare the cost of making alum versus buying it.

Richard and Diane are considering making their own alum from discarded aluminum cans. In order to determine if it's worth the time and trouble, we need to compare the cost of making alum with the cost of buying it from the grocery store.

To make alum, we need aluminum cans and a source of sulfuric acid. The cost of the aluminum cans is not provided in the question, so we will focus on the cost of sulfuric acid. From the information given, we know that the 9M H2SO4 solution costs $8 per liter.

We need to calculate the amount of alum that can be made from the given quantity of cucumbers. From the equation:

2 KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2 H2O(l)

We can see that we need twice as many moles of KOH as H2SO4. Using the molarity and quantity of KOH solution mentioned in the question, we can calculate the number of moles of KOH:

Molarity of KOH solution = 1.5 M = 1.5 mol/L

Amount of KOH solution needed = x L (unknown)

Number of moles of KOH = (1.5 mol/L) * x L = 1.5x mol KOH

Since we need twice as many moles of KOH as H2SO4, we can determine the number of moles of H2SO4:

Number of moles of H2SO4 = 0.75x mol H2SO4

Now we can calculate the cost of H2SO4 needed to make the alum:

Cost of H2SO4 = ($8/L) * (0.75x L) = $6x

On the other hand, the cost of buying a 50 g jar of alum from the grocery store is given as $5.79.

Based on these calculations, we can compare the cost of making alum versus buying it:

If $6x (cost of making alum) is less than $5.79 (cost of buying alum from the grocery store), then it would be worth the time and trouble for Richard and Diane to make their own alum from discarded aluminum cans.

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Richard and Diane should calculate the cost of making their own alum from discarded aluminum cans to determine if it's worth the time and trouble.

Richard and Diane should calculate the cost of making their own alum from discarded aluminum cans to determine if it's worth the time and trouble. First, they need to calculate the cost of materials for making alum. They can find the cost of KOH, H2SO4, and ethanol per liter and the cost of alum per jar. Then, they need to calculate the volume of alum solution needed for pickling cucumbers and determine the cost of the required amount of alum solution by multiplying the volume by the cost per jar. Comparing the cost of buying alum from the store and the cost of making alum will help them make an informed decision.

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The length of DC is 10 units and the measure of angle ∠C is 74°.

To find the length of DC, we can use the concept of similar triangles. Since triangles ΔABC and ΔDEF are similar, the corresponding sides are proportional. We can set up the following proportion: AB/DE = BC/EF = AC/DF. Plugging in the given lengths, we have 5/3 = BC/6. Cross-multiplying gives us BC = 10 units. Therefore, the length of DC is 10 units.

To find the measure of angle ∠C, we can use the fact that corresponding angles in similar triangles are congruent. Since angle ∠B = 53° in ΔABC, we know that angle ∠F = 53° in ΔDEF. Therefore, angle ∠C = 180° - 53° - 53° = 74°.

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Answer:

15.8528 grams of oxygen will be obtained.

Explanation:

[tex]4C_3H_5N_3O_9\rightarrow 6N_2+12CO_2+10H_2O+O_2[/tex]

Mass of nitroglycerin = [tex]4.50\times 10^{2} g[/tex]

Moles of nitroglycerin:

[tex]\frac{4.50\times 10^{2} g}{227.08 g/mol}=1.9816 mol[/tex]

According to reaction, 4 moles of nitroglycerin gives 1 mole of oxygen gas.

then 1.9816 mol of nitroglycerin will give:

[tex]\frac{1}{4}\times 1.9816=0.4954 [/tex] moles of oxygen gas.

Mass of oxygen ;

[tex]0.4954 mol\times 32 g/mol=15.8528 g[/tex]

15.8528 grams of oxygen will be obtained.

The maximum mass of O₂ that can be obtained from 4.5×10² g of nitroglycerin is 15.86 g

The balanced equation for the reaction is given below:

Next, we shall determine the mass of C₃H₅N₃O₉ that decomposed and the mass of O₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of C₃H₅N₃O₉ = (12×3) + (5×1) + (14×3) + (16×9)

= 36 + 5 + 42 + 144

= 227 g/mol

Molar mass of O₂ = 2 × 16 = 32 g/mol

From the balanced equation above,

908 g of C₃H₅N₃O₉ decomposed to produce 32 g of O₂.

Finally, we shall determine the mass of O₂ that will be produce from the decomposition of 4.5×10² g of C₃H₅N₃O₉. This can be obtained as follow:

From the balanced equation above,

908 g of C₃H₅N₃O₉ decomposed to produce 32 g of O₂.

Therefore, 4.5×10² g of C₃H₅N₃O₉ will decompose to produce = [tex]\frac{4.5*10^{2} * 32}{908}[/tex] = 15.86 g of O₂

Thus, the maximum mass of O₂ obtained from 4.5×10² g of nitroglycerin is 15.86 g

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Final answer:

Ethanol and 1-decanol both possess a hydroxyl group and are classified as primary alcohols. They also both have the capability for hydrogen bonding, influencing their physical properties.

Explanation:

Both ethanol and 1-decanol share common structural features. Firstly, they both possess a hydroxyl group (-OH) connected to an aliphatic carbon chain, which categorizes them as alcohols. Ethanol is a two-carbon alcohol with the molecular formula CH3CH2OH, while 1-decanol is a longer ten-carbon chain alcohol with the formula CH3(CH2)9OH. The similarity lies in the fact that for both ethanol and 1-decanol, the hydroxyl group is positioned at the end of the carbon chain (primary alcohols). Additionally, both alcohols have the potential for hydrogen bonding due to the presence of the hydroxyl group, affecting their physical properties such as boiling point and solubility in water.