For the reaction 2Fe+o2 -->Feo how many grams of iron(ll) oxide are produced from 479.6 grams of iron in an excess of oxygen (Fe=56gmol, O=16g mol)

Answer:

tree

Explanation:

so easy and obvious

Answer:

[tex]1.2x10^{24}g[/tex]

Explanation:

Hello!

In this case, it is possible to treat this problem by using a proportional factor which indicates one iron nail equals 2.0 grams:

[tex]\frac{2.00g}{1nail}[/tex]

Now, for an amount of 6.022x10²³ nails, the corresponding mass will be:

[tex]6.022x10^{23}nail*\frac{2.00g}{1nail} \\\\1.2x10^{24}g[/tex]

Best regards!

Answer:

20 mole of co2

Explanation:

I hope this helps

Answer:

8.397J/mol is the activation energy for the reaction

Explanation:

The graphical form of Arrhenius equation is:

ln k = -Ea/R*(1/T)+lnA

Where k is activation energy

Ea is activation energy

R is gas constant (8.314J/molK)

T is absolute temperature

And A is the pre-exponential factor.

The slope of the plot is -Ea/R:

-Ea/R = -1.01x10⁴K

-Ea/8.314J/molK = -1.01x10⁴K

Ea = -1.01x10⁴K*8.314J/molK

Ea = 8.397J/mol is the activation energy for the reaction

Answer:

1.403x10²⁴ molecules

Explanation:

In order to calculate how many molecules of CO₂ are there in 102.5 g of the compound, we first convert grams to moles using its molar mass:

Now we convert moles into molecules using Avogadro's number:

Answer: Theoretical yield of [tex]SO_3[/tex] produced by 8.96 g of S is 33.6 g

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]

[tex]\text{Moles of} S=\frac{8.96g}{32g/mol}=0.28moles[/tex]

The balanced chemical equation is:  

[tex]2S+3O_2\rightarrow 2SO_3[/tex]  

According to stoichiometry :

2 moles of [tex]S[/tex] produce =  3 moles of [tex]SO_3[/tex]

Thus 0.28 moles of [tex]S[/tex] will produce=[tex]\frac{3}{2}\times 0.28=0.42moles[/tex]  of [tex]SO_3[/tex]  

Mass of [tex]SO_3=moles\times {\text {Molar mass}}=0.42moles\times 80g/mol=33.6g[/tex]

Thus theoretical yield of [tex]SO_3[/tex] produced by 8.96 g of S is 33.6 g

Answer:

Septic tanks.

Explanation:

Wastewater typically refers to a body of water that has been contaminated through human use in homes, offices, schools, businesses, etc. Wastewater are meant to be disposed in accordance with the local regulations and standards because they are unhygienic for human consumption or use.

Generally, many homes use a floor drain in their bathrooms and toilets to remove wastewater in order to mitigate stagnation and to improve hygiene. A floor drain can be defined as a material installed on floors for the continuous removal of any stagnant wastewater in buildings.

Wastewater flows into a septic tank once it is released into a floor drain or from the water closet through the use of a pipe such as a polyvinyl chloride (PVC) pipe, which directly connects the to the septic tank.

It is used by nearly one-third of the U.S. population, septic tanks are frequent sources of biological contamination, such as fecal bacteria and virus contamination, of water supplies, and often the owners of these systems dispose of detergents, nitrates, chlorides, and solvents in the system leading to potential downstream contamination if not managed properly.

Hence, the wastewater and biological contamination should be removed from septic tanks when they are filled up through the use of a pump.

Explanation:

A. AgNO3 + KCl --> AgCl + KNO3

Reaction is already balanced.

Type = Double Replacement

B. H2O + SO3 --> H2SO4

Reaction is already balanced.

Type = Synthesis

C. KI + Cl2 --> KCl + I2

Balanced Equation = 2KI + Cl2 --> 2KCl + I2

Type = Single Replacement

D. NaHCO3 --> Na2CO3 + H2O + CO2

Balanced Equation = 2NaHCO3 --> Na2CO3 + H2O + CO2

Type = Decomposition

E. Zn + HCl  --> ZnCl2 + H2

Balanced Equation = Zn + 2HCl  --> ZnCl2 + H2

Type = Single Replacement

F. BaCl2 + Na2SO4 --> BaSO4 + NaCl

Balanced Equation =  BaCl2 + Na2SO4 → 2NaCl + BaSO4  

Type = Double Replacement

G. C3H8 + O2 --> CO2 + H2O

Balanced Equation =  C3H8 + 5O2 → 3CO2 + 4H2O  

Type = Single Replacement

H. Al + CuCl2 --> AlCl3 + Cu

Balanced Equation =   2Al + 3CuCl2 → 2AlCl3 + 3Cu  

Type = Single Replacement

Answer:

6.76 mol/L

Explanation:

A chemist prepares a solution 0.607 kg of calcium bromide by measuring out of calcium bromide into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's calcium bromide solution. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Step 2: Calculate the moles of solute

The molar mass of calcium bromide is 199.89 g/mol.

607 g × 1 mol/199.89 g = 3.04 mol

Step 3: Convert "V" to liters

We will use the conversion factor 1 L = 1000 mL.

450. mL × 1 L/1000 mL = 0.450 L

Step 4: Calculate the concentration of calcium bromide in mol/L

[CaBr₂] = 3.04 mol/0.450 L = 6.76 mol/L

Answer:

5.80200 Kg / L

Explanation:

400 Joules

Work done is measured in Joules (J) remember that

Answer:400 joules

wut do u need

Explanation:

7.5 ppm

Given

1000 g of water

0.0075g of Ca²⁺ ion

Required

the concentration in ppm by mass of Ca²⁺

Solution

ppm = part per million

solvent = water ⇒ ppm = 1 mg/L(water density is 1 kg / L) or mg/kg

Convert g to mg of  Ca²⁺ ion :

0.0075 g = 7.5 x 10⁻³ g = 7.5 mg

Convert g to kg of water :

1000 g = 1 kg water

So the concentration of Ca²⁺ ion :

= 7.5 mg / 1 kg

= 7.5 ppm

Answer:

Empirical formula = CH₃O

Molecular formula = C₂H₆O₂

Explanation:

The empirical formula is the simplest whole-number ratio of atoms present in a molecule. To find the ratio of atoms we must find the moles of C, H and O. With the molar mass we can find the molecular formula:

Moles C = Moles CO₂:

12.65g * (1mol / 44.01g) = 0.2874 moles of C * 12.01g/mol = 3.452g C

Moles H = 1/2 moles H₂O:

7.768g * (1mol / 18.015g) = 0.4312 moles of H₂O * 2 = 0.8624 moles H * 1.008g/mol = 0.869g H

Moles O:

Mass O = 8.919g - 0.869g H - 3.452g C = 4.598g O * (1mol / 16g) = 0.2874 moles of O

Ratio of atoms -Dividing in the lower number of moles-:

C = 0.2874 mol C / 0.2874 mol = 1

H = 0.8624 mol H / 0.2874 mol = 3

O = 0.2874 mol O / 0.2874 mol = 1

The molar mass of this formula is:

1C = 12.01g/mol

3H = 3*1.008g/mol = 3.024g/mol

1O = 16g/mol

12.01+3.024+16 = 31.034g/mol

Twice this formula gives the molecular formula = 31.034g/mol*2 = 62.068g/mol ≅ the molar mass of the compound. That means molecular formula of the compound is:

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Answer:

7.5 g of hydrogen gas reacts with 50.0 g oxygen gas to form 57.5 g of water.

Explanation:

Here we have the check if the mass of the reactants is equal to the mass of the products.

Reactants

[tex]7.5+50=57.5\ \text{g}[/tex]

Products

[tex]57.5\ \text{g}[/tex]

The data is consistent with the law of conservation of matter.

Reactants

[tex]50+243=293\ \text{g}[/tex]

Products

[tex]206+97=303\ \text{g}[/tex]

The data is not consistent with the law of conservation of matter.

Reactant

[tex]17.7+34.7=52.4\ \text{g}[/tex]

Products

[tex]62.4\ \text{g}[/tex]

The data is not consistent with the law of conservation of matter.

Only the first data is consistent with the law of conservation of matter.