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3 years ago

How many moles of oxygen are required to produce 37.15 g CO2?

2 answers:

8 0

Carbon = 12.010. Oxygen = 15.999 x 2 15.999 x 2 = 31.998 + 12.010 = 44.008 \frac{37.15 grams * 1 mole CO2}{44.008 grams}

Nostrana [21]3 years ago

4 0

1.407.........................................................

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Do you think it is important for archaeological sites to be protected from development once they are discovered? What are the pr

Oxana [17]

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2 years ago

Given 4.80g of ammonium carbonate, find:

V125BC [204]

Answer:

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

1) Number of moles of the compound:

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g)/(96.09 g/mol) = 0.05 mol.

2) Number of moles of ammonium ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄)₂CO₃ = (2.0)(0.05 mol) = 0.1 mol.

3) Number of moles of carbonate ions :

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄)₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄)₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄)₂CO₃ = (1.0)(0.05 mol) = 0.05 mol.

4) Number of moles of hydrogen atoms:

Every 1.0 mol of (NH₄)₂CO₃ contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

∴ The no. of moles of H atoms in 0.05 mol of (NH₄)₂CO₃ = (8.0)(0.05 mol) = 0.4 mol.

5) Number of hydrogen atoms:

It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

Using cross multiplication:

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains → ??? atoms.

∴ The no. of atoms in 0.4 mol of H atoms = (6.022 x 10²³ molecules)(0.4 mole)/(1.0 mole) = 2.4 x 10²³ molecules.

8 0

3 years ago

Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the

rusak2 [61]

Answer:

Option d: C₈H₉NO₂ = acetaminophen, analgesic

Explanation:

% composition of compound is:

63.57 g of C

6 g of H

9.267 g of N

21.17 g of O

First of all we divide each by the molar mass of the element

63.57 g / 12 gmol = 5.29 mol of C

6 g of H / 1 g/mol = 6 mol H

9.267 g of N / 14 g/mol = 0.662 mol of N

21.17 g of O / 16 g/mol = 1.32 mol of O

We divide each by the lowest value, in this case 0.662

5.29 / 0.662 = 8

6 / 0.662 = 9

0.662 / 0.662 = 1

1.32 / 0.662 = 2

Molecular formula of the compound is C₈H₉NO₂

7 0

3 years ago

A potassium bromide solution is 8.30% potassium bromide by mass and its density is 1.03 g/ml. what mass of potassium bromide is

pshichka [43]

Answer: The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

Explanation:

We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.

To calculate the volume of KBr, we use the formula:

$Density=\frac{Mass}{Volume}$

Mass of the solution = 100 grams

Density of KBr solution = 1.03g/mL

Volume of the solution = ? mL

Putting values in above equation, we get:

$1.03g/mL=\frac{100g}{Volume}\\\\Volume=97.08mL$

Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.

In 97.08 mL of solution, mass of KBr present is 8.3 grams.

So, 41.2 mL of solution will contain = $\frac{8.3g}{97.08mL}\times 41.2mL=3.522g$ of KBr.

Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

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2 years ago

The levels of organization within an organism are atom,

OleMash [197]

Answer:

The answer is: Molecule and Organ system.

4 0

1 year ago

Read 2 more answers