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3 years ago

How are you finding the moles of the unknown substance when you are only given the grams of the unknown molecular compound?

Chemistry

1 answer:

Natali [406]3 years ago

4 0

Answer:

Determine the moles of unknown (the solute) from the molality of the solution and the mass of solvent (in kilograms) used to make the solution. Determine the molar mass from the mass of the unknown and the number of moles of unknown.

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Which statement correctly compares sound and light waves?

Mrac [35]

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Light waves carry energy parallel to the motion of the wave, while sound waves carry energy perpendicu

Explanation:

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3 years ago

Alkanes are hydrocarbons containing only single bonds. Acyclic alkanes have carbon atoms arranged in chains, whereas cycloalkane

sashaice [31]

Answer:

156 Hydrogen atoms

Explanation:

<u>Any acyclic alkane has a molecular formula that can be expressed as</u>:

CₙH₂ₙ₊₂

Where <em>n</em> is any integer and the number of carbon atoms. For example, Propane has 3 carbon atoms, this means it would have [2*3+2] 8 hydrogen atoms, resulting with a formula of C₃H₈.

An acyclic alkane with 77 carbon atoms would thus have:

2*77 + 2 = 156 hydrogen atoms

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3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

vladimir1956 [14]

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ $\frac{5.60}{2}gallons = 2.80 gallons$

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = $(M__1}) = Density*Volume$

$= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 11659.06grams$

On the other hand; the mass of water = $(M__2})= Density*Volume$

$= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}$

$= 10577.95 grams$

Molarity = $\frac{massof solute*1000}{molarmassof solute*massofwater}$

= $\frac{11659.06*1000}{62.07*10577.95}$

= 17.757 m

≅ 17.76 m

∴ the boiling point of the solution is calculated using the boiling‑point elevation constant for water and the Molarity.

$\Delta T_{boiling} = k_{boiling}M$

where,

$k_{boiling}$ = 0.512 °C/m

$\Delta T_{boiling}$ = 100°C + 17.56 × 0.512

= 109.09 °C

6 0

3 years ago

As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis wit

jeka94

Answer:

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced 0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles = 0.0003663 moles

Mass of K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%

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3 years ago

What does a high number on the pH scale say about a solution?

erik [133]

Explanation:

A is cortect answer dsfa

8 0

1 year ago