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2 years ago

WHO CAN HELP ME?!?!?!?!

Chemistry

1 answer:

Lady_Fox [76]2 years ago

3 0

Answer:

322.1262g

Explanation:

This question is mole to mole reaction but the answer should be in gram.

4moles of Al gives 2 moles of Al2O3

3.1581 moles of Al gives x moles of Al2O3

Cross multiply

4x= 3.1581×2

4x=6.3162

X=6.3162/4

X=1.57905moles

To convert to gram

Moles ×molar mass

Molar mass of Al2O3 is 204g

Which means 1.57905×204= 322.1262g

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PLEASE ANSWER!!!VERY IMPORTANT!!! I WILL RATE BRAINLIEST!!!!

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3 years ago

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What is the pH of a solution made by mixing 30.00 mL 0.10 M HCl with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solu

nirvana33 [79]

Answer:

pH = 12.15

Explanation:

To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.

HCl + KOH → H₂O + KCl

We need to determine the moles of each compound

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl

HCl + KOH → H₂O + KCl

3 m 4 m -

1 m 3 m

As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.

1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume

[OH⁻] 1 mmol / 70 mL = 0.014285 M

- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15

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2 years ago

What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit

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Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

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2 years ago

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How

labwork [276]

First convert grams to moles using molar mass of butane that is 58.1 g

3.50g C4H10 x (1 mol C4H10)/(58.1g C4H10) = 0.06024 mol C4H10 

Now convert moles to molecules by using Avogadro’s number

0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 mol C4H10) = 3.627x10^22 molecules C4H10

And there are 4 carbon atoms in 1 molecule of butane, so use the following ratio:

3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 molecule C4H10)
= 1.45 x 10^23 atoms of carbon are present

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