How many moles of sodium will react with water to form sodium hydroxide and 4.0 moles of hydrogen?

30 POINTS! ----- How many moles of oxygen gas are needed to completely react with 145 grams of aluminum? Report your answer with

gladu [14]

First a balanced reaction equation must be established:
$4Al _{(s)} + 3 O_{2} _{(g)}$ → $2 Al_{2} O_{3}$

Now if mass of aluminum = 145 g
the moles of aluminum = (MASS) ÷ (MOLAR MASS) = 145 g ÷ 30 g/mol
= 4.83 mols

Now the mole ratio of Al : O₂ based on the equation is 4 : 3
[4Al + 3 O₂ → 2 Al₂O₃]

∴ if moles of Al = 4.83 moles
then moles of O₂ = (4.83 mol ÷ 4) × 3
= 3.63 mol (to 2 sig. fig.)

Thus it can be concluded that 3.63 moles of oxygen is needed to react completely with 145 g of aluminum.

4 0

3 years ago

Hydrofluoric acid, hf, has a ka of 6.8 × 10−4. what are [h3o+], [f−], and [oh−] in 0.710 m hf?

STALIN [3.7K]

Answer:

[H₃O⁺] = [F⁻] = 2.2 x 10⁻² M. & [OH⁻] = 4.55 x 10⁻¹³.

Explanation:

For a weak acid like HF, the dissociation of HF will be:

HF + H₂O ⇄ H₃O⁺ + F⁻.

[H₃O⁺] = [F⁻].

∵ [H₃O⁺] = √Ka.C,

Ka = 6.8 x 10⁻⁴, C = 0.710 M.

∴ [H₃O⁺] = √Ka.C = √(6.8 x 10⁻⁴)(0.710) = 2.197 x 10⁻² M ≅ 2.2 x 10⁻² M.

∴ [H₃O⁺] = [F⁻] = 2.2 x 10⁻² M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.2 x 10⁻²) = 4.55 x 10⁻¹³.

6 0

3 years ago

A cylindrical glass tube of length 27.4cm and radius 4 cm is filled with gas. The empty tube has a mass of 254.3 g. The tube fil

Temka [501]

Density of the gas is 3.05 × 10⁻³ g / cm³.

Explanation:

Volume of the cylinder = π r² h

where r is the radius and h is the height of the height or the length of the glass tube.

Here r = 4 cm and h = 27.4 cm

Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³

We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.

Mass of the substance = 258.5 - 254.3 = 4.2 g

We have to find the density using the formula as,

$$ Density = \frac{mass}{volume}$

Plugin the values as,

$$ Density = \frac{4.2}{1376.6}$

= 3.05 × 10⁻³ g / cm³

So the Density of the gas is 3.05 × 10⁻³ g / cm³.

3 0

3 years ago

Fill in the blanks with appropriate options given in below:

jarptica [38.1K]

Answer:

1) -COOH

2) -NH2

3) hydrogen bonds

4) dispersion forces

5) -CH3

6) hydrogen bonds

7) negative

8) negative

9) positive

Explanation:

Alanine has a -COOH and a -NH2 group available to form hydrogen bonds with water molecules.

Although there are some potential dispersion forces between the terminal -CH3 group of alanine and hexane molecules, we expect the hydrogen bonds between alanine and water to be stronger.

Stronger intermolecular attractive forces between alanine and water lead to a more negative ΔHmix and more negative (smaller positive) ΔHsoln for water than for hexane.

4 0

3 years ago

Which mixture would be best to use as a wash for removal of the alkylcatechol from skin exposed to poison ivy?water and baking s

Rudiy27

The question is incomplete. Here is hte complete question.

The components of poison ivy and poison oak that produce the characteristic itchy rash are cathecols substituted with long-chain alkyl groups. The image is attached.

If you were exposed to poison ivy, which of the treatments below would you apply to the affected area? Justify your choice.

(a) Wash the area with cold water.

(b) Wash the area with dilute vinegar or lemon juice.

(c) Wash the area with sopa and water.

(d) Wash the area with soap, water and baking soda (sodium bicarbonate)

Answer: (d) Wash the area with soap, water and baking soda.

Explanation: Alkyl is hydrophobic, which means it doesn't "mix" with water. To make it more soluble in water, soap helps to dissolve the chemical compound. As shown in the image, alkylcathecol is mildly alkaline. Baking soda is a salt that also helps dissolve it by ionizing with the -OH group, which makes it more soluble.

Although lemon juice and vinegar are acids, they are weak acids and so, won't have a strong effect on the alkylcathecol.

As said before, alkyl is hydrophobic, so plain water won't affect in any way the cathecol.

6 0

3 years ago