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fenix001 [56]
2 years ago
13

Solve

Chemistry
1 answer:
stiv31 [10]2 years ago
7 0

The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the following Charle's law equation:

V1/T1 = V2/T2

Where;

  • T1 = initial temperature
  • T2 = final temperature
  • V1 = initial volume
  • V2 = final volume

  • V1 = 9L
  • V2 = ?
  • T1 = 325K
  • T2 = 450K

9/325 = V2/450

325V2 = 4050

V2 = 4050/325

V2 = 12.46L

Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

Learn more about volume at: brainly.com/question/2817451

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3 years ago
How would you prepare 2.00 l of a 0.25 m acetate buffer at ph= 4.50 from concentrated acetic acid (17.4 m) and 1.00 m naoh? the
zvonat [6]

The reaction of acetic acid with sodium hydroxide is:

CH_3COOH + NaOH    CH_3COONa + H_2O

The ratio of A-/HA is calculated as follows:

According to Henderson Hasslebach equation:


[A-]/[HA] = 10^p^H^-^p^K^_a

= 10^4^.^7^6^-^4^.^5^ = ^2^.^1^8

The total concentration of HA and A- = 2.0 L * 0.25 M = 0.5 mol.

[ A- ]+ [ HA ]= 0.5

[ A^- ] = 0.5 – [ HA]

[ HA] = 0.156 mol and [ A- ]= 0.343 mol

Total of 0.5 moles of acetic acid is required:

0.5 mol HA * 60.0 g

HA = 30.0 g acetic acid

Conversion of 0.343 moles of the acetic acid to acetate can be performed by adding NaOH

0.343 mol HA * (1 mol NaOH/1 mol HA) * (1000 mL/1 mol NaOH)

= 343 mL  

Thus, 343 mL of 1M NaOH is required

3 0
3 years ago
Bombardier beetles inject boiling hydrogen peroxide, or H2O2, into the victims. It then decomposes into hydrogen and oxygen gas.
insens350 [35]

Answer:

0.12 g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

H2O2 —> H2 + O2

Next, we shall determine the mass of H2O2 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:

Molar mass of H2O2 = (2×1) + (16×2)

= 2 + 32

= 34 g/mol

Mass of H2O2 from the balanced equation = 1 × 34 = 34 g.

Molar mass of O2 = 16×2 = 32 g/mol

Mass of O2 from the balanced equation = 1 × 32 = 32 g

Summary:

From the balanced equation above,

34 g of H2O2 decomposed to produce 32 g of O2.

Finally, we shall determine the mass of O2 produced from the decomposition of 0.128 g of H2O2. This can be obtained as follow:

From the balanced equation above,

34 g of H2O2 decomposed to produce 32 g of O2.

Therefore, 0.128 g of H2O2 will decompose to produce = (0.128×32)/34 = 0.12 g of O2.

Therefore, 0.12 g of O2 was produced.

8 0
3 years ago
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