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nataly862011 [7]
2 years ago
5

A redox reaction always involves

Chemistry
1 answer:
Sonbull [250]2 years ago
3 0

Answer:a

Explanation:

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A sample of Ne gas has a pressure of 654 mmHg with an unknown volume. The gas has a pressure of 345 mmHg when the volume is 495m
Lemur [1.5K]

Answer:

The initial volume of Ne gas is 261mL

Explanation:

This question can be answered using Ideal Gas Equation;

However, the following are the given parameters

Initial Pressure = 654mmHg

Finial Pressure = 345mmHg

Final Volume = 495mL

Required

Initial Volume?

The question says that Temperature is constant;

This implies that, we'll make use of Boyle's law ideal gas equation which states;

P_1V_1 = P_2V_2

Where P_1 represent the initial pressure

P_2 represent the final pressure

T_1 represent the initial temperature

T_2 represent the final temperature

P_1 = 654mmHg\\P_2 = 345mmHg\\V_2 = 495mL

Substitute these values in the formula above;

654 * V_1 = 345 * 495

654V_1 = 170775

Divide both sides by 654

\frac{654V_1}{654} = \frac{170775}{654}

V_1 = \frac{170775}{654}

V_1 = 261.123853211

V_1 = 261mL (Approximated)

<em>The initial volume of Ne gas is 261mL</em>

7 0
2 years ago
What is 7% as a decimal?
Colt1911 [192]
.07! you divide 7 by 100%


6 0
3 years ago
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug
OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 =  log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] <em>(1)</em>

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = <em>0.035M</em>

Thus: [CH₃COO⁻] = 0.100M - 0.035M = <em>0.065M</em>

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles +  2.646x10⁻³ moles = <em>8.071x10⁻³ moles of CH₃COOH</em>. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = <em>0.296</em>

4 0
3 years ago
How many moles of oxygen are required to react with 12 moles of FeS2?
Radda [10]
19 moles of O2 probably
3 0
2 years ago
What is the pOH of<br> a<br> 2.6 x 10-6 M H+ solution?
scZoUnD [109]

Answer:

pH = -log 2.6 x 10-6 M

pH = 5.585

pOH= 14 - 5.585 = 8.415

8.4 as 2 sig figs

8 0
2 years ago
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