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Vedmedyk [2.9K]
4 years ago
7

A sample of an ideal gas at 1.00 atm and a volume of 1.82 L was placed in a weighted balloon and dropped into the ocean. As the

sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 15.0 atm , what was the volume of the sample? Assume that the temperature was held constant.
Chemistry
1 answer:
Anvisha [2.4K]4 years ago
3 0

Answer:

The answer to your question is Volume = 0.12 L

Explanation:

- Data

Pressure 1 = 1 atm

Volume 1 = 1.82 L

Pressure 2 = 15 atm

Volume 2 = ?

Temperature = constant

- Formula

- To solve this problem use the Boyle's law

            P1V1 = P2V2

- Solve for V2

            V2 = P1V1 / P2

- Substitution

             V2 = (1)(1.82) / 15

- Simplification

             V2 = 1.82 / 15

- Result

             V2 = 0.12 L

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The combination of oxygen with other substances to produce new chemical products is called
Flauer [41]

Answer:

                 The combination of oxygen with other substances to produce new chemical products is called <u>Oxidation</u>.

Explanation:

                     Oxidation reactions are defined as,

In terms of Inorganic chemistry:

                 (i) <u>Removal of Electrons: </u>

                            Example:  Mg  →  Mg²⁺  +  2 e⁻

                (ii) <u>Addition of Oxygen:</u>

                            Example:  2 Mg  + O₂  →  2 MgO

In terms of Organic chemistry:

                (i) <u>Addition of Electrons: </u>

                            Example:  Cl₂  +  2 e⁻   →   2 Cl⁻

                (ii) <u>Addition of Hydrogen:</u>

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4 0
3 years ago
150 ml of 0.1 m naoh is added to 200 ml of 0.1 m formic acid, and water is added to give a final volume of 1 l. what is the ph o
N76 [4]

Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles

Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles

Here, the limiting reagent is NaOH

The reaction is represented as:

HCOOH + NaOH ↔HCOONa + H2O

Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles

Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles

Total final volume is given as 1 L

Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M

[HCOONa] = 0.015/1 = 0.015 M

pKa of HCOOH = 3.74

As per Henderson-Hasselbalch equation

pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22

Therefore, pH of the final solution = 4.22


                       


3 0
3 years ago
A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm
Ulleksa [173]

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100

Percent error = \frac{0.1}{63.2}\times 100

Percent error = 0.00158\times 100

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100

Percent error = \frac{0.2}{63.2}\times 100

Percent error = 0.00316\times 100

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3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100

Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100

Percent error = \frac{(0.5)}{63.2}\times 100

Percent error = 0.00791\times 100

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Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%




7 0
4 years ago
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Valentin [98]

Answer:

16 amu

Explanation:

6 0
3 years ago
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Dmitry [639]

The correct answer is option a, that is, it gets broken down.  

A set of metabolic reactions and procedures, which occurs in the cells of organisms to transform biochemical energy from nutrients into ATP, and then discharge waste components is known as cellular respiration. At the time of cellular respiration, a molecule of glucose gets dissociated slowly into water and carbon dioxide. With it, some of the ATP is generated directly in the reactions, which transform glucose.  


4 0
3 years ago
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