Which reaction is an oxidation-reduction reaction?

How do you know that a sealed calorimeter is a closed system?

3241004551 [841]

Answer: Option (c) is the correct answer.

Explanation:

When a system is open then there will be exchange of energy between the system and surrounding.

Whereas when a system is closed then there will be no exchange of energy, that is, thermal energy will not flow into the atmosphere.

Thus, we can conclude that a sealed calorimeter is a closed system because thermal energy is not transferred to the environment.

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3 years ago

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A rectangular block has the following dimensions: 3.21 dm, 5.83 cm, and 1.84 in. The block has a mass of 1.94 kg. What is the de

spin [16.1K]

The density of the rectangular block in g/mL is 7.0.

<u>Given the following data:</u>

Mass of block = 22.8 gra1.94 kg

Length of block = 3.21 cm

Width of block = 5.83 cm

Height of block = 1.84 in.

To find the density of the block in g/mL:

First of all, we would determine the volume of the rectangular block by using the following formula:

$Volume = length$ × $width$ × $height$

<u>Conversion:</u>

1 in = 2.54 cm

5.83 in = X cm

Cross-multiplying, we have:

$X = 2.54(5.83)\\\\X = 14.81 \; cm$

$Volume = 3.21$ × $5.83$ × $14.81$

Volume = 277.16 cubic centimeters.

<u>Note</u>: Milliliter (mL) is the same as cubic centimeters.

1000 grams = 1 kg

Y grams = 1.94 kg

Cross-multiplying, we have:

Y = 1940 grams

Now, we can find the density:

$Density = \frac{Mass}{Volume}\\\\Density = \frac{1940}{277.16}$

<em>Density </em><em>= 7</em><em>.0 g/mL</em>

Therefore, the density of the rectangular block in g/mL is 7.0.

4 0

2 years ago

A wide range of molecular compounds dissolve in water. this allows water to act as a medium for moving molecules into and out of

Sergio [31]

Lipids cannot dissolve in water.

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3 years ago

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Submit your answer for the remaining reagent in Tutorial Assignment #1 Question 9 here, including units. Note: Use e for scienti

djverab [1.8K]

<h3>Answer:</h3>

The mass of excessive water (H₂O) is 40.815 kg

<h3>Explanation:</h3>

The Equation for the reaction is;

Li₂O(s) + H₂O(l) → 2LiOH(s)

From the question;

Mass of water removed is 80.0 kg

Mass of available Li₂O is 65.0 kg

We are required to calculate the mass of excessive reagent.

<h3>Step 1: Calculating the number of moles of water to be removed</h3>

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Mass of water = 80 kg (but 1000 g = 1kg)

= 80,000 g

Therefore;

$Moles of water = \frac{80,000g}{18.02 g/mol}$

= 4.44 × 10³ moles

<h3>Step 2: Moles of Li₂O available </h3>

Moles = mass ÷ molar mass

Mass of Li₂O available = 65.0 kg or 65,000 g

Molar mass Li₂O = 29.88 g/mol

Moles of Li₂O = 65,000 g ÷ 29.88 g/mol

= 2.175 × 10³ moles Li₂O

<h3>Step 3: Mass of excess reagent </h3>

From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)

1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH

The ratio of Li₂O to H₂O is 1:1

Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
This means, Li₂O is the limiting reactant while water is the excessive reagent.

Therefore:

Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles

= 2.265 × 10³ moles

Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol

= 4.0815 × 10⁴ g or

= 40.815 kg

Thus, the mass of excessive water is 40.815 kg

7 0

3 years ago

Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su

Free_Kalibri [48]

Answer: