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2 years ago

_Cuo +H, → _Cu + _H,0

Chemistry

1 answer:

Elena-2011 [213]2 years ago

3 0

<h3>Answer:</h3>

CuO(s) + H₂(g) → Cu(s) + H₂O(l)

<h3>Explanation:</h3>

Assuming the reaction is the reduction of CuO by H₂
Then the balanced equation for the reaction is;

CuO(s) + H₂(g) → Cu(s) + H₂O(l)

The equation shows the reducing property of hydrogen gas, such that hydrogen reduces metal oxides such as copper(ii)oxide to the respective metals.
The law of conservation requires chemical equations to be balanced so as the mass of reactants will be equal to that of products.

In this case; there is 1 copper atom, 1 oxygen atom and 2 hydrogen atoms on both side of the equation and thus the equation is balanced.

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Which of the following characteristics describes aerobic organisms.

horsena [70]

Answer: Option (3) is the correct answer.

Explanation:

Aerobic organisms are the organisms which survive and grow in the presence of oxygen.

When oxidation of glucose occurs in the presence of oxygen then it is known as aerobic respiration.

In aerobic respiration, food releases energy to produce ATP which is necessary for cell activity. There is complete breakdown of glucose in aerobic respiration that is why more energy is released. Therefore, aerobic organisms become active.

Thus, we can conclude that characteristics very active, efficient use of energy describes aerobic organisms.

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3 years ago

Compare and contrast the concepts of average mass and relative mass. Which one is more accurate, and why? Why is the one you did

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Answer:

Relative and average atomic mass both describe properties of an element related to its different isotopes.

Explanation:However, relative atomic mass is a standardized number that's assumed to be correct under most circumstances, while average atomic mass is only true for a specific sample.

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3 years ago

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Emily floats a note across a river to Dylan instead of directly passing it from her hand to Dylan's hand. If, in this scenario,

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Answer: The Note

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2 years ago

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How the molecule’s properties affect the physical properties of a molecule

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The intermolecular forces, such as hydrogen bonds or van der Waals attractions, which draw one molecule to its neighbors, govern a substance's physical properties. Due to the relatively weak intermolecular forces of attraction, molecular substances typically take the form of gases, liquids, or low melting point solids.

<h3>How do the intermolecular forces affect physical properties?</h3>

The forces that bind two molecules together are known as intermolecular forces. Intermolecular forces have an impact on physical properties. Strong and weak forces both exist; the stronger the force, the more energy is needed to separate the molecules from one another. As intermolecular forces increase melting, boiling, and freezing points rise.

The following intermolecular forces are listed in order of strength:

Van der Waals dispersion forces
Van der Waals dipole-dipole interactions
Hydrogen bonding
Ionic bonds

It would take very little energy to separate two molecules if they are connected by van der Waals dispersion forces. On the other hand, it requires a lot more energy to separate two molecules that are joined together by ionic bonds.

To know more about molecules refer to: brainly.com/question/1819972

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1 year ago

An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient wate

stiv31 [10]

Answer:

2710.2g/mol

Explanation:

Step 1:

Data obtained from the question. This include the following:

van 't Hoff factor (i) = 1 (since the compound is non-electrolyte)

Mass of compound = 33.2g

Volume = 250mL

Osmotic pressure (Π) = 1.2 atm

Temperature (T) = 25ºC = 25ºC + 273 = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Molar mass of compound =.?

Step 2:

Determination of the molarity of the compound.

The molarity, M of the compound can be obtained as follow:

Π = iMRT

1.2 = 1 x M x 0.0821 x 298

Divide both side by 0.0821 x 298

M = 1.2 / (0.0821 x 298)

M = 0.049mol/L

Step 3:

Determination of the number of mole of compound in the solution. This can be obtain as follow:

Molarity = 0.049mol/L

Volume = 250mL = 250/1000 = 0.25L

Mole of compound =..?

Molarity = mole /Volume

0.049 = mole / 0.25

Cross multiply

Mole = 0.049 x 0.25

Mole of compound = 0.01225 mole.

Step 4:

Determination of the molar mass of the compound. This is illustrated below:

Mole of the compound = 0.01225 mole.

Mass of the compound = 33.2g

Molar mass of the compound =.?

Mole = Mass /Molar Mass

0.01225 = 33.2/Molar Mass

Cross multiply

0.01225 x molar mass = 33.2

Divide both side by 0.01225

Molar mass = 33.2/0.01225

Molar mass of the compound = 2710.2g/mol

Therefore, the molar mass of the compound is 2710.2g/mol

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2 years ago