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otez555 [7]
9 months ago
11

What type of bonds are shown in this diagram?

Chemistry
2 answers:
Dmitrij [34]9 months ago
6 0

Explanation:

no diagram is available in this question , so I am sorry not for replying this answer

qaws [65]9 months ago
3 0

Answer:

What type of bonds are shown in this diagram?

A: covalent bonds

B: ionic bonds

C: hydrogen bonds

D: metallic bonds

(answer) metallic bonds

In what type of bonds do atoms join together because their opposite charges attract each other?

A: metallic bonds and covalent bonds

B: metallic bonds and ionic bonds

C: ionic bonds and covalent bonds

D: ionic bonds and hydrogen bonds

(answer) ionic bonds and hydrogen bonds

What types of bonds are shown in this diagram?

A: covalent bonds

B: ionic bonds

C: hydrogen bonds

D: metallic bonds

(answer) hydrogen bonds

Which statement best describes the types of bonds shown in the diagram?

A: an ionic bond; the hydrogen chloride molecule has an electrical charge

B: an ionic bond; a hydrogen ion is bonding with a chlorine atom

C: a covalent bond; the hydrogen atom’s two electrons are being shared with the chlorine atom

D: a covalent bond; the hydrogen atom’s single electron is being shared with the chlorine atom

(answer) a covalent bond; the hydrogen atom’s single electron is being shared with the chlorine atom

Which of the following bonds is the strongest?

A: hydrogen bonds

B: metallic bonds

C: valence bonds

D: covalent bonds

(answer)

Explanation:

UwU

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Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

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