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3 years ago

What are the uses of zinc

2 answers:

5 0

Zinc is also used in alloys such as brass, nickel silver and aluminium solder. Zinc oxide is widely used in the manufacture of very many products such as paints, rubber, cosmetics, pharmaceuticals, plastics, inks, soaps, batteries, textiles and electrical equipment.

Travka [436]3 years ago

5 0

One is for making little shacks

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Explain how using dimensional analysis could have prevented this crash

irakobra [83]

Answer:you keep it safe

Explanation:

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3 years ago

Which are characteristics of natural selection? Select three options.

Luda [366]

Affects populations,
Occurs when there is genetic variation
Selects organisms with certain traits to survive.

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2 years ago

An airplane flies at a constant speed of 920 km/h. How long will it take the plane to travel a distance of 1150 km?

rewona [7]

Answer:

1.25 hours or 75 minutes or 1 hour and 15 minutes

Explanation:

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2 years ago

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Two substances with empirical formula HNO are hyponitrous acid (M = 62.04 g/mol) and nitroxyl (M = 31.02 g/mol).(b) For each spe

Natalija [7]

Below is an attachment of the Lewis structure with the lowest formal charges.

The formal charge is the fictitious charge that an atom in a molecule would have if the electrons in the bonds were evenly distributed among the atoms. The nonbonding electrons on a neutral atom are subtracted from its valence electron count, which is then followed by the number of bonds that bind it to other atoms in the Lewis structure, to get the formal charge. This is another way to put it. When hyponitrous acid is oxidized in the atmosphere, nitric and nitrous acids are produced. By reducing a nitrate or nitrite by sodium amalgam in the presence of water, hyponitrite salts have been created.

Learn more about formal charge here-

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6 0

2 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

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3 years ago