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katen-ka-za [31]
3 years ago
5

How many orbitals are available in a “d” orbital configuration?

Chemistry
1 answer:
ratelena [41]3 years ago
8 0

Answer:

e.)5

Explanation:

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If you made the sugar and corn syrup solutions you heated, the sugar made the solutions' boiling points higher than that of pure liquid water.
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the hcp ordered 0.2 mg of methylergonovine, and the vial contains 0.8 mg/ml. how many ml of methylergonovine should the nurse dr
olasank [31]

The volume (in mL) of methylergonovine the nurse should draw up in the syringeis is 0.25 mL

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

With the concept of density, we can obtain the volume of methylergonovine. Details below

<h3>How to determine the volume </h3>

The following data were obtained from the question:

  • Mass of methylergonovine = 0.2 mg
  • Density of methylergonovine = 0.8 mg/mL
  • Volume of methylergonovine =?

Density = mass / volume

Cross multiply

Density × volume = mass

Divide both sides by density

Volume = mass / density

Volume = 0.2 / 0.8

Volume of methylergonovine = 0.25 mL

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2 years ago
How many moles are in 19.82 g Mg?
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I think it’s 24.305? Not sure
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3 years ago
2. How many grams of zinc will be formed if 32.0 g of copper reacts with zinc nitrate? Copper(I)nitrate is the other product.
Yakvenalex [24]

Answer:

16.46 g.

Explanation:

  • It is a stichiometry problem.
  • We should write the balance equation of the mentioned chemical reaction:

<em>2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).</em>

  • It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).
  • We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:

<em>n = mass / molar mass</em>

  • The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.

<u><em>Using cross multiplication:</em></u>

2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.

0.503 mole of Cu produces → ??? mole of Zn.

  • The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.

∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.

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