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3 years ago

How are the properties of organic compounds determined?

Chemistry

1 answer:

Harman [31]3 years ago

4 0

Explanation:

The physical properties of organic compounds typically of interest include both quantitative and qualitative features. Quantitative information includes a melting point, boiling point, and index of refraction. Qualitative properties include odor, consistency, solubility, and color.

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The validity of scientific theories and laws depends on evidence. Which of the following statements of evidence supports a law r

Ksju [112]

Answer:

d the amount of matter in a closed system is the same at the start of a reaction as at the end of the reaction

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3 years ago

Read 2 more answers

Which type of bond does this molecule show?

Gekata [30.6K]

The correct answer is D. magnetic bond

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2 years ago

Copper reacts with silver nitrate through single replacement.

-Dominant- [34]

Answer:

Part a) 0.0104 moles copper(II) nitrate.

Part b)

i) 0.0418 mole Cu

ii) 0.0209 mol Ag NO₃

Explanation:

<u>1) Balanced chemical reaction (single replacement):</u>

In a single replacement reaction a more acitve metal (Cu) replaces a less active metal (Ag)

Cu + 2 Ag NO₃ → Cu (NO₃)₂ + 2 Ag

<u>2) Mole ratio: </u>

1 mole Cu : 2 mole Ag NO₃ : 2 mole Ag

<u />

<u>3) Moles of Ag</u>

n = mass in grams / atomic mass
atomic mass of Ag: 107.868 g/mol
n = 2.25 g / 107.868 g/mol = 0.0209 mol Ag

<u>4) Moles of copper(II) nitrate:</u>

Set the proportion using the mole ratio:

2 mole Ag / 1 mole Cu (NO₃)₂ = 0.0209 mole Ag / x

Solve: x = 0.0209 / 2 mole Cu (NO₃)₂ = 0.0104 moles Cu(NO₃)₂

That is the answer of part a: 0.0104 moles copper(II) nitrate.

<u>5) Moles of each reactant</u>

i) Cu:

Set a proportion using the theoretical mole ratio

1 mole Cu / 2 mole Ag = x / 0.0209 mol Ag

Solve for x: x = 0.0209 / 2 mole Cu = 0.0418 mole Cu

ii) Ag NO₃

Set a proportion using the teoretical mole ratio

2 mole Ag NO₃ / 2 mole Ag = x / 0.0209 mole Ag

Solve for x: x = 0.0209 mol Ag NO₃

6 0

3 years ago

PLEASE ANSWER I AM BEGGING

TiliK225 [7]

Taking into account the definition of dilution:

you have to use 8.23 mL of a stock solution of 7.00 M HNO₃ to prepare 0.120 L of 0.480 M HNO₃.
If you dilute 20.0 mL of the stock solution to a final volume of 0.270 L , the concentration of the diluted solution is 0.518 M.

<h3>Dilution</h3>

Dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where