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den301095 [7]
3 years ago
15

when a solution of sodium chloride is added to a solution of copper(ii) nitrate, no precipitate is observed. Write the molexular

equation that describes this behavior,
Chemistry
1 answer:
zhuklara [117]3 years ago
6 0

Explanation:

1.

Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)

2.

Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)

A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.

3.

Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)

2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)

Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)

2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)

4.

The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.

Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.

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Rutherford, geiger, and marsden’s experiment demonstrated that the volume of the nucleus is roughly what fraction of the volume
trasher [3.6K]
Rutherford, Geiger and Marsden's experiment proved that every atom has a nucleus and that this nucleus is of positive charge and contains the most of the mass of the atom. 0.005% of the volume occupied by the electrons is the volume of the nucleus.
4 0
3 years ago
Calculate the molarity of a solution that contains 3.00 grams Na2SO4 in 25 mL of solution.
Naddik [55]

Answer:

M Na2SO4 sln = 0.8448 M

Explanation:

  • molarity (M) [=] mol/L

∴ mass Na2SO4 = 3.00 g

∴ volume soln = 25 mL = 0.025 L

∴ molar mass Na2SO4 = 142.04 g/mol

⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol

⇒ M Na2SO4 sln = (0.02112 mol/0.025 L ) = 0.8448 M

3 0
4 years ago
True or False? Not all compounds will dissolve in water.
In-s [12.5K]

Answer:

False

Explanation:

Polar molecules and ionic compounds dissolve in water, not nonpolar molecules

4 0
3 years ago
How many grams of iki would it take to obtain a 100 ml solution of 0.300 m iki? how many grams of iki would it take to create a
GalinKa [24]

0.300 M IKI represents the concentration which is in molarity of a potassium iodide solution. This means that for every liter of solution there are 0.300 moles of potassium iodide. Knowing that molarity is a ratio of solute to solution.

By using a conversion factor:

100 ml x (1L / 1000 mL) x (0.300 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 4.98 g

Therefore, in the first conversion by simply converting the unit of volume to liter, Molarity is in L where the volume is in liters. The next step is converted in moles from volume by using molarity as a conversion factor which is similar to how density can be used to convert between volume and mass. After converting to moles it is simply used as molar mass of Kl which is obtained from periodic table to convert from mole to grams.

In order to get the grams of IKI to create a 100 mL solution of 0.600 M IKI, use the same formula as above:

100 ml x (1L / 1000 mL) x (0.600 mol Kl / 1 L) x (166.0g / 1 mol Kl) = 9.96 g

3 0
3 years ago
Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at ≥ 160 ppm. Then draw th
-Dominant- [34]

Answer:

1. Butyraldehyde; 2. but-3-en-1-ol

Explanation:

1. Peak ≥ 160 ppm

The formula C₄H₈O shows that the Index of Hydrogen Deficiency = 1.

It could be caused by either a ring or a double bond.

A peak at ≥ 160 ppm strongly indicates a C=O group, so the rest of the molecule can contain no rings or double bonds.

There are no other heteroatoms, so the compound most be an aldehyde or a ketone.

One compound that meets these criteria is butyraldehyde, CH₃CH₂CH₂CH=O (see Fig. 1.)

2. No rings; all peaks < 160 ppm

If all peaks are < 160 ppm, there can no C=O group.

There is no ring, so there must be a C=C double bond.

There is no other unsaturation, so the O atom must be present as an alcohol or an ether.

One compound that meets these criteria is but-3-en-1-ol, CH₂=CHCH₂CH₂OH (see Fig. 2).

4 0
3 years ago
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