What is the product when magnesium reacts with nitrogen? Mg(s) + N2(g) →

Identify and calculate the number of representative particles in 2.15 moles of gold.<br>

LekaFEV [45]

Answer:

$1.29 * 10^{24}$ particles of gold

Explanation:

To convert the number of moles of any substance, in this case gold, you need Avogadro's number.

Avogadro's number is always $6.022$ × $10^{23}$

$2.15$ moles Au × $\frac{6.022*10^{23} particles}{1 mole Au}$ = $1.29 * 10^{24}$ particles of gold

5 0

3 years ago

Which statement about relative motion is not true?

AURORKA [14]

Explanation:

laws of physics which apply when you are at rest on the earth also apply when you are in any reference frame which is moving at a constant velocity with respect to the earth. For example, you can toss and catch a ball in a moving bus if the motion is in a straight line at constant speed.

7 0

3 years ago

How would you test a colorless crystalline compound to determine if it was a hydrate?

Cloud [144]

A hydrate is a substance where in it contains water and other constituent elements. To know whether if that compound was a hydrate,you should record its mass, then put it in a test tube and heat it with a Bunsen burner. If the compound is a hydrate, the water in the compound will discharge in the form of water vapor. At the next 5-10 minutes, remove it in the test tube and weigh it up again. If the mass is now fewer, that means that there was water existing that has now evaporated, and the compound was a hydrate.

6 0

4 years ago

Would you expect the wavelength of maximum absorbance for [Cu(NH3)4]2+ to be greater than or less than the wavelength of maximum

DENIUS [597]

Answer:

Less

Explanation:

Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:

The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.

The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand

A partial spectrochemical series listing of ligands from small Δ to large Δ:

I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO

Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.

7 0

4 years ago

The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0

3 years ago