One of the most popular club drugs. Originally patented as an appetite suppressant, this

Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

KATRIN_1 [288]

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent

8 0

2 years ago

A gas occupies the volume of 215ml at 15C and 86.4kPa?

kherson [118]

Answer:

About 0.1738 liters

Explanation:

Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

$0.8527\cdot 0.215=n \cdot 0.0821 \cdot 288\\n\approx 7.754 \cdot 10^{-3}$

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

$1V=7.754 \cdot 10^{-3} \cdot 0.0821 \cdot 273\\V\approx 0.1738L$

Hope this helps!

3 0

3 years ago

Atoms of metals tend to1) lose electrons and form negative ions2) lose electrons and form positive ions3) gain electrons and for

timofeeve [1]

Answer:

2) lose electrons and form positive ions

Explanation:

Metals are generally electropositive elements due to the fact that they lose electrons to their non-metal counterparts and hence, form CATIONS or positively charged atoms. Non-metals, on the contrary, gains electrons and become negatively charged i.e form anions. These ions combine to form stable ionic compounds.

This electron-losing characteristics of metals make them have properties that includes: good conductors of electricity and heat, being lustrous etc.

4 0

3 years ago

An element crystallizes in a face-centered cubic lattice. If the length of an edge of the unit cell is 0.408 nm, and the density

V125BC [204]

Answer:

Au

Explanation:

For the density of a face-centered cubic:

$Density = \dfrac{4 \times M_w}{N_A \times a^3}$