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grigory [225]
3 years ago
12

Why is copper (i) sulfate Cu2So4^-2? not copper (ii)?

Chemistry
2 answers:
const2013 [10]3 years ago
5 0
Copper (I) oxidation state is 1 Cu2So4
copper (II) oxidation state is +2 CuSo4
copper (i) also give up one electron so you need two of them to react with the sulfate ion (which has charge of -2)
and also all metallic ions have an multiple oxidation levels corresponding to the number of electrons they can exchange or loose
Hope this helps
ArbitrLikvidat [17]3 years ago
5 0
<h2>Answer</h2>

This is because Copper (I) has +1 charge and Copper (II) has +2 charge.

<h2>Explanation</h2>

The chemical formula of an ionic compound consists of cation first, followed by anion. Both cations and anions are incolved to make compound to be nutral or charge entity for example Cu2SO4 is cuprous compound contain Cu (I) where two Cu give only one electron as  

Cu is cation and SO4 is anion

SO4 has a valency of -2

So, Cu2 has a valency of +2

But in Cu (II), The single copper gives two electron to SO4 and form CuSO4 compound named as copper sulphate.

These metalls are called transition metals who can change their oxidation states


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Which is least likely to be reduced? <br> A. Zn^2+ <br> B. Fe^3+ <br> C. Cu^2+ <br> D. Fe^2+
alekssr [168]

A. Zn²⁺

<h3>Further explanation</h3>

Given

Cations of several elements

Required

The least to be reduced

Solution

If we look at the voltaic series:

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe²⁺-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Fe³⁺-Ag-Pt-Au </em>

The electrode which is easier to reduce than the hydrogen (H2) electrode has a positive sign (E red= +) and is located to the right of the voltaic series (right of H)

The electrode which is easier to oxidize than the hydrogen (H2) electrode and is difficult to experience reduction has a negative sign (E red= -) and is located to the left of the voltaic series (left of H)

Or you can look at the standard reduction potential value of the metals in the answer options, and the most negative reduction E° value which will be difficult to reduce.

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6 0
2 years ago
One mole of oxygen gas is at a pressure of 6.00 atm and a temperature of 27.0°C. (a) If the gas is heated at constant volume unt
Umnica [9.8K]

Answer: a) 900 K

b) 1200 K

Explanation:

According to the ideal gas equation:

PV=nRT

P = Pressure of the gas = 6.00 atm

V= Volume of the gas = ?

T= Temperature of the gas = 27°C = 300 K    0^00C=273K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas  = 1

V=\frac{nRT}{P}=\frac{1\times 0.0821\times 300}{6.00}=4.10L

a) To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=6.00atm\\T_1=300K\\P_2=3\times 6.00=18.0atm\\T_2=?

Putting values in above equation, we get:

\frac{6.00}{300K}=\frac{18.0}{T_2}\\\\T_2=900K

The final temperature is 900 K

b) The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 6.00 atm

P_2 = final pressure of gas = 2\times 6.00atm=12.0atm

V_1 = initial volume of gas = 4.10 L

V_2 = final volume of gas =  2\times 4.10 L=8.20L

T_1 = initial temperature of gas = 300K

T_2 = final temperature of gas =?

Now put all the given values in the above equation, we get:

\frac{6.00\times 4.10}{300}=\frac{12.0\times 8.20}{T_2}

T_2=1200K

The final temperature is 1200 K

5 0
3 years ago
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