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Alex787 [66]
3 years ago
7

What is the oxidation number for CI? -2 +2 -1 +1

Chemistry
1 answer:
jeka943 years ago
8 0

Answer:

-1

Explanation:

Chlorine is present in group seventeen. It has seven valance electrons. it is the member of halogen group. These are six elements Fluorine, Chlorine, Bromine, Iodine, Astatine. Halogens are very reactive these elements can not be found free in nature. Their chemical properties are resemble greatly with each other. All halogen member need one electron to complete the octet and thus show -1 charge.

Properties of chlorine:

1. it is greenish-yellow irritating gas.

2. its melting point is 172.2 K

3. its boiling point is 238.6 K

4. it is disinfectant and can kill the bacteria.

5. it is also used in manufacturing of paper, paints and textile industries.

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1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
For the chemical reaction 2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq) write the net ionic equation, including the phases.
Crazy boy [7]

Answer:

2H⁺(aq) + 2OH⁻(aq)  --> 2H2O(l)

Explanation:

2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)

We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.

The ionic equation is given as;

2H⁺(aq)  +  2Br⁻(aq)  + Ba²⁺(aq) + 2OH⁻(aq)   --> 2H2O(l)  +  Ba²⁺(aq)  + 2Br⁻(aq)

Upon eliminating the spectator ions; The net equation is given as;

2H⁺(aq) + 2OH⁻(aq)  --> 2H2O(l)

8 0
3 years ago
If the density of an ideal gas at stp if found to be 0.0902 g/l, what is its molar mass?
lisabon 2012 [21]
P=0.0902 g/l
v=22.4 l/mol   (stp)

M=vp

M=22.4 l/mol * 0.0902 g/l=2.020 g/mol

M=2.020 g/mol
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