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2 years ago

Which half-reaction correctly shows the oxidation of iron

Chemistry

1 answer:

pishuonlain [190]2 years ago

8 0

Answer:

What is the oxidation half reaction for iron?

The two elements involved, iron and chlorine, each change oxidation state; iron from +2 to +3, chlorine from 0 to -1. There are then effectively two half-reactions occurring. These changes can be represented in formulas by inserting appropriate electrons into each half-reaction: Fe2+ → Fe3+ + e.

Hope this helps..

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How many neutrons does 3919K have?

Igoryamba

19 protons and 19 electrons but neutrons are vary

5 0

2 years ago

Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -

kogti [31]

The given complex ion is as follow,

[Ru (CN) (CO)₄]⁻

Where;
[ ] = Coordination Sphere

Ru = Central Metal Atom = Ruthenium

CN = Cyanide Ligand

CO = Carbonyl Ligand

The charge on Ru is calculated as follow,

Ru + (CN) + (CO)₄ = -1
Where;
-1 = overall charge on sphere

0 = Charge on neutral CO

-1 = Charge on CN

So, Putting values,

Ru + (-1) + (0)₄ = -1

Ru - 1 + 0 = -1

Ru - 1 = -1

Ru = -1 + 1

Ru = 0
Result:
Oxidation state of the metal species in each complex [Ru(CN)(CO)₄]⁻ is zero.

3 0

3 years ago

Read 2 more answers

If 45.0 mL of ethanol (density = 0.789 g/mL) initially at 8.0 ∘C is mixed with 45.0 mL of water (density = 1.0 g/mL) initially a

strojnjashka [21]

Answer : The final temperature of the mixture is, $22.14^oC$

Explanation :

First we have to calculate the mass of ethanol and water.

$\text{Mass of ethanol}=\text{Density of ethanol}\times \text{Volume of ethanol}=0.789g/mL\times 45.0mL=35.5g$

and,

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.0g/mL\times 45.0mL=45.0g$

Now we have to calculate the final temperature of the mixture.

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ethanol = $2.42J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ethanol = 35.5 g

$m_2$ = mass of water = 45.0 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of ethanol = $8.0^oC$

$T_2$ = initial temperature of water = $28.6^oC$

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

$35.5g\times 2.42J/g^oC\times (T_f-8.0)^oC=-45.0g\times 4.18J/g^oC\times (T_f-28.6)^oC$

$T_f=22.14^oC$

Therefore, the final temperature of the mixture is, $22.14^oC$

3 0

3 years ago

Find the molecular mass of MgSO4

rewona [7]

120.38 g/mol is the answer

7 0

3 years ago

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At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

algol [13]

Answer:

820 K

Explanation:

As per Boltzman equation, kinetic energy (KE) is in direct relation to the temperature, measured in absolute scale Kelvin.

KE α T.

Then, the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be 3 times such temperature.

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

Second, convert 0°C to kelvin:

T (K) = T(°C) + 273.15 K= 273.15 K

Then, 3 times gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

7 0

2 years ago