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nika2105 [10]
3 years ago
13

What is the oxidation number for iodine in Mg(IO3)2 ?

Chemistry
2 answers:
sukhopar [10]3 years ago
8 0

<u>Answer:</u> The oxidation state of iodine atom in the given compound is +5.

<u>Explanation:</u>

Oxidation state is the number that is assigned to an element when it gains or looses electrons. If an element gains electron, it will attain a negative oxidation state and if the element looses electrons, it will attain a positive oxidation state.

For the given chemical compound: Mg(IO_3)_2

We take the oxidation state of iodine atom be 'x'

Oxidation state of magnesium atom = +2

Oxidation state of oxygen atom = -2

Evaluating the oxidation state of iodine atom:

+2+2x+6(-2)=0\\\\x=+5

Hence, the oxidation state of iodine atom in the given compound is +5.

navik [9.2K]3 years ago
7 0
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as 
   Mg(IO3)2
   MgI2O6
As we know that
Mg has +2
O has -2
So,
   (+2) + 2I + 6 (-2)=0
   2 + 2I - 12 =0
   10+ 2I =0
    10 = 2I
     I =5

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Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

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Answer:

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Explanation:

Hello there!

In this case, according to the given information for this calorimetry problem, we can proceed as follows:

1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

1cal=4.184J\\\\1kJ=1000J

Then, we perform the conversion as follows:

600.0cal*\frac{4.184J}{1cal}*\frac{1kJ}{1000J}=2.510kJ

2. Here, we use the general heat equation:

Q=mC(T_2-T_1)

And we plug in the given mass, specific heat and initial and final temperature to obtain:

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