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SashulF [63]
4 years ago
15

Molar mass of osmium

Chemistry
1 answer:
lidiya [134]4 years ago
7 0

Answer:-

190.23 gram mol-1

Explanation:-

Molar mass is the mass of 1 mole of that substance. It's unit is gram / mol or gram mol-1.

For an element it is equal to it's atomic weight.

The atomic weight of Osmium is 190.23

Hence the molar mass of Osmium is 190.23 gram / mol

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Nitrogen gained 4 electrons.

Because Nitrogen's redox number went from +6 to +2, it must have gained 4 electrons (-4) in order to achieve this number. Thus, Nitrogen is reduced.

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3 years ago
The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq
Pavlova-9 [17]

Answer : The molarity of I_3^- in the solution is, 0.128 M

Explanation :

The given balanced chemical reaction is,

2S_2O_2^{-3}(aq)+I_3(aq)\rightarrow S_4O_2^{-6}(aq)+3I^-(aq)

First we have to calculate the moles of Na_2S_2O_3.

\text{Moles of }Na_2S_2O_3=\text{Molarity of }Na_2S_2O_3\times \text{Volume of solution}

\text{Moles of }Na_2S_2O_3=0.260mole/L\times 0.0296L=0.007696mole

Conversion used : (1 L = 1000 ml)

Now we have to calculate the moles of I_3^-.

From the balanced chemical reaction, we conclude that

As, 2 moles of S_2O_2^{-3} react with 1 mole of I_3^-

So, 0.007696 moles of S_2O_2^{-3} react with \frac{0.007696}{2}=0.003848 mole of I_3^-

The moles of I_3^- = 0.003848 mole

Now we have to calculate the molarity of I_3^-.

\text{Molarity of }I_3^-=\frac{\text{Moles of }I_3^-}{\text{Volume of solution}}

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

\text{Molarity of }I_3^-=\frac{0.003848mole}{0.03L}=0.128mole/L=0.128M

Therefore, the molarity of I_3^- in the solution is, 0.128 M

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3 years ago
According to the fossil record found in these sedimentary layers, what conclusion can be drawn about the movement of life to lan
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3 years ago
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What is the ratio of the concentration of sodium acetate to the concentration of acetic acid at ph 5.75?
GREYUIT [131]
To get the ratio of the concentration of sodium acetate to the concentration of acetic acid we are going to use H-H equation:

when:

PH = Pka + ㏒[salt/acid]

when the Ka of acetic acid = 1.75 x 10^-5 

so, we can get Pka from this formula:

Pka = -㏒Ka

       = -㏒1.75 x 10^-5

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and when we have PH = 5.75, so by substitution:

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